Differential-Equations-by-Zill-3rd-Edition-Solutions-Manual(ciitbooks.blogspot.com)_text.pdf

1Introduction to Differential Equations Exercises 1.1 1. Second-order; linear. 2. Third-order; nonlinear because of (dy/dx) 4 3. First-order; nonlinear because of yy'. 4. First-order; linear. 5. Fourth-order; linear. . 6. Second-order; nonlinear because of sin y. (Sy/dx2^ 7. Second-order; nonbnear because of 8. 2 Second-order; nonlinear because of 1/r 9. Third-order; linear. . . 2 10. First-order, nonlinear because of y x /2 11. From y = e~ 12. From y = 8 we obtain y 13. From y = we obtain y' 1 e 3* + lOe 2* = 0, = -\e~ x l 2 6 6 5 5 . + y= ~e~ x l 2 + e~^ 2 = + 4(8) = 32. Then 2y' we obtain dyjdx From y = - — -e -20i we obtain $+ (3e = 20j, 31 = <&//<i( 0. + iy = = 3e 3r + 20e 2a\ Then so that y' ^ - 2» = 14. . + 20^) - 2 20t 24e 24e- , 20t (e 31 + lOe 21 ) = e 3r so that 2 0(g-^- 20£ )=24. at 15. From y = 5tan5x we obtain y' — 25sec 2 5i. Then 2 y = 25sec 5z = 25 1 16. From y = (y/x + ci) 2 we obtain y' = 2 (l .+ tan 2 (./c + 5ar) ci) = 25 + (5tan5x) 2 = 25 + 2 j/ . /2y^, so that (^ + Cl) \ 17. From j/ = - sin x — ^cosx 2 + lOe x we obtain ?/ = -cosi + 2 y' + y — ^cosx + 2 ^ sin -1 x — lOe ^ + ^ sinz — lOe *. Then 2 ^ sins — ^ cosi + 10e -1 ^ = sinx. — 1 Exercises 1. 18. First write the differential equation in the form 2xy 2 x y 19. 20. + V2 = 1 we obtain 2xy + {x + ci First write the differential equation in the so that ~2y/x From y = + x = 2x~ % = 21. Implicitly differentiating i/ = = y' = 2 j/ + (x ci {x 2 —x ^ (y') + xy' = 3 ifx>0 „ Smce if*<o' = xlnx we obtain P= 24. Differentiating y' we obtain y x > ~~ (l + + = + x — c\/2y. Then 4y (2x , = see that ifx>0 if ,<o, } {-2., Then \y\ at ) ?/ a 2 ae at - 1+bae*' 2 —— 25. Implicitly differentiating In (2 - 26. Differentiating y W = e - x = x 2 , -co < x < itiBaPP-ent that v — —y = cc, and f-~ , =VM- 1. ab Cl e at acje at 1 ln(2 — x) + „. h [ bcK** —r 1 — ln(l — x) = f we ^ (3lj£. obtain ct^C — 1- i). edt + c^e x we obtain X 2 - e-^V - Ixer 3? [ Jo y' = we obtain fccie"') bCl e we y f»(l+fci^)-^j ^= , y. ifx<0 , = 1 + lnx. ac\e at / (l dp Then 2x~ 3 C V + y(y'f= -^ + ^ = -=y. if * l 1 c ij ' From y Implicitly differentiating 0. = —l/x 2 we obtain y' = —2y/x. From y V 23. — 1 y 0. form so that 1, 2xy' r— (x, vlfl't-x, + 2y} y'. we obtain 1 — 2yj y' (x 2 + e t2 dt - 2c-l XfT x* - 1 - 2xe~^ [' e^dt - 2cixe~ x2 - Jo Substituting into the differential equation, we have 1 y + 2xy = 1 - 2xe~ x * f e^dt - 2c xe" x' l + 2xe~ x * Jo -x - y — — 2 27. First write the differential equation in the form y —+ — x ^ z (x m y) 2 and + 2c\xe~ x% = 1. 2 . xy Tfj/1 -r dt j JO implicit differentiation gives 2ci (x + y)(l + y') Then cAx + = v/x xv' , xe 2 y) —V „ — xe y ' x implies + ev I ' x . Solving Exercises we obtain for y' y + y)- 2c\ (x 2s e<J/z FVom y + y' - y" 29. = FVom y that y" c,??1 = \2y = Sx e -6y' + c2 e~ ix we obtain y = we obtain + = 13y From y - e^+se 21 we obtain 31. From j/ = coshx 32. From y = C\ 33. From y = 35. + From y = ci we obtain c\ + I 3e ix cos 2x = ^ = 3e y' 2t — 2e 3x +2ic 3ir and = y' = — 5ci sinhi ix = and y" 2x and y" sin cosx + -1 x + tan 2). we obtain x 9cie — From 3/ = that x 2 y" 37. From y 38. 39. From y = — — 9ci y'» _ y" - +2X 2 From u* = From y = + 3y" cos 3s c2 9c 2 cos 9y = In 3y' — = -2bc\ cos 5s, so that y" {x + c-iY , = — 1 + sins ln( sec x + + y = tanx. and y" = 2c2S -3 , 1 f0, < , = - sin(lnx) + cos(lnx) and — 0. = 0. iy') 2 tan x) and x^-^ so that y" + 25y = + so that y" y. j! = + 2^OX 0. —- cos(hix) — — sin(lnx), so 3a: = + 2s In £ 3s and y" = 5 + 21ns so that x 2 y" r + 4e we obtain y' = 3ci cos 3s — 3c 2 sin 3s + 4e + 4e s and y'" = — 27ci cos 3x + 27c2 sin 3s + 4e^, — 3xy' + 4y = 0. x , so that , 0. dx j/ +V x<0 ' „ 2 sin 2x, so = x 2 ex + 2xe x y" = x2ex , + 4xe x 2e x and , y'" = xV + Sse 1 + 6e x , so -y = 0. X (-x 2 Vie1* 4"|-4— +4y - 0. + sinhs = coshx obtain y' Then y" y' s + 4x 2 we obtain dx 2 ( y' obtain — sin 3a; + 9j/ _ From y = c\x dx* 42. + 3s ci sin — 0. From y = x 2 e x we obtain 1 41. + 2y = xy' — x 2 + x 2 In x we y" that y'" 40. scos(lns) we obtain - lGcae- 4 *, so that + so that (fcr 36. 2x 5e 31 cos 2x Se^+tee 2 * and y" + cj = — cis -2 y' + coshx —-— and y" = 1 + tans) we ^= sin5x and y" we obtain y = c2 ln(sec cox y' we obtain cos x ln(secs = tans + y" + sinh x cos 5s From y — — 4c-2 e~ 0. 30. 34. 3ci<! ' x 2 - xy ^ 0. cos 2x In Is - ix 1 _ +y x 28. 1. x > s<0 = y' + c\ c.\ in x + 8s, y" = cis -1 + 8, and y'" = — C\X~ 2 2 we obtain y , = f-2s, x > 1 2s, we obtain y , ' = so that a;<0 f0, < x<0 i so that .. 3 , (y xy ,,9 V— —2y = f°/ 0. x<0 , so that Exercises 1.1 43. = From y y' = + cx c2 we obtain 2kx so that xy' + (y y' = 2 1 ) — c so that xy' + (y') 2 = cx + 4k) — 2 y implies that x k(l — + Then 0. = — 1/4 A: = From y V- fcx 2 we obtain produces a singular solution. 44. From y = cx y' = 4- \/l 4- e2 -xjy. Then 1 we obtain y # for y xy' 0, = r + c so that xy' — = -x + (y') + ^/l + 2 45. + y2 — implied by x 2 By = —1 inspection, y is ^ together with y 1 a singular = 2 From x 2 + y 2 y. = 1 we obtain 2 w — =!/. 1 + -7= = 2 y is (y') vy 2 The condition -1 < x < 1 " 0. Note that solution. this is the "solution" obtained by computing the limit as c approaches infinity of the one-parameter family of solutions. 46. The = function y f 47. 48. right = From y hand -V4 - limit is we obtain x —2,) and hence = me" 11 y' — 0. does not exist at x =m and y" 2 mx Then y" e . m = 2 and m = — for all x, From y = e™ we obtain for all not continuous at j/ = — (the left hand limit is 2 0<x<2 , (m _ 2 )(m - > ml > y' x2 m 2 e mx _ 5me mz + fe m* = Since e mx Since e -2 < x < is I and the s/T^, i Thus y 3. mi and rae = y" m 2 e m;E e . 2x and y Then y" 5y' 3)e = e + ml 3x = implies 0. are solutions. 4- lOj/ 4- m 2 e mx + IQme™ + 25e mi = (m + 5) 2 e mi = x, m = Thus, y = e 51 a solution. 5. = 6y 25y = implies 0. is m~ l = mx and y" = m(m— l)x m_2 and substituting into the differential equation we obtain m m m(m - l)x - x = (m 2 - m - l) x m = 0. Solving ro 2 - m - 1 = we obtain m = (l ± i/o") /2. 49. Using y' Thus, two solutions of the differential equation on the interval 50. Using j/ + 4y and y" . m(m Thus, y 51. — m= x~* and y It is easily y?, —4, - arO+v^)/^ = m(m— l)x m ~ 2 and substituting into the differential equation we obtain = [m(m — 1) + 6m + 4]x m The right side will be zero provided m satisfies = mx"1-1 x 2 y" + Sxy' 0<z<ooarey = 1) —1 and two = x _i shown that 2c!X4-3c2 x 2 and + 6m + 4 = m 2 + 5m + 4 = (m + 4)(m solutions of the differential equation 4- 1) = 0. on the interval < x< 00 are . y\ 3/3' = x 2 and j/2 = x 3 arc = 2a + 6c 2 x so that solutions. If x 2 t/3' - 4xy'3 are solutions. 4 4- y3 6y3 = cjyi + c2 y2 = cix 2 4- c2 x 3 then = 0. Hence cxj/i, c2 y2, and yi 4-3/2 Exercises 1.2 52. shown that easily It is = y\ x 2 and = 5/2 x 3 are solutions. If = y = ciyi 2cix + 2ci — 1 then y 2c\ so 2 (uM + 2U- = that zy' + 2c\^y 2c x x 2 = ~x + 2 y' 53. (a) y ~~ + so that = ~^-x 2 ji ? and for c 2 / and + —^— = so that xy' + 2^y. = The sum on the body of the forces acting , , mo — , second law we have s , kv mg — is from g „ Using , = k/R 2 = , , (b) a „ „ , , we m— = find k = gR 2 gR 2 dv dr — , dr dt 4. (a) = 3/ ;/i cm , t The sum of the forces acting ma — — kv m -t-t at 1 d^r k dr dt 2 m (b) Letting 1 6. By = is on the kv where k k or —kv 2 — mv 1 dt &= _ a is — + = 1 then 2 are solutions. 1/ 2 — —1 or y a constant of proportionality, From a. 3, a = = k ^= gR 2 dr rl = ma — SR 2 gR ___ = 2 d2 r or . 0. kv, so by Newton's second law, ^5-. Thus, using v l = ~. r and r — = kv t1 dt v, the equation is part (b) becomes Problem = m ^Xr2 or dt mv = a, which is the 100, and 1. = + dt* equation 1 dt E(t). at we obtain R^- + ^-q = is + Ri = we obtain E(t) dt ~=— dt g j/ 1/1 + r2 and ^ dt differential and 2x 2 Kirchoff's second law The = -x 2 /2 + = -c 2 x + mg. dv satellite is C at 7. i/2 (c) dv „ = 0oru- From Problem . _ gR dt R = equation i + (-C2/2)a: a then y' c2 y2, dt 5. Since - . <^r J t > u -^j and part (a) we obtain Part (b) becomes (c) c 2 y2 acts in a direction opposite to the motion. dv m— = from Newton's second law we obtain (a) If 1. = j, ° 1 dt 3. ^ c2 dt 2. If 1. Thus, none of and the minus sign indicates that the resistance „ Newton c, (b) no real solution Exercises 1. y for C] Aw E(t). J2gh. Using .4o = jt ('-—') \ ' = 127 32 this becomes dh dt - 0.6tt/36 ' 100 8nVh rvr-— 64/t = - --„„,- - - —- Vh tt 6000 750 . — 36 , ^= 10 2 = ' Exercises 1.2 8. The differential equation and 5 = is ^=_ 0.6j4o ^f— 0.6tt/576 dt= The differential equation — = — —A—w — To 32. find Aw we x solve 2 0.6(8) = - \l2gh : is "4(576} We . v at and g tt = = 7r(2) 2 = 47r, + (5 — h) 1 — have An = 2 A^ ^ "480 ^- 25 where a: - VlO/i ft 2 we obtain dft 0.6^/144 _ A A w = ir(10/i - ir(10A - /i /l /g^r 2 ^= 1 = - 30A(H> ) 10. The differential equation is A'(t) = kA(t) where k 11. The differential equation is x'{t) = r 12. Equating Newton's law with the net forces fcx(i) 144 represents the is From h. Thus ). 2 - — (——] \12/ iv radius of the circular area of the surface of the water whose depth x = f^r) 32 this becomes <to 9. A§ = /^ft. Using > 31>A (10 - ' h) 0. > where in 1 A) 0. — the x- and ^-directions gives and 2 d y m— j = — 13. respectively- Prom Newton's second law in the x-direction we have 2 d x m— dt2 md 2 y A 15. To = respectively, and k > 0, see from the figure that tan20 = , , Id 11 r- . rfi is Ady k(a~x)(j3—x) where a and rfi = 20, tan <j> 1 —^ = 2tan0 - tan-^ x x L down -, so - = j/ y —- 2ldx/dy) 1 f 7T I L-~~ and i (dx dy) 2 6 ^ ~dt are the given — — tan — — 9 \2 ax , dy lcl to the x-axis. x dy — — and y = 1 . better understand the problem extend the line tan?i dx —dx = — — v dt , x'(t) differential equation and B, , = ~ mg ksm0a = ~ m9 ~ k = ~ mg ~ ~ vTt -k¥ The , we have In the y-direction 14. „ —kcos6 — —k - 2 ) = amounts of chemicals Then we cot G. Now 'V y — (dx\ „ ( dx\ = +2y \dy ) \dyj — L i. Exercises 1.2 16. We have from Archimedes' principle upward It force of water on barrel then follows from Newton's second law that = weight of water displaced = (62.4) = (62.4)7r(s/2) y x (volume of water displaced) 2 —^-^ = —15. 6tts g dt 1 g 17. = w 32 and ~ &i ' m gives Dividing by 2 15.67rs y. 2 or i( dt 2 w iere ' ^-f 4 dt M the mass of the earth and k\ is = --4£ of motion with his law of gravitation, k\M = gR j/B is , 3 = w q wnere where k , = k\M. The constant k = gR 2 or k . we obtain a constant of proportionality. is is R is the gR 2 where , y the earth. This follows from the fact that on the surface of the earth y 2 15.67T8 + the weight of the barrel in pounds. is By combining Newton's second law m ^f" = = = If t is =R radius of ^" = m9< so that ^l - the time at which burnout occurs, then y(0) the distance from the earth's surface to the rocket at the time of burnout, = R + yg, where and y'(0) — Vg is the corresponding velocity at that time. 18. Substituting into the differential equation we obtain —(mo — at)g = du — at)— + M—a) (mo or dt i (mo dv — at)— = i , ab — m^g + agt. at 19. By the Pythagorean Theorem the slope of the tangent line is y' ^ — . 2 20. (a) We have M r = \l» M •So 3 -dr and = ^SR 3 Then . r , M r where The 3 w2 = differential = is 2 part (a) we have d r — m —-^ at 4 dA — = k(M — dA — = ki(M - A) A). dt 22. The differential equation is dt -^ and 3 k-^j- equation r Mm/R _ 2 d r . From rr = ma = m-pr and at* 21. T it Mm_ (b) M 3 - y2 k^A. —k mM mM _ r R Aa <Pr or — ^ dt* — kM = 2 r —w r R6 , Chapter Review Exercises 1 Chapter Review Exercises 1 2 1. First-order; ordinary; nonlinear because of y 2. Third-order; ordinary; nonlinear because of tAnxy. 3. Second-order; partial. 4. Second-order; ordinary; linear. 5. From y = x + tunx we obtain y' + 2xy From y = c\ have 6. = 2+x +y 2 2 so that x y" 7. + From y — cie y'" = cje and x xy' +y= + - c2 eT — sm2x + 1 sec 2 x tan obtain = — [c2Cos(lnx) %c i e 2x we obtain so that y'" = cosh2x we obtain = y' - cye x — -y' 2y" + C2B~ X y = x2 10. y = e 11. y = \x 2 12. y = 2 13. y = e* p 14. = V* 15. y = sinx, y 17. For all x2 - x 18. If is 19. \x\ — - < 1 > -1. That 2 and \y\ also true for The differential > \x\ 2, > 2 Aw ff 16. y > 2y is, x —1. Avoiding < or x > 2 < then (dyfdx) and equation r corresponding to is \y\ < 2 x we ci sin(lnx)j and c\ sinflna;)] left- y" — , c\e x + c 2 &~ x + Ac^e 2*, — 16 cosh 2x so that — e \Qy = 0. 5x x and right-hand derivatives, we then must have 1. and the ^~ — —^-\/2gh. dt Aw v satisfies + 2<?3e 2x differential equation has no real solutions. This 2. r = dh dt 20. 16 sin 2x = values of y, y 2 sec + 2y = 6. 9. = + tan 2 x = 1 0. + cosx, y — + C2 sinflnx) + ci cos(ln x) — From = Using x cos(lnx) [ci y' 8. y x. . + Qe* x + 3 x = and y" 2 — —j C2e~ x x = 1 + sec 2 x, + C2sin(lni) we cosfrnz) y" y' . rr- _ " From Newton's second law we obtain Thus We r have An = -. 4 To y— _ _25y^S 9 1/4 = \mg — 2 Aw we note that the radius Then " loir/* 3 / 2 u— mi 2 8 find ^ and A w = —^r-- 4^/25 V at — = or at 16 f1 > — VSu). ' For /(i. For f(x. Thus the solution in the entire plane. y) = y/xy we have = — we ^ region where x 4. 2 2 . For f(x. 10.1 = —-2x y (x 2 ^ . = 2. Thus ay 2 j/) For f(x. (Also. and y = and y = which is 2 3 x 2 = e^ ^ dy ^ \ 1-1 *. . 3. For fix. For f(x.2 First-Order Differential Equations Exercises 1. For f(x. + y we x ™= solution in any region where y 6.) . differential equation will have a unique - —1.? cosy we have x ha ~2x (y-x) 2 . y) in y 2 ^ we have df — = —y~^. The solution is y = 0. Thus the - differential equation will have a unique + y2 f solution in any region not containing (0. Thus the differential equation will have a unique differential equation will have a unique x. u) = have ~=- For f(x. — -3xV —— 9/ — = j or y 2. For /fx.y) — df — dy = where y < x 3 a.y) = (x - l)e v/l' x ~ l) we have unique solution in any region where x 11. where y > or df — = —x dy 2 siny. 3/ = dy 9/ we have 7— jx^ + y dy — Thus the -r- solution in any region where y For f(x. 8. any constant multiple of x 2 unique by Theorem 2. y) — 5. ^ 0. r 5 2 ^ Thus the 1. . Two solutions are y = = 12. . Two solutions are y 13. we have ^ 7 < equation will have a unique solution equation will have a unique solution in any equation will have a unique solution in differential 2 2x y * Thus the . the differential equation will have a unique solution i any rectangular region of the plane where y 2. > 2.y) = . is a solution. Thus the = have 2 7. Thus the . we have s 4-y 2 —x— —y differential differential . y) —+—Xy = we have y-x solution in any region 9.0).1. Thus the differential equation will have a .y) 0.y2 ) —2 < y < equation will have a unique differential (4 —2. - 9 and and that df/dy is df/dy = y 2 /\Jy 2 . For y c. Exercises 2.4). —3). — 15. 2 + 1 value problem on the interval — 1 < x < is Theorem = 0.y) = not continuous at y cx we have cx a solution since 16. its partial the differential equation f(x. The differential equation is not guaranteed to have a unique solution at (5. 1). dx we obtain y o = \{x + l) 3 + c. 10 This yields = x c\e^ \ \g{y)\ = \g(y)\ e^ +e = f(x) + c. < 1. df/dy continuous. 18. Since (a) 1 -\-y 2 = y' it is and c. is not a solution on the a solution of the is initial 1. y = tana: y — tana: initial condition.3). The differential equation has a unique solution at (1.9 discontinuous for \y\ < 3. 3. 2. Letting c\ From dy = sin 5a: da: From dy = (x + l) 2 encounter an expression of the form In will sides of the equation. y for all values of piecewise defined function is not 0. y = tana. we we exponentiate both — ±e e^ c T '. satisfies the differential tt/2.y) Theorem 1 a solution of xy The 2. different iable is so Although 1. = 2 yjy a.-e -31 dx we obtain y = \e~ + c. — ±e c we we obtain y obtain g(y) = —-cos 5a.1. has a unique solution through every point in the plane.1 — does not apply. We • We further note that then apply 2.1 Exercises 14. The differential equation is not guaranteed to have a unique solution at 20. = y is derivative with respect to y are continuous everywhere in the plane.y) = — \y — 1| is is is All of these solutions satisfy the initial condition y(0) = 0. from which we see that y not different iable at x — (tanx) = ax (b) Since equation and the sec 2 Since (d) Since tanx x = and tan x tanO discontinuous for \y\ < 3 = undefined for x is and continuous on — 1 < For Problems 17-20 we identify f(x. (2. (c) = 1. o 3x From dy . —2 < tt/2 < 2 and tanx interval -2 < x < 2. A function satisfying the differential equation and the initial condition }{x.2 In many of the following problems solve for g(y) implies g(y) 1. + c. 19. The differential equation is not guaranteed to have a unique solution at (-1. 17. 2. To = e c e^ which . 2 From —^— dy = .2 4. = smxdx we dy 2y^j or y 2 k dr = obtain 2 \-y . From (~K2 Vy From 19.dx we obtain y+i x In \y c.P| = kt . \y\ = In \y\ 4 In = |x| +c or y = cjx 4 2 +c or y = Cie —a: .+ we obtain y — xl x +6 + x 5 In + II c. or 2 + y2 = ci (4 + x 2 ).y + In U |y (4x + az 2y a 2y + = fcf + ln|y| IWI = 3 — 4x + +c or 2 dx we obtain we obtain 5 = = —. \y\ y ) 11 From From + obtain In obtain 3e~ = j—pdx l\ dy +2+ (2y 21.dy yj » ) -r* dy y + 3) + 2e 3x = dx we obtain ye - —+—x^ 1 iy In +1= + y 2 = — cosx + c. I c. dj/ 2 dy \* 20. 17.Exercises 2. = — 2xe~ x + 2e~ x + c.+ 13. (e~ x ) y { = + From 2y dy 18.-x 3 9 1 1 1 3 1| = — — + c\. + e~ x + ^e~ 3x = ey + y2 = In |4 | —+ + x2 + c y — tan"" — In |x + 1| + c. From y 2 dy = ( —x£ + —~\ dx we obtain -y3 — . 2/ 9. \- x y* 10. 5 kT'. x -dx = + x —. | 1 x c. (+c so that In j^—p = *+c . From y 31 e~ * — 2 — — dx we = X ~ dS = we obtain <ix ln|2 +c Ixl = —3 + 3xln Ixl + cix. — x From —= dy — —rz dx we obtain y~ 2 — + = 11 In Ixl +c — 11. From ^. From ^ 2 16. From 23. obtain In |Q + j^Tp) Solving for 3 i ^= we have - 70| weobtain P= 11 In Q . + c. From or dQ = kdt we p ^ p2 ——= = dP = cie'. \dx From dy = = From dy = / 1 5 + \ 2xe~ x dx we obtain y From dy 7. From e -2 "^ = e^dx we 14.dy 4 = — dx 8. J 1 22.+ c or 2 ^y - 3 + 2 -—-^ 2 5) 1| 2 2 x lnxdx we obtain ^- — — + = - 1 y + In |y + X — ln|x| . xy 3 or c. _:c . From . From — dy — —2x dx we obtain II In we obtain =x+ dx we obtain y i + 1/ \ 1 6.+ \x 3 x x) 12. 5. y obtain y 2 = x ~h 1 = \dx we l\ . + c.70 = cie |P|-ln |1-. From ye"dy 15'. we obtain C\x. + c. = xdy 2 - 6 sin lj = = — tan 3x sec 2 3xda: we From tan ydy = xcosx dx we obtain (e^ 3 = -cos 2 xdx = --(1 + cos2x) dx we 4 cosy or 28. From 25.+ ' l) ln|y+l| 32..2 24. t+2 + c. 2 2 x-3 / . From — ^ (2cos-! cot 36. = . dx we obtain j.—— — - Exercises 2. = sin x cos « 4.-51n|y-3| 34. = In \N\ x cos y — cos x sin y esc 2j/ dy = cos x dx. Fromey — y. 2 j/ - e" + 21n|y-l|=x + 51nlx-3| + c - c. = } ^ — cos^y-l dy x'y .sin 2x + c 26. ) + c. dx or cosx From 30. dy sec y— obtain = — . + (e* = dy e + sin2x + c\. From <fy - From y — -dx From 2y dy — 33. ——+ ie'+ 2 . obtain 2 1 +4 t ^ we obtain - = - 1) . dx cosx 2 cos x 31. 2x 2 sin x cosx = 29. dx we obtain y 2 dx or ? ? 2 ' (l V sin x dx or cos-'y dx we obtain = or dx we + —^—] 3/ 2. y From —- ^-r j/ = — cos x + 1) — I- sin dx dy 2 sin y cos y — „\ = -2sinxdx we obtain ye y )dy ' In sec y | = | xsinx (e y + cosx 4- I)" = 1 * + c.sec 2 3x + c.cosy = -ix . + ^-y = ^ln|l+x|-iln|l-xl + c.esc 2 ydy = sin xdx we obtain c. (e l 1 + l)" (l+y 2 )^ = dx or f —j— . From — (2x + 3 = - \dy +3 y 1 From y^tl dy y - = 1 y 35. + [ 1 + 4/ V = x-51n|x + 4|+c f 1 H = (l ] dx we obtain + x + c. From ^dN = (te t+2 - —^— dy = 5— dx or sint/di/ x sec-1 esc j/ we obtain l) df From = — 2i/dy — r^r~ dx cos" ix e 2y 27.cos x sin y we Then - In | esc 2y 2 12 — cot 2y| find sec ydy = sin x + c. obtain g ~ or cie cie'~». ^— ) dy = ^ 5— dy = dj/ ^— \dy=(l [l ^ dx or ) + c. = 2 sin y cos x t .— = a: 2 ?/-3y + 1/2 2 a. -~-dy = + ye~ y + e~ y = 2x . From + cos x 1 (1 is e~ y + e")(l + cosx) = (l + e M )(l + cosi) 1 42. Then 2udu = dx and x. The 4.2 37. +1= 2x 2 + \/2' c or y yV + 1 2 <Jy =t— -t 2 + -1 = x tan -1 dx or ^ (—^— 2\y-l - + 4y x — y = ~ + c. = find ci tan'^ + tan" 43. !2 = 2x 2 + c. . from rfw {e x make r = / — — / r- +c y — or y + y2 — sin -1 sin .„ 2 1 = dx ^— ^ f 2Vx-l x + find c find 2 /2. i/ the substitution 2udu = -j— — = 1 r / u2 = —2du— = dx we obtain . = 3e -(t-i) we x = tan ^4y — we 3 2 cie'""' ^ Using x(7r/4) = J 1) ci. = ^. . = 4idi we — = 3e The c obtain i)df we obtain is . + 2 In (. FVom l + . 2 —— 1 ttt 1 tan -1 e x + c. To integrate dx/ {x -Jx) dx r ——— Thus. 4y is Using y(0) = y = 3e — or = is '-' 2 find c = V2.„ = (2{/) we €^ = we * 2 + FVom -dy = (1 2y = = — —1 tan _1 7r/4. FVom dy we obtain -x . The — dx we obtain lJ we . 21n y . — dy . From tan or 2 -1 ' 2 solution of the initial-value problem _! *j x + J/ c + c or dx we obtain solution of the initial-value problem 44. 1 solution of the initial-value problem = 4dy we 2y solution of the initial-value problem In solution of the initial-value problem 46. Irtfl (z8) I find Ci dy . = — 3tt/4. x sin 41. — \ / u+1 +c = 21n(^+l) + c. FVom — lnfl + cosx) = obtain e" 4.+ 1 dx we obtain Y+ ci 1- f ) c. ^ = 2. From dx 2 obtain tan * y2 - dy 1 = - x2 — 1 -l is + _1 9 tan = a.— sin dx we obtain J^ . From xdx = 38.Exercises 2. = From —zdy + ex jr dx e~ x + 40. The * 45. lie ^ = dy . 1 = 1 Using y(0) — tan Using y{l) 1 —x l —+ — dy we dy — - + c\. y/4-x* + 39.The Using y(l) = 3 1 /2-i/2 = 1 .75 = 1) 2 In (\/x~ + l) +c or + e y )' y/y+l=ci (v^+l). = 0theny = solution y obtain y y 3 \y- 1 x solution of the initial-value problem -~ + -^—) dy = dx we y-3) | — I -4. - In |i = ~—y~ ^ = (it. Setting c - 100 l^tan The lOy 5 In and y -1 10(y — 1) = 1 + c = 1 + — tan — \ solution - 11 = is x y and + c. = 1 satisfies we obtain 1. 6x ce . + = — 01 1 is 14 dx. \ + -^.(0) From The . —+ = x . we obtain C ^ 56.— . 3 then » 1 -I- . 3 Iy — 1 1 — = lnlyl In +c la:] = or y — 1-cix Another . = l/2theny = |a. Separating variables — tan X + the initial-value problem.(0) = (c) Ify(l/2) In - |1 2 _ 2 + 3 e 1 theny = 1. -. 1 = —— — \y\ x. x 1 —+ = i <£r we y / 1 y = = ltheny = Ify(l/3) is + 2 = + c. Separating variables . = l\ solution of the initial-value problem — 1) = —1 we 48. By inspection a singular solution is y = 0. Using y{2) I r 51. 0. — = = 2y| obtain In y y = i is = or y 1 efa „ a inspection a singular solution and y In 1 + 61 " 2 By — x = 2 we find c = 0. The singular solution 54. ^. The 55. Exercises 2. +c 1| we obtain ~) ~~ f ~~2 (a) (c) 50. 52. = c = Then x + c —— —— X ^ ^ we obtain = y — 1 el1 0. 1 49.2 \y-l\~ In The 47. e c\e~ l ^ x Using .| = + % ^ - - c or 1 - is 1 2y is xy x = 2y c or c\e~'* = -Ae~ 2x . In \y From - From 1 From \ dy — = find ci . \y + 1 = - y y is —+ or - solution of the initial-value problem ~ dx we obtain iy Ifi. xy = = -1-1 ^. — dy ] = (a) If 3/(0) - (b) If 2. Setting z we obtain ^ = y - solution = 1)^+ and y ^— Then 1) we obtain c — —100.01 1 - -rs = dx. (y + In T 1. Setting x X — .In \x + 1| find ci = = e _1 The 3 (b) If 1/(0) = 3. Using y(0) = 5/2 we 2x + - or y = 2e~ . 53. Separating variables t. then y = 0. Let u 1 tan 58.y). i 4. 2. Since f(tx.Exercises 2. cos-' tanii 4. . du/dx du/dx so that = 1 tan(x a. Since f(tx.1 5 In is cfa/x. = (txV jt.. 2. Multiplying u . the function is homogeneous homogeneous. Exercises ° 1. Thus 1. Let we obtain 1 dy/dx - 1 = 1 + e" or e~ v du = c.2x .ty) = 5. Then — + dx </u or -^= du = dx. Thus \u2 dx.ty) — Since f(tx. Then ^+2 =2+ 1. y).) 2 t2 f& V)< the function the function is homogeneous of degree homogeneous of degree is 3/2. = +c x 2 ti and 2(x + dy/dx. x = du/dx ™ — dx du u — dy/dx du/dx dy 3 so that 5 so that and — + c\ and 2\/w . ty) = sjtx + ty (4tx + 3ty) = 3 2 t ^ f(x.y).tan it sec u) du 2 v + + sin u 1 = Thus dx. Then + y) — sec(x +3= = udu — 2 tan « or cos 2 + y) = sin2(x + ci Ax Thus dx. fx 2. or —+—u^ £ \ —- = - dx tan (a: + c) — x — udu = or = ifu ^— 1 + 1/] + dx. Thus + c. Let u 2^/u = 62. Let u or u 2 ~u 2y = x = x + «y = = i (1 +y x - = 2x - lOy The solution = + 1 2x + ci = 1 + 61. or -e^"^ = 5 x + 1 = sinu or = dx or (sec = j/) x + rfu = dx. Since f{tx> ty) = 3 cos +y 1 -f(x. 2 + y) 2u 0. ty) — 3. and £ + y + sin 2u = so that du/dx sin«)/(l 1 = y so that du/dx + sin 2(x + y) = = + c). Thus dx. and (x + 5 In so that 1 + coru = + 2x + = c + isin2it-x + cor 60.ty) = 3 (tx) + 2(tx)(ty} 2 = - 3 t f{x. of degree 0. y). ' c. +c — or tan(x 1 1 + = 1 u2 = or c) x. sinu) we have — secu = x + c = - 10y-9 Then dy/tt Then + dy/dx. the function is homogeneous of degree 3. Since f(tx. the function is not X sin is of degree 2 4- x = f(x. —1. (x (tx) 6. Since f(tx. Let u -e~ u = x = = — - y +c y x 2x + 4x + ci. Let u by x + y u u 59.3 and y = = 57.x+ - 11 ~ dx tan(x Then dy/dx. the function —= ("cos —X — - x+y for any - x -\- y homogeneous n. ux we have {x 12. ty) — + inx 2 — ^nix J (i^^ 3 In ty Since f{tx. . ty) — 9. 11.y). n. not homogeneous.Exercises 2. Letting y homogeneous of degree y = + + + In y) = 2(ln( |:r| + = + xdu = + l 2u 2u\ = c x 2 + 2xy = ci. ux we have (a: 4- ux) dx + x(udx + (1 + 2u)(ix —x + In 13.ty) \nt 8. the function — ux)dx + x{udx + xdu) — dx 4- x du = dw = f- a: a: = In |x| +m= c In \x\ +y = ex. Since f(tx. # n t (x tne function +y+ l) 2 for is any homogeneous of degree —2. the function ^ ne functi on ^ i s is -j$f{ x < f). Since f(tx. Letting y = ^j(tx In ty + 2 l) for an y n f(x. Letting 0. Since f(tx. In ^ xdu) + [1 x = vy we have + y (fv) + vydv + 1 {y (j/ — (w In |w - 1| - 2v - 2vj/) <fy = + = l) 2 1) i— + In v — 16 1 dy y |j/| = c is not homogeneous. ty) = 10.3 2 7. vy we have y{vdy + ydv) y dv 2{vy + y) dy = - + 2) % = (v tit) + v lnjv = ux we - 2 —= <ft/ 2 y + 2| - . Letting y . Letting x = + \ny — x/y-l V .3 ^-1 In (x 14. ux we have 2 (u x 2 + ux 2 ) dx + x 2 (udx + xdu) - (u 2 + 2u)dx + xdu = dx du x In \x\ + ^ In |u| - u(u ^ In ju + 2) = + 2| = c .+ In 15. + 2j/ = ci?/ 2 have 2 [u x 2 + ux 2 ) <te - 2 + xdu) = x {udx (is — 3 <iu = (is <iu 2=0 —= In |z| H y In = c +-=c lnld 16.y) In | a: - y\ -y= c(x .y). Letting y c In \y \ = c - In \y\ = c a. [a:j +x= cy.Exercises 2. have (ux .3 u+2 ci \x x = 2 x y 17. du - ln\x\ + Inz ^ln (u 2 ^+ 2 ln(z 2 18. ux we have (i + 3ux) dx (u (3x 2 - + ux){u dx + + X In |rc| dx l) + x(u + ti 7 lj = 3) du = r + w 3 du = — In |ti+l| = c - Cl (u-l)(u + 21n |u — x du) + l) z(u-l) 2 !1+ ( !/ 18 1 -x) 2 = c 1 (y + x).x)dx (u 2 + x) (ti dx + x du) = (ux + l) + dx + x(u du 1) + — + -~u — + dx 1 . Letting y = iiiwe cj(y + 2x).= c ci 1 ci. Letting y = + l) = _ 1 U-' a: = 1 + tan" lj+2tan !/ _1 = ^ = + y 2 ) +2tan~ . .Exercises 2. 2 c)' + c have x\j\ + u 2 dx-x 2 du = du dx In III — i+ In u + u2 y/l + + 1 21. Letting y = ux we have —uxdx + (x + {x +x + xdu) = ~Jux)(udx ) + u 3/2 dx = du u! \ x -2u.c \/l + u2 = c\x 2 + x2 = c\x y?/ x — vy we have 2v 2 ydv - (v 2u -In 3 It. 3 + l] 2 + .\2/3 = c 2 . Letting y = ux we In \x\ = 2y[x]y - = 4x.^ + lnH+lii|x| = l In \y/x\ + y(\n\y\ 20. Letting c = . dv lj 1 - In dy = dy — = \y\ .Exercises 2.3 19. . - l] . have x 4 + uV) dx - 3 + x du) = 2x ux(u dx 2 u2 - l) = dx . Letting y — ux we - I I2 set + I U a- . ux we have 3 (u x 3 + 3 a: + u2x3 } + xdu) = dx . Letting j? = 2 = c c. ..u 2 x 3 {udx (1 4- w 2) da: - u 2 x du —+ dx ~ x In In txl |a.\f u2 - 1 )lnM+x 3 = C (» -x a 2 have (x 2 + u 2 x 2 ^ dx - + xdu) = 2 ux {udx = dx — ux du dx u du — x In |a:| — ^u 2 = 2la\x\-(y/x) 4.+ 1 tan tan an = - u2 = -1 -1 u= c .Exercises 2.| 20 —u+ .3 25 + £T Letting y — ux we =c 2 y 3 + y°) =c2 y s x* 12.2xudu —x — —1u du dx (u 2 In 2 (K :3.= c. Letting c In |x| \x\ - n .3 25. Letting t/ = tii*e |x| + cosu| = c y x cos — x = c. Letting y — ux we 2 e~ u + u 2 x 2 ) da: - ux 2 (utte + x du) = — ux du — e~" dx —i -«c x In j/ = ux we = a. x)e y t have (ux + x cot u) rfx — x(u dx + x du) — cot u dx — x du = —x — tan u du — In 28.4e" 2 "rf{/ = |e 2 "-4In |y| = e^'v .we" + (y - du = e" = c x = ex.8]n\y\ 26.r 27. have (.Exercises 2. Letting x = vy we have y(vdy + ydv) - + 4j/e" 2 ") (vy dj/ = ydv . In | have ux In u dx — x(u dx + x du) = (it In u — u) dx dx x — x du — du u In u — u = . . + ux 2 - = u 2 x 2 ) dx + ux 2 {udx + xdu) (1 + u) lix + xudu = —x + uu +dur = dx i In |x| +u— + In |u u + 1| 1 = c = cie = ciie"/* a. 30. du = = + u2 ) = c 1 1 +u 2 =c ie tH1 " .3 - ln|x| In | = c = c\ x = Cl lnu = cjx s = xe - lnu 1| x — lnu 29. Letting y = ux we (x 3 have + ux 2 + 3u 2 x 2 ) dx - (x (l In Ixl 2 + 2ux + u2 ) - 2 dx ) - (udx i(l + x du) = + 2u) rfu <ix 1 + x 1 + v?T tan" 1 u - In f 1 2u .Exercises 2. Letting y = ux we 1 I i n V x _ l + In 2 + 1 1+C2X . have 2 (a.u x^if + x )^™ 2 22 1 *!*. ^ + 1 a. 2xdu 2du dx ln|x| + xdu) = u(u + 2 In \u + x(u x(y + x) 2 = + ciy 2 . solution of the initial.Exercises 2.3 31.^ln In |x| + y3 = Sx 3 + u2 +u * we find ci = 1/2.value problem (l = y have (x Using y(-l) = 3 + ln|x| c 4 — ci = ci 2 {y solution of the initial-value + * 2 ). Letting y 2 we = ux we find c\ = 8. . problem we have {Zux 2 + u 2 x 2 ) dx (u 2 2 2x (u dx + u) cfz £ - 2 In |w| . The 2 + 2 2u x 2 ) +u 2 In udu |x| 1 . Letting y is 3a: dx-ux 2 (udx + xdu) = eta: 1 c\J?. The = +u 2 ) = (l l = 3 . 2 . Letting y = ux we have + u 2 x du = dx dx Vu 2 du = Q x 1 + ^u 3 = In |x| 3r3 = Using 32.ux du = dx ) x 33. . + _ q l) 1\ l) = c = ci 2 is 2x 4 = y 2 + a. 3 Using j/(l) 34.Exercises 2. Letting y 4x(y + \Jx + x) vy we have vy 2 (vdy Using is = 1 we = ux we find ci = The 1. Letting x = —2 we = = find c\ 1/4. Letting x = = we find c = ~ e" = c x = c. e»' —1. dy + y dv) + (y cos 3/<ii> — vy) dy = v + cos dv sec v t> (fy = = -\ V In sec v I + tsnv\ + In \y = \ / sec y \ 24 x — y I- tan z\ — — ] v) c c\ is In \x\ = e^1 . x 2 + y2 have (x + uxe") dx - xe u (u dx + x du) = = dx — xe" — e "du = x - In |x| ln\x\ Using y(l) 36. The solution of the initial-value problem vy we have 3/(1. The solution of the initial-value problem + ydv) — 2 (v y 2 djf = 1 dj/ = In Ivl = c = Ciy + vy\Jv 2 y 2 + y 2 ^ ydv — + \jv 2 — In \7 + x2 3/(0) 35. 1 + y l =ciy 2 solution of the initial-value problem is . The du 2a: dx 2du a: Ji'' In 39. Letting y = ux we have 2 (u x 2 + 3uz 2 ) dx - (ix 1 + ux r) (udx + xdu) = — u dx — x(4 + u) du = — dx 4 + u — + du = X u In \x\ + 4In |uj . Letting y = 1 we ~ ux we find cj = e. Using y(l) 4 = c = r= \x\ H t — c solution of the initial-value problem is a. The 3 ) da: - 2a? (udx = ua: = rA x 3 e~y^is = u4 ~ x^e i v l x . = e _x2 /!/ 2 +i/2 . vj y 37. +u= xu 4 y Using y(l) 38.a: / .— Exercises 2. + xdu) = we find ci = e'^ 2 . we have (x — ux — v?t 2 xj dx + (i + v^a:) (udx dx + xdu) = + x (l + V") du = ^+ In x (l + du = + u + ^u3 ^ 2 = c O 3 2 3a: / lna: + 2 'ix^ y + 2y 3'2 = 3 2 c.+ tan -1=2. Letting y -ti have u 3 da: - = c\e solution of the initial-value problem (uV + 2ua.3 (zr x\ sec . Letting x = vy . Letting x = e we find ci = — e. The solution of the initial-value problem (Note: Since the solution involves yfx 40. -1. The solution of the initial-value problem + y) In \y\ = y - (x + y) 26 or (x + y) In \y\ = is -x. + - yln 41.Exercises 2. ) l) 2 dv (v + v + x +y \) is 1 we find c {x = - dy = dy = y .3 Using y(l) = 1 we find c\ = 5. The solution 3 ci. in . tf = c 1 y = dy 2 1 j/(0) c vy we have y (n dy Using =Q y + In |y| = In |ln |f || = ve we have y(vdy Using y(l) is + In \y\ = c. yki . x > and we do not need an absolute + ydv) + vy(\nvy — Iny — y du = 1) + ulnudy = <fo w In v dj. of the initial-value problem + y du) + 2 (u y 2 + m/ ydw + 2 + (jj + 2 t. 2tdt = '-* 1.l|1+a|+c I Vx r 3/2 <lt —d« = « du - is . — ux we Letting y have + {^/x dx — x{ u dx y/ux J + x du) = du (tc 1 a: + 2.v -1- 2\/l --f = = + 2^1 -x/y = c c.Exercises 2. Letting x = = we find c\ — 1. solution of the initial-value problem \n\y\ + 2yJl-x/y=V2.3 L2.e solution of the initial-value problem is vy we have 2 vy- \Jy -vy 2 - y(«d?/ J vT— v In In Using &(l/2) = 1 we find c = \/2 . x The + 2^/y = I u= + 2.1- +c 2 c. The |y| + ydu) = — y dv = dy dv y Vl ./u = nil in \x\ = = f / —-= /rfa Using y(l) 3. sin0)(sin 0dr + r cos0df>) = 0. sin9)(cos9dr Simplifying [M(cos 9.[rM(cos 9. l)dy + yM(v.9) cos9 28 Q . l)(vdy 46. 2/ = r cos (? and ?/ = r sin In |x| = c — In |x| = c. From i j — differential M(vy. y) dy = 0. JV(t>.y) = [vM(v. r sin 9 = r n M(cos9. we have sin 9) cos 9 + AT (cos 6.y) find c = n y M(v.rN(cos 9. sin #) the differential equation becomes r n M(cos9. sin 0) sin 5] dr dr r Af(cosfl. du 1) = dv = Q = 0. Letting y have x(u dx + x du) — (ux a: x cosh u) dx du — cosh u dx = sech u _1 tan 1 tan = Using 45. y) = M(r cos 0. sin. 1) we obtain dx = 1) + is tan -1 ^sinh - ^ = In \x\. 1) + 2 y N(v. y)(v dy n y M(v. — —= — ciu solution of the initial-value problem = vdy + ydv obtain = The 0. we From i = vy we Using M(vy. = N(r cos 9. sin 9) cos 6 ^ _ + N(cob 8.r N (cos 0. sin 9) cos 0] d9 = . sinfl sin 8) sin . 1)] dy vAf(u.Exercises 2. — + n N(v. N(v. sinf) and y) ) n N(cos 9.3 — ux we 44.r sin 0) = T Wfx. and simplifying we have 1) + ydv) + y M(u. sin9) sin 9 M(cos9. equation becomes N(vy. -r sin fldfl) + r n A^cosfl. 1) cos <?dr — r sin and dy — sin fldr + r cos fldfl. Using Af(i. 1) (sinhu) ^sinh and the + y dv) + and N(vy. sin ff) . y) ^ . we obtain = lj dy x \y |j =i"/(l. Nx The . y) .4 1. h'{y) = 2 Let A/ fx is = = siny siny — ysina: and — ysini we obtain 1 2 = xsmy + ycosx — ~y c. = = 3y + 7. and zosx = 4 -2i/ + xcosy — . -c ( .y n 48. y / = isiny + ycosx + frf!/). 1 y x 3 v / 4. 3. Mx dy dx we If . = 2x-l = ^x + 4xy+h(y).x M = 2x + y and N — — x — Then M = and N Let M = 5x + 4y and N = 4x .Exercises 2. so the equation = JVj. Let Af /i'(jf) 2. then by homogeneity f(x. The From/^ . 3 solution is x2 . let 1 j ^ u = y/x. The solution so that /i'(y) = = Prom /a 5i + 4y we obtain ^x +4xy . |) = tf).. Using M(x.8y so that M = 4 Let 6y. + yn N n y or M M^ iy' l) N(x/y. -Sy = 3 . = and JV = 3y+7sothat Afj. = —1.ljdx .2i-l we obtain / = x 2 -x+h(y).2yA = 2 is My = not exact. and — = sinx = 2 ^y . Exercises 2.4 = yn M 47. Using the chain rule for we obtain and du 9j/ du dy \x/ du Then = nx n /(l. c. 1) ( partial derivatives.u). u) = nx n f (l. and h(y) = -y 2 + 7y. 3 + -y 2 + 7y = c.y) = x"f (l. From solution . 1 j and N(x. War. is cosy —y. / 9. Let = / 6. The 0. h'(x) = 1 + ini. so that 2xj/coshx we obtain / 2 3 is From . h'{y) = 0. Let / . and h{y) 2 3 2 = A^.. M- + h(y). 2 3y = = is ~x 4 = 0. .1/f + $x 17.4 5.2xex we obtain . Let Nx = + xy 3 = c. = The 0. = Nx 4xy The solution 1/x + — 2y = y 2 x y is = From . Let - = h{y). x 2|fsinx From fx = x 3 + y3 we obtain . + N y/x 2 and = = and h{y) equation M 4 so that is = 1y. h'(y) JVX solution x In y so that JWU / 14. 3: From fx = 3x 2 y + e" we 3 2 is x y + xe v — y — c. 30 JV. 8.3 and x y — 3x 2 2 = Let A/ JV^ 7. From /x = y . 0.2x 3 . and 15. = and Afy) l/y + Afj. /e = iV. Let = y =y 3 - fx solution 10. = \ay.1/ + 9x and JV = x ^ so that M = 3xV = we obtain / = \x y . Let JW 3xy 2 not exact. Let Jtf = Zx y + e" and JV = x + le" . The ye~ xv and JV. 11. 2 (l 1 solution = and h(y) 0. ^'(j/) sinh ar = 0. V(jf) - y 6x 2 and h(y) 2ie I and - M = I-3/1+ if and JV so that — 0. is not exact. Let M = xy From fy = solution is + 1 lay = -2x/y 2 = solution Af„ = I = x2 is Nx + 2 3\n \xy\ sinh x e" = +y 1 -3/j/ + x so that h'(y) = 0.2y so that Afj. Let M 3 - . + 2y cosh x = and h(y) = JVX .^x = -X 2 /y 2 2x/y and JV h 13. 3xy 2 so that = = 2j/cosx so that JUV 2 2 / = xy 3 + y cosx . is = = = JV M = y\ny . The 2 1/x = JV 4. M— + lnx + y/x and JV = -1 + lnx so that My = l/x = Nx Prom /B = -1 + lnx we obtain = -y + y\nx + k(y). M = r ^ .6x 2 . — 2 - we 3 + 4j/ = c.2xy so that Afy Nx = and 2x ~ The equation 2y.\ arctan(32:) + h(y). M„ = 3 1 cosh x and = 1 = JVX . and h(y) - . = JV. The solution is — + y lnx + ilni = c. From h'(y) = 0. and k(y) = x\nx. Let x and . The = 1 -3/x + y we solution obtain is c.x we 3 2 +y xy — - sinx sinx 2 Jf is J/ 2 x3 + y 3 obtain = and JV h'(y) M = + 0. = -2y 2 x . j/ Let A/ / l 3 ) y 3 2 3 ) — arctan(3x} = c.2xe + 2e* = c. From fx y .0.3 In N = e y + 2Tj/cosha. + 3y 2 c.. The . JL/ y The N — x so that = . and h(y) x 3 ^3 = 2x/y we obtain cy.e'^ and N = equation 12.^x 2 = cos 2 = ^x 4 + xy 3 + h{y). 2 + + xj/ 2 cosh 1 = ey = ey + xj/ 2 coshx + A/s = 2a:?/ /((j/). The c. The solution is xy ..Exercises 2.c M=x 2 — y 2 N and 2yx2 = h'(y) — 3 sin 3x. = — 3sin3x — 3x M cos3x so that 2 2y x y obtain 1/x 2 and not exact. h'{y) 2y. = "ix + 3 2 v — — — — = y The solution obtain / x y + xe + h{y). / = x + y + xy - 16.31n|x[ + xy + h(y). 4X3 — — 3j/sin3. From . = xy-2x 3 -2xe x + 2e* + h(y). = Let 2y2 x . The solution fx l/x 2 + y2 ) - My) = so that 2 y/ (x + y2 ) - ey — c. and h{y) . solution — 2xy + is From TV^.y/ (x 2 + y 2 ) . Let A/ 21. If y(0) = 1 then c = 3 and the solution of the initial-value problem is e x + xy + 2y + ye" — e" = 3. h'{y) = 2y . Let 0. = ^y 2 The /i(y) . h'(y) then c = — M y = —1.Exercises 2. Let is 2 y sin x . we obtain The solution is - y If c. and ft(y) = y -y. and h(y) W^. 3 + y so that .3 and TV = 2y + 5 so that My = = TVX From = 3xcos3x + sin3x . —2xy+h(y).15x 3 23. 20.y + 2y 2 e xy2 we obtain / = ysin 2 x — xy + 2e xy2 = h{y) The 0. . and h(y) — 2y + ye y — y. M = tanx . solution M = 4x y . The = x 2 + 2xy+y 2 general solution is 4/3 and the solution of the initial-value problem is - M = e x + y and TV = 2 + x + ye x and From M„ = 1 = TVj. 2(x + y) = ye y .4 M = -2y and TV = 5y-2x so that M 18. h'{y) = 2y + 5.sin x sin y and TV = cos x cosy so that M Let = — sin x cosy = + cos x sin y + h(y). and h(y) = y 2 + 5y. k'(y) = y /x = tanx solution x sin y we obtain / = sin In sec x\ is | + = cos x sin y secx| In j TVX . Let M v /= — 5x 3 — M — l/x - 4 2 In \x\ + y = c. The solution is x + 2x y + y . 26. — y 3 . My). Let — h'(y) 19. From = — y. h'{y) c.2sinxcosx - 1 sin + Axy z e 2 x xyl + Axye^ so that + Aye 1^ = Nx .y + 2y 22. The solution is xsin3x — 3x + y 2 + 5y = c.y and 2 TV — xy + 2e = xy2 = x 4 + 3y 2 .3 we obtain / = xsin3x . = 3xcos3x + sin3x. = -2 = y and 5y.x 2) 2 (y --- / {x 2 2 + y2) = arctan x [ - ] + Nx so that TW„ 3 and TV = From /a . \y/ In Ixl 25. h'[y) = ye" = 2 — — — + 2xy + y 2 and TV = -x 3 -I- + xy 2 — y = x 2 y + xy2 l/x 4- . From fx = e x + y we obtain / = e + xy + h(y).x 4 x y is 24.1. From and h(y) 0. =^ y(l) = 1 [ — | + ye y — e 8 \yj 1 so that hly). 2 = 4x 3 + 4xy we obtain Let M = 4x 3 + 4xy and TV = 2x + 2y . The c. h'{y) = 2 + ye". fx = 4x y — 15x — y we obtain / = x y — 5x — xy + h{y). arctan 2xy + a: 2 - we obtain / = -x 3 + x 2 y + xy 2 + -x 3 + x 2 y = 4x 3 - h'{y) 2 + x/ ye y and x M=x — 3y 1 2 = Nx . The general solution is x e + xy + 2y + ye" — e y = c. 2 jjy = /j = -2y we obtain / = c. From fs = 2y sin x cos x . l/x 2 . Let 2 xy2 e and = -x + TV My . M = 2ysinxcosx . . = 0.. Let = 3 xy 4- h(y).3x + h{y)..y = c.1 so that My = 4x = Nx From 2 i 2 4 2 2 f = x + 2x y + /i(y). cosx - ysinx and 4- is — +ln obtain \y\ = c. If y = 8. = 6xy 3 we obtain solution of the differential equation M = -l/x a = Ax From /x = -y/x 2 we is . y y 34. Equating 35.5 we 2 = 3y .+ h(x) so that = 2y 1/2 18x?/ 2 = x 1/2 + ^ ^ A^.5x + x 2 —y= c. and 2 3 2 The general solution is y sin x — x y — x + y In y — y — c.Exercises 2.The ^(t/) ' Nx and the solution it/4 = . M Equating M 32.y. = M 2kxy 2 — siny we obtain (z< ») 2 +x ' c. - 1. /. (x 2 In |x 2 + y|+/i{x) From j. so that ft'(y) + iOxy 3 Sy 2 Axy + 2xy + 1/x ti( x ).4 M 27. The = e fc xy 2 +» " Let N(x. lny. solution of the differential equation 32 so that 1 + The y = — 5.ycosx + y sinx general solution 31.)" . and 18xy 2 — s'my and M{x. Let = Ay = obtain / 4xy + 28. 1. . Since /x = /j. y(l) = 1 3 5 — s = — 4 2y — 7 is = A^. The general solution is . + xy 2 + . h'{y) 6y h(y). A. Let . = 2 3i y 3 M 4xy + e x Nx = and A^ - xe x " = ^+ + 5 = + h(y).y) + is — — tan 2 = — xy cos xy — siuxy + 4% 3 My = ye*" 36. h'{y) — lny.value problem is y sin x — x y — x + y In y — y = 0.y) = y 1 y. + 3y + 3y 2 — 5i M = x/2y 4 N= and 3-2 then c 29. Let / = - — 2 J. From /x a-2 = —5/4 M= 2 = ^.x2 ) M = y 4 = Nx Prom = 4y + 2x .Let = h'(y) = 6xy 3 and = — + h(y). Let M=y = fx 2 + J/ 2 -1 xy 2 — ycosx — tan y is the initial-value problem A/j. Since /x = + y2 ) 1/ (l we obtain / = ly 2 . Equating N = 2xy . + Ax h'(y) = y{— 1) .1/2 x'/ 2 M xy — ycosx + 4fcxi/ 3 2 3j/ N(x. = -y/x 2 {x 2 '' 2 A' and h'{y) = A'x = M i- 1 /2 (a: If -1 y(0) y and A^ + ke x 3 + 4j/ Ay N= 3 . = y-^V/ 3 + = of 4 we obtain / 9x 2 y 2 so that 2y = — 20i/ 3 ~ xycosxy — s'mxy we obtain we obtain eXV = J^~2 then c — —1 — we obtain k we obtain / and l/y B = = —1 — ^ 1 + y)' + h'(x). = y + y) ' and + = y 4 . 30. y) 1/x so that = 1 My + sinx = . Let / = 38. From = -tan ~' V. Let cosx — 3x 2 y — 2x and 2 4/ A' 3/ j/. and h{y) — 2 then c = 8 and the solution of the initial-value problem 5 so that /y 6y 3 = and so that 1 M„ = — 2x/j/ 5 = The and the solution x general solution problem of the initial-value — 4y is 2 x/2y4 we obtain ^ ^3 - If c. (3?/ 3 = ^4 + Hv). Equating 33. / N= and 5 2 x 2 — 5x + 4xy is + 4xy — 2 2x - + - q l = 2ysinx — x 3 + ln so that Ms = 2j/cos£ — 3x 2 = A^. = 9. If y(0) = e then c = and the solution 2 3 2 of the initial. From 2 2 3 2 2 = j/lny — /e = y cosx — 3:r y — 2x we obtain / = y sin x — x y — x + h(y). = 37. = 10. = by -co < x < 2. + 4y = . 2 — For ]f+3x y for an integrating y' y' / 5dl e is oo.5 M = —x 39. — and h(y) 0. FVom 2 2 3 = 0. JV = 2x 2 y so Nx A^. For for 5.Exercises 2. an integrating factor = +-y= < y' = e~ 5x — dx so that = 5x y] fe 1 and y = ce 5x for J factor is ef 2dx = e 2x 4~ so that dx 2x \« = y] < _2x and y = ce and w = . —y = e x an integrating factor -oo < x < CO. h'{y) 2 e = 0. 2y -oo < x < 6.y ) / (x + 2xj/ + ) and N 41. For for 4. solution of the differential equation c(x 43. j/ My = -Axy/{x 2y / = x2 + y2 = x H + y} 3 = = A^.+ for J oo. _ Exercises 2. For y' is —oo < x < is x 2 an integrating factor oo. /i'(y) x+y + y) M — —g(x) and = = -1. o 3 . y coax — 2x 2 ys'mx + 4xycos:c = xe x = The 2ye x + = and h(y) 0. Identifying N = h(y) we from see that exactness follows M v = = Nx . e'f^ = is e / 3l2liE e~ x so that = — [e -:I j(] 3 e^ so that e dx y = = 1 and y 3 = xe x x 2 eI and y + ce x for 1 — —+ce~ x . and 3 x + x = c2 2 2 = (j/ 2 + 2xy . FVom solution of the xye x = 2xy 2 3x 2 we obtain + 3a: 2 and that My = 4xy = A^.x 2 ) / (y 2 + 2xy + x 2 ) so that Let M = (x 2 + 2xj/ . Let From /y = 2x y cosx we obtain / M /y = = xe x equation xye x + 2yex + is M = 2xy x 2 y 2 cos x 2 x y e + =x = c. The solution of the differential equation is / = x y + x + k(y). 0. h'(y) = 0. + e .5 1. solution of the = x The + y 2 e x = c.an 3 —oo < x < oo. For y' - + — oo < 3. Let 2 -+- V 42. +y = e integrating factor Z/x an integrating 3x is = e/ 4<tc e 41 le^y] so that ox factor an integrating factor is e/* 2 ^ = is *e < = x 2 so that e* so that = -e ~ [e*j/] 2 e 4x ax ce _4t 3 = [x y] = 41 3 J 1 3x and y and y = = 31 ^e 4 - + cx~ 2 + ce~ x for oo. and k(y) The -y. For x < 7. 2 N = ye x and we obtain / differential equation is N ~ 2x 2 ycosx so that My = 2xy 2 cosx and 2 differential 40. For y' x < oo. xe 1 +y xy<? + h{y). From /E [x 2 + 2xy + y 2 - 2y 2 ) /(x + y) 2 we obtain 2 h /i(y). Let smx + 2 y'i = 2 — h'{y) + 2ye x so that jWs x + h(x). 2y 1 and y 11. +— + x2 y = + y sinx an integrating factor x ^ 1 < for c + — — In x + 00.% an integrating factor —t + 11 — + — x = -2y an mt dy 2y — —x = an dy ce 2xr for is e'^ 2dx _0° < x < integrating factor y = +— y — ~ x sinx \J\ 15. For y . For j/ = -x y + -|^-y = = < < x< ior +e + e1 x x —-—1 y = 1 1 + x. an integrating factor + x2 = -| (l is is e j>/0+z 2 )l<fr = go tnat 1/2 + x 2 ) + c (l + x 2 ) eJ^^ 1 ^^ = 1 for + -co < x < e* so that [1 ax 1 for < x< is + e x y\ = and eJP*VC**-i)]<k = x* - 1 so that [(x 3 - l) yl = 00. For y = y' — dx -x\j\ for is = x3 + -l for + ccosx (cotx)y [(sinx) y] = ~ e~ 2x so that _2ir [e = y] x 2 e~ 21 + 5e _2x 00 ef^ 2 ^= y l i2 — ay so that fy 3 2 xl = -2y ' — f 1/2 L and 00. [(sec x) y] = sec 2 x and . cos x x 14. x an integrating factor 2 sinx cos x and y e/tanstte = sinx -+- is ccscx eJ for 34 coix <te = <x< sini so that it.Exercises 2. For if y x 00. — —-y 2 + cy~ 12. uX 7r/2. + 2xy — For y' x 3 an integrating factor ef 2xdl is — — \e^y dx so that = x 3 x2 e and I 1 y 1 = -x 2 — 9. an integrating factor sec = and y an integrating factor -co < x < + (tanx)y = sinx 18. „ . For = ~ [^f] dx 2?i x 4~ is _ gec ^ go na ^ (. is e S dy — e y so that dy is e/t 1 /*)*** ''xl fe = ye * and x = — y — 1+ce" J L = iso that [xy] = xsinx and ax 00. For . Fory' and y 17.5 8. For y' 16. For y' - + ce -x + —~ y = <x< 10. 00. For y' 1 ^2 y an integrating factor —00 < y < 13. factor 1 r tic e/' / ' is = x so that for = x 2 +5 1 1 2 = ~-x -7. —00 < x < for -i an integrating X1 X for 1 00. x an integrating factor — tt/2 < x < 2 cos 00. For t/ 4 4 \x y\ e -J>/(l is r = x+ i) e ( -* go that (1 d — dx \. is e/P+W*)]-** is e = xe 31 so that ^ [xe 3x 3/] = 1 and oo.a^j = so that e 2* ^ 00. 22 . For = - y xj 23.tor H [xe y] = j.2T 1 y'+ 21. For ox 29. ce _I \ 4~ ax / x(x 1 ^ + (l = 1 yJ for < x< x < oo. and i/ —+— x+ 2x 3 - - + ce 1 e /ll+(2/^)l^ is „ x+ -1 < x < oo for 1 = x2 ei ^.^3 + 3x = 2 y x 4 ~ i 3 In e~ = y < x< for CMtxdx J x ] xx + 1 x esc x an integrating factor y) 1 1 y . 5 = x an integrating factor 7~— + xj V - 20.r 2 x2 w' + + -~\ + x = 24.-s H e> 1 4 y2 ce _B y s2 < for w an integrating factor is < e is In j sec x\ e^ l+(2/y ^ dy = is = sec 2 x = _ (i C0S x} 2 + cosx 4-c for so that — dy < x < 2 fv e"xl L tt/2. For = -x- .Exercises 2. For 3/ + \(x = For 1 . = J oo. e"^ 3 ^ dx ~ x~ 3 _3 — tJ = —-— and x+l (2 so that [x i/l (ix —1 < 1) + cx 3 jj = ~.* 7 <x < for i = x 6 . For y x 2 - an integrating factor 1 pJW x is ) dx = — dx x 4 so that a: y 111 — -x + cx = -a. l)e . fWvWv = «~ 4 so that 4 fi/'" xl = Ay and .an integrating factor ce~ ix for H [(sinx)«l /[aein 1 + ^ ~ dx sinx so that n/2.x* and J oo.— -e y 2 cosx) +- 1 2 = sec + ccscx secx ( y e x sin2x and xex y = -r 1 2 and y 26. For l)e < X< T= (l + ^ an integrating factor *. For y = 28. x* — ^H. x y' + \(x + y x l)e y] ^ — -x = ^1— an integrating = 2x and (x = 4t/ 5 + lje 1 ?/ = x2 factor +c an integrating factor for is J\(x+2)/(x+V]dx -1 < x < e~ = ( x + i)^ so t^t oo. 3 y tanx(l = |x 2 = y] -cosx) an e« tan x integrating factor — sinx — cosx) 2 = and y(l an integrating factor — + . *sin2x an integrating factor = — -cos2x + c for is < x < is eS e f^ + ( lS x )\ dx = xe* so that oo.5 + -y = 1 19. For y' + = so that 2smx COS 3^ J- ^- + ^ dy \ | and x — [(1 27. For y'+(cotx)y 25. Jt + 37. 2 an integrating factor a xe x J L <x>. For y' 31.y = . For c 3/ + ^ do —+ x \e = y] e x + -5 y is 1 e ef^y+ is ^ e^ 2 y < .4 for )\ c(z eSW^- is 1) for -1< lOcoshx an integrating lOtcoshiJe"" 1131 and y = 10 { x x< + 2) 4 -2 < x < ^=-+ 3 1 + c(x+ 1 = so that 00.5 x = 2y 6 + cy4 = = ze 1 — + ce 1 1 1 1 y y 2 = ey + -5 -x 1- for = x ] r rf —00 < a: < is e x —d so that is = J"*** e < for y < = 35.an integrating factor + e' = ^xe x + 33. For + \{x i/ 39.— Exercises 2. For 2 - ey -e" !T r seed = [r(sec0 36. For 1/ 3/ oo. For 7a. factor -21 e + e~ x + ce~~ x = y - ex e~ x lnfe1 an integrating . cosS an integrating factor + tan0)] = (2i < for 2 - „ ex 00. (I_1 &= so that e dt d_ = Pe {At 2 - 2)e' "' and P = 2 + ce'"'* -00 < for i < 00. factor is + ce _sinhi: 36 e/ coshllil for = -co < e a ' nbx x< 00. = eJ = 2 so that y) . For e^ 2 ^^ = x 2 is <x< for sinhx an integrating factor \ = +c —00 < x < for ) — + \2y + dy and x 2 . y] = + = l) an integrating factor and y = |(» + 2)" 1 an integrating factor + is JW^ x+2 dx + 2).+ -ce _ y dx 34. an integrating factor 2 + x In x 00.i 1 =1 + lV .(e x + In x) 30.IIP = + sin0 1 -2 At e~ x — and + e' x so that [e = j/] (l ^ — e ^ and ^^ v dy = ye y * so that ye y *x = dyl 2ye» dy = a 2 — ayL> = y so that \y x] and ey and 00.lnx - 1 +y= y' 32. For ^ y' [e" 4 2) y\ t — 1 . For < for r(secf? is sec eJ ^^ = + tan#} - an integrating factor is 9 gee $ + tan# -cos0 + e-^ 2 '" 1 ' so that c for -tt/2 < < tt/2 .ex + ^. so that dx . x\ so that 1 1. y x \x y) dx oo.l)y = + + 2)- {x = 5(i + 2) 2 —y = x — + and (x = y x(x (coshi)y inha. xx and i 2 !/ y y < + . For — — x+ 2 438. t L -oo < for R 44. For 1/ ?/ = < y < t = oo. ^ £ [ie^l so that ax for = + ce~ 2y J _2x so that = /W^< = e «A cos 2 x an integrating factor — -5x g = 46. If y(l) + (tanx)?/ = sin 20e5x and « ey . For —- jfcT -7- y dy 50 ce ftt 1 = e y(0) and <? if] = xex — x = and —+ F. If yll) = 10 then c = 21 + 1) w] = hix and 21 x + 1 an inteeratine factor is e^ l^ x da ^ — x so that ~ dx \xy\ 1 1 — e x and y xx = —e + — 11 .2« 2 2 . = eJ™* is hx e 3x 31 — /L 2 then c = —2 and y = 4 ) an integrating factor is e~f 2dx 2a .\e i 20 an integrating factor -oo < i < — 2y and y = 3 xe * ^ dt 43. If an integrating factor e is 2y 2 47.Rt = (e = | ^ ^ L + /it i oo. 1 = ) x/ and y + ce" 51 and 3 -x — is e J" t_fc)lfc = e~ fc( — so that + -00 < f < 00. at T= = . For y' y l + 150e fct _fc( and . — + Ce.5 + 2x = Ze ^ ay 40. If y{\) — then c = -1 and 50 V = y = [7\r fct l at + 4 .lS so that = —1 c so that e -7e l5 = [(sees) — and y [e ax -50fc an integrating factor foT +- x [e _ *6 x cos x sin — cosx. For I hix — + x + u' + — v — — ex x 1 x x + 1 is e/[V(*+i)H* = ^ + 1 so that [(x dx 1 for <x< 00. If y(Q) — y' 42. For = 00.E/R and i = I so that / tflnirfx -I^ Adx = = -7 then c= -7 IfQ(O) 00. c = hix "x+1 x+lx+1 and v V = 50. For x - e J. 1 x- 49.ix 2 e 2x + ce 21 for -oo < x < integrating factor ~ — — x = 2y an dy is e — 2e~ 5x < 45. For e = = 4~ U^y] dx so that J. For for —oo < < y an integrating factor ef 2dy is = e 2y so that 2y -f.—\e~ mlL (i \ H and 2 . . = andQ = Q| L cosx and ce x * J . If = . If i{0) integrating factor = e 2 then c + ccosx 1 cos x 4 -00 < x < jm = then c io iQ = /(Vs^J/ e is 5 then c = — jt/2 < x < for = -49/5 and x is e ir/2. + 5y = 41. For 1/ + —+— x y = x 1 + an integrating factor X 3. — [xVyl dx so that e -50fce J L 1 x = 2x J x . til + 49 = = —1 = _2at J \ S ecx then .e~x y T(0) = 150 then an integrating factor c is = e 150 and /(l+2/*)<k T= £ x c 1 —^e <x< for 00. If j/(0) y for = n.—y. For and x 3y 3e J l for = x\ dy oo. oo.Exercises 2. we must have Ue 6 -lle.esc x. 38 = -y + - for y g0 that = -3 3. For y' + 2y = 2 = — ie 2 then c sec 2 t8JI1 = 8 = and z dy and y = 1 y(0) = 8 2 y + ce~ tani /(a. If 1 -y+ - x an integrating factor = = -1 and y = . we must have 1-1 f(x) an integrating factor x>3. For — (a: 52. If y(0) ci = and = C2 < a.21 f 57. ^ so that then c = -4 and .) an integrating factor is e then ci = — 1/2 and 21 = x-2 s inx y so that ( is z)dx sec -7. C2. je^ — 3. If y(l) 1. z(x - = 2)y + 2 then c = - 2 and y e an integrating factor -e* e H e^ 2/xi x is — 2 1 = = - 2 )\ dx - — for < a: = — tt < — + -x — ay oo. 1. + ci. 1.Exercises 2.[yx] y' +y = = 1 /(a:) then For y' + 2xy — e tani an integrating factor e is 37 tt/2. If y(—ir/2) - an integrating factor 1 e^ is 1 54. so that fe* C2 = and X > . for continuity and T . If j/(3) = 6 then c an integrating factor x < 0. For y j/ = 53. c ^ x i-2 2x = 2 and y /™ tKitc = e is then 1 = y < y< ^dx .5 < x< for 51.4e. C2 < > a. \ -e* for continuity = . For y — dx 00. 0<i<1. a: = < 2e so that 1. . If t/(5) + (sec tana: fe 2 x'jy - yl sec y=l. so that +c u 0<x<l- I I C2.i + e is x > x2 1. y and x ? = ^= y(0) [(sin x) y] so that X > pC If = dy = <x< for -tt/2 < f 56. .so that 2) (cotx)y ccscx = go. For + 2 y' y = cx for 2 < . For so that .tani 55. From y' xx + — y = iy x3w = x3 + 2. — — y j/ e 31 y' y y' — ^1 + 5.. obtain = —H we obtain —+ 4- —^ w = —1. we must have C2 = 1 so that 2(1+^) Exercises 2. °S r<1 fi£?- 2x = —^y +x 1 2 0<z< . ' 1 so that fix If y(Q) 7=^ 1. An integrating factor is x 3 so that An integrating factor is e x c or 2 and 1 . For + y' { an integrating factor +x is 1 = then ci = and * > . An integrating factor is x so that . An integrating factor is e~ 3x so that ax =x+ y~ 3 and y + w = -e ^ dx .*. we obtain y u> = i+ — y^ we 1 = ~xe x + e1 + c or y + ^ ax we obtain = -re 1 + ce — — y = — -^y 2 a. From im = y' In + +c = if y 3 and 3x y -1 = y + = = u> -3 _. x so that ^ — 3w = — 3x.6 = If 3/(0) 2 then c\ = 3/2 and 3 = -e + 1 we must have for continuity C2 \{ + \e. it) or \e~ xe 1 so that xe*w and y = xy 4 xe~ 3x = From + and = 1 + cx -3 x 2 w = — -e 2* + c Z) From e" y' 3 c or 2 or y _1 y —x c _1 ce 3x . we obtain -iu = — x . . From e z 4.6 1. u. so that Z z for continuity a 0<x<l. Exercises 2.x \ 58. An integrating factor is c 1 h x ^1 -e _a \ x ^ + —w = . = — In x H x 1 . + ci. n trom u 3 2 / x = -^w y or From £ 9. An . Identify Pfx) = e integrating factor 16. = 2 + /2 2ef and uj = -3x 1/2 + cor = -2. and R{x) x 3 so that x 3 w .. Exercises 2. x — Then . integrating factor = 1. ^— u= or = -1. An integrating = + u. u= = -3x 2 + 4x 3 / 2 . = 1 so that e Q(x) 2l w = — -e + x 2e x . {0 j = ^u. Identify P{x) j f j. 2. = — = —y h yx x so is 5 7 and 2 An /2 .. ^ + f-. -1 factor is xe 2:13 15. An -^-w 2x Then 1. y 2e* - x x 2e )w = -1. y Thus.6 integrating factor + xl 1 7. = 4 and y3 + 4)uj = Then x. and R(x) w = -^e 31 + c integrating factor 14. 2 dx = is 1/3 -1/x. i 1 x^ = --x 5 + c x% w —^—^ = r so that is ~ + ^-4xjw = = — 1 2 I- u.e V/2. 5 fxe x2 ~x dx ./2 W= e 3 x /2 Prom = yx 10. 3 X /2 y' + y = y~ l. and Q{x) — ^ ax = ~Zx 2 + cx 3 ' 2 -1. (1 + Thus. +c or ^+ -1. +c Thus. Identify P(x) — and y" = c ^ K r integrating factor . w— + 1 5 y x e^l^w = -ye" so that x- y + xl = --x -1 + cx~ 8 3 and + cot ay 1 . +corr' = 2-j + ce~f /2 + c fl + x 2 )/ v 1 5 we obtain — x 2 so that e e . =- J" = xe ce~ sx = 12 ""dx If . y .. then c 1 (-1 Thus. An x \ ~l integrating = 1 + u. 11.then ay 2 = x x ^+ we obtain = x + ce -3x/2 3/2 = — ^l/~ 2 — x" 3 ' 2 w is —w = = y(l) If . An 6 I- -3 2 3 factor — ax . integrating factor 2 = y{\) 4 = = ~x. - tu -1. x 2 (? l so that is integrating factor = If . Q(x) 3l An . Identify P (x) = 2x so that 2x is . = 2- From 3/ so that 2 3/ e 3x w— y 3 ^ w = and 2x 12. Identify P(x) is ° y + = x^ lilt we obtain — . 2. —& X ~ X 4 = An + 2x)w = (-1 — —\x^ + cor u — --x + cx -3J and R(x) x~ s / 2 + u. e Q(x) x sec 3 x. = —9. or 1/ c or u and R(x) = . r- e" '* is = -1 then c and ~+ 2x 2 (?(x) = 1/ 1/a. Q(x) = 2 x x so that e ~ w = -4/x 2 is . l -j— + (— tan x + 2 tanx)to = — 1. +c ce x 1. Then An -x.+ xj Then 1 -2. 6 = j + 7fi -to/2 j. c or y .3/2 . that —5 x~ = --x _1 + 3 integrating factor 4 then c 2 = = y y 3 3 we obtain = - 1 x. y Thus. + ^ dx Then = = 40 1. and jT 5 8. Identify P(x) x2 ~ x 13. 2x = and w — —e x + = — tanx. An = — e x + u. . The singular solution The singular solution . y . Identify P(x) x so that sec is a: 1 + given by x f{y') where f(t) = —y and y — Int. A . Let is = xy1 Q(x) — — 9. Let y is — — R(x) — y\ 37 5. y = cx — e ! t given by 24. = f(t) x = 2t~ 3 and y is . ^— + (5 — 4)u. 2 xy 4- ccosx] -l family of solutions 2 = j/ 27x 2 2 -t- — Ax + t2 A and = —t y = cx — is cx +1— In c. + R(x)yf\ + [Q(a> + 2 Vl R(x)u + R(x)u 2 + } dy\ du dy . w= so that 19. Identify P(x) | + u. Q{x) given by x is given by 21. Let y is 6. y — is e = is 1 = = solution 20. The singular solution In c. family of solutions 4f is y — cx — <?. 2 . tan x 17.6 integrating factor w secx = — In Thus. family of solutions or y = — ^(2 + + f(y') where /(() = — e'. Let y i/ = = solution 23. = \nt— = 1 A family of solutions ory = ln('— — [P(x) — is y = cx + 1. = — = -2 + u. The singular solution cx . The singular ma:. A family of solutions x — e and y = — e' + Je or y = In x — x. = 2 (~ 2 . Let y + tana. 3t~ An —2. + c or +c or u= —a: where ~ t~ xy' + xy' filf) — + 3t 2 f(y') u = [— cos x In sec a. Let y is — or 6. — —4 — 2t integrating factor for Ini or = — t3 A = A t. y . j/ = family of solutions 3 ~ + (6-6)w = -1 ax = —3 + u. singular 2 4) 1 xy + tana:] integrating factor for — or 4y or 27y = yi Thus. 1 so that e x w 18. 3 . | where and y = /(() = where 2t 3 f(t) and 1. is y + c~ 2 . An -3. Thus. —x + c = 1 — In = = 1 — u and y where /(f) given by x is l and 1.} + c ~ + f(y') xy' sec a. + f{y') given by x 22. is y — + c2 The + Ac c .e 1 Exercises 2. c) = 0. becomes Thus. Let u = lny. and (y') 2 + 5 = 0. + c + 5 = 0. If y = is WW — then ^ = w ax = and y — cx. — Ru o (Q + 2yiR)u -. x 2 u = 2xlnx. This equation or In y e~ x du) 42 — is x* +u= ^ dx + c.2x + 2. Then — = ax with integrating factor e 1 . Let u = ye*.6 26. Let u = „ e iV . the equation becomes . This equation and = ^e1 + ce~ K = — ue~ x dx + e~ x du. y') and 29. + u){—ue~ x dx + 2 I and and the equation becomes u 3. and the equation becomes or (1 + u) du = u z dx. Then y = ue~ x dx ~ ax = ie 1 ue~ x and dy + = 21nx 2 c ii] or (1 [e x u] = e + ce" 1 21 x u = x 2 e 2y = 2xlnx . ax (b) Write the differential equation in the form (y solutions so that dx obtaining F(t. (a) + y 2 . by the chain 0.Q(x)y Assume that — [Ft{y F{» - — 2 P(x) = by differential equation 27.Assuming = F(y . this ^ 0. rule. [x y dx Thus — dx rfii - . we have ^ Ft ~WW " - y' 3 xy') = + c . .cy.Q(x)w' W Then. — x 2 Thus . the parametric equations form a solution /(?/) Exercises 2.2x 21nx 5— — .— u Exercises 2. s) w = u _i let + f(y') = xy' /(f) + f{i/) into the differential equation ic tf(t) we find f(t)-tf'(t) an is + f(y') we obtain + /(c).7. Assume du that ax 1 If y 28. (y ndu dw — = —u — ax — + (Q + 2yiR)w = —R. 3 cx) cx + f(c) then y the identity cx 30. £ = ^ = xf = = + /((). 1. e*. is linear . F= c )l x?/. 1 ) — 1 FYom i = = 1 /{() — tf'(t) — ~ */'{*) + of y = V+ = j> xy and substituting c = -/'(t) and y Substituting into y ' ' and w" .P(x)w = 0. - which xy' we obtain identity. „ Then du „dy — = 2e — and dx ay 2s linear with integrating factor . j/ = By (a) a family of 2 = + /(c) = Multiply the given . = Since x f for /"(() -/'(*). c) — and F(y (f ™ c — 0. _I so e . Let u =— ln|y| +x= x +c yln \y\ = e~ x + cy. V (2 2 + e~ u )(udy + ydu) + 2(l -u)dy = -+*- 2. An integrating factor so ^[e*u] 8.c. equation becomes = 2x dx. The equation becomes dx +w+ 1 = uV T — + u = uV*. ^ = 2™'.e 1 /*' + x . An.-x a + x - 1 + ce". i/6 ZL = so that x = uy and dx u dy 4- The equation becomes y du. Separating variables and integrating we find — — + In |u| = x —~ + +c 4. Let w= x +u + x4 -e _u = x 2 + find so that -r- = 1 dx + This is Let u — a = => + u) = cl y 2 is so that 4ui 3 di + x 4 du. . The 3 c or -r- .4x w = 2x ^ dx 1 dx =0 dv — du + — = xig y and dy 4ux 3 Integrating dy y(2e =» = ux 4 so that -^7 = |s/| +1 du H +u e~ u + or u =_ —y+ 1 (j? + so that x ~) e* c du dy — =— dx dx . * u= _ (xV . dx or j w = u _1 to obtain dx w— e -3*.Exercises 2. + c. = + e""rfu or rfx x 0. -e~ y t xi = 5 e" x2 e . Let c 2t. Let u — = c we 6. and integrating we . The equation becomes ~ +u= — ax 2 — 1. 2e u 2eu -- Writing this in form + «| +ln In |2e" 5.xe 1 + e x ) + c => 2 y The equation becomes du — =u— dx 1 + xu+l du or dx 2 u = xu . y find p ^2e*/» a Bernoulli equation and we use the substitution integrating factor 7. is . cosh x so that du is u= c or wi dx y — = — —h2ylnx dy x dy cfti so that — = xe" —dy— = y dy = — 2u = x 2 ey =*• iz' = =f- e The equation becomes 2 15. 1 (1 du -r- — 2x —u or dx — c = ln(tau y) x H — = u dx The equation becomes 2xy. x efy. Let u = x3 y3 = => . Separating variables and integrating we have du + u2 11. y x so that u' -2 u = x +y integrating The = e y y'. x d — ax 10. -— = so that 2 y dy — tany dx sec rfu u — mitany} u = x2 w => . so . = -xe x -^-\e w\ w — I- dy — dx i xx — H . 2x xuj — = x -— dx dx 2 so that 1 u —+w = to obtain dx —x. Let = r An 2.x + ce -1 ) -1 2 + = x . An integrating e = x 2 ln III + cx 2 . Separating 2x 3 - 91n or u' x + c. — ye y ti ye" — ev 44 2 y In x = ye" — e" = 0. _u sin x.=> u = x2 = 1 + y'.1. Let x y and integrating we have variables 12. — dx 13.u = (1 ax 9. c = li . dx 1 u tan = 3x y ^ + 3x y ^ dx dx 3 so that tan = du 2 2 3 . -x + ce -1 )^ . 2 1 =1+c l or ey . . Let u + c. = 2x ^ dx 3 — 3. Separating variables and we have sin y „ xu' + cx 2 In Ixj x J so that v! e 14. An so . . 1. . = 2 fx" ul 1 The equation becomes = . or c cos y cosh I+y x du dy y dy dx we have = + c =^ . The equation becomes ^x 3 6x -- 2 =s> it = 2x 3 - + 9In|x| ==* x 3 y 2 c x Let u factor = is e. |x| u— x. = y The equation becomes x .— 7 Exercises This 2. solution Let w ~ = v u — sinxdx du e u = — cosx + — sinysinhxdx + sin ^ cosh x = e. Let — 1 (ill dx — u I dx integrating factor 2csc2y so is x. The equation becomes y du c. The equation becomes du —— = y-— Separating variables = — cosx + + c. a Bernoulli equation and we use the substitution is integrating factor x e is x w = -x+l+ce~ x =s. l-2xlnx. y This — — 3* 2 ) = is ^Cix 1 + a Clairaut equation 2 4f + 3 . Eliminating the x\ 5 /2 24 / ==> u xa + C2. (l + cfj x + ~~5\~^) = v! = tan(x 1 + u2 + ci) . I' tl" 21.= -dx => 18. Let = _ = « 1 J that u' => is | =i* 20. which = -in n\ i C1 .— 7 u Exercises 16. x => . if U^f—+ ci ^x + 1/ci = xv! 1 + x + (u') 3 + + (l + cfj. + —w = we obtain cos(x + Cj)| + C2- . . 3 — t t ( \cix ci a 1 / 1. v! The equation becomes x 2 w' j/'. A . „ = -2x 1 w] L An —1. is . so x 1 X y 3 singular solution 22. Let u jd y' so = that u' — we obtain — dx =u d f jx i 2 iix = it j/ so | x w— du dx .x2 . u — C\X is _ ^/C j _ The equation becomes _1 -j . Let du Ihe equation becomes xdx _.. . 1 —— = — 3t 2 = 1 + 4 -- tan ^ 2x J family of solutions y". — —xe + cx =5- = cost/ xe 1 — ex. -1 becomes u=-2 + ci =* y' u' = -u - = tan(ci - => i) Thus separable. so is A. The equation y' so that u' 9 u — ~x e .x 2 +C3 Bernoulli. parameter we obtain u 1 -1 + -|i2 ^ given by x is integrating factor c = w> i In Icii + = x. .x2 c\ 1 -= integrating factor The equation becomes y". u = t/ so —^i— = 1 + u1 that u' dx =^ = i/ = 2 ±\/c i . u z The equation becomes u .2 ™-2 — =f> tt = A 1 — 1 _ U y' so that with /(() + t it' — y". Let An . | Using so 2 x + u2 = x\ and u = 3 /2 ) '^^ = x + ci Separating variables we obtain 0. is j/ = In cos(ci - x)\ | The equation becomes y". . 1/x. Let = ii 1 y so that u tan = u' du ax _ 1 01 - x u = — xe . Separating variables we obtain j/ = — In I cj. ~ y". Bernoulli.X _ c\x + 11 + c2 / I u „ 1 3 u = — is x ci . Let . which 1 — — u = u2 u' which . Let An du dy — = smy — dx dx = —cosy so that u integrating factor = u —1 u X -^L. is + c% Using the 'X' substitution w = u _1 ax -M = -x=> » = 11 d = 19. dx 17. 2. > = y cio. In u = In +c —U = -<tr du 1 = xu' . u— dy V + =s- dy = dx —y + 4 1 and another solution (y') +u —- 1 — y* The equation becomes y . We / [I = 0. Let u = du 11 =u so that y 11 = dy T 27. =* cicosi I- = 2yu y = c\ ainx we obtain + C2- Separating variables 0. = tan0 . Let = 1/ —= = In |u| = du — dy ti u' + ut&ax = + c => In cosxj j so that y if The equation becomes y".) dx it = 25.. / — s-r . 24. need to solve + ir J = u u — dy y 2 = 1- (1 + / or (1 11 => C] . dy we obtain du u* 26. Let u = y' so that u' = u Separating variables 0. = so that v! y' The equation becomes y". Separating variables and using the substitution u tjz d$ co$6dQ Vl - = x => x-1 46 = x =f- sin# = a..7 Exercises = 23. . C2£ +C3. —(tana.. Let u 2.. — du The equation becomes w— . 1 = — dx |l => c V then ydy ->\3/2 h t/it = u — If ci 1 J = + „2y ay 2 = ^ ] J = we have — = dx => Let u -y = — x + c% is = y' so that u' = y". u= a. The equation becomes /u v" du ^- . Separating variables we obtain u. y c\y = — 1 2 —— )dy = T # + ci ci _ Separating variables we obtain ciy-l 0) I — / j/ y =^ dx (forci + du = «. d$ = x 3.. a. 2.. j/o 1. .. yn (x) y3 (x) = * + yn-i(*)- + x + ^x 2 yi(x) As n — we = 1 y„{x) . = y 1. j/ = Identify — n (x) eT * = 0.Ix 4 »(*) = ix 2 . = 1- find y 2 (x) oo. . 3. = Identify x for = n 0.i/n-i(i)) = 1 + x + x 2 + jU 3 = 0.£n-i{')) = 2ty„_i((). ... .±x*. 2e Picard's formula 2 + ^x 3 + ^x 4 = 1 + = 1 + x + x 2 + ^x 3 + -^x 4 + x + x ^x z . = -x 2 -2 Picard's formula is *Hn-l(0* find = ^x 2 .gz 3 . 2.7 Exercises 2. 2. . — —1 — x + » Vo 1. = l-x »(x) = oo.8 1. 5 .8 Exercises 2. j/n(x) Identify xo 1. 1 n As n 3. /(*. 2. + ^x 2 . !fl(x) As n —* /(t. + x 1 . 3.X + ^x 2 . 2. and /(t.X + ^X 2 ra = . = — » 1.. and /((. . 3.. 3. Iterating . . Picard's formula find yn (x) for = -yn -i[t). yo e * = 1. Iterating . = 7i = i«(x) = = l + 1 a.j/„_i(()) we =t- 2fj/n _i(f). = 4. and 1. is x t/i(x) Identify xn for . we Iterating . 0. Picard's formula j/„(x) = 1 we is is + 2 £tyn ^{t)dt find + x2 !fl(x) = nix) = 1 + x 2 + ^x 4 y4 (x) = 1 + x 2 + ^x 4 + ix 6 + ^x 8 0. .jU 3 + ^x 4 .ix 4 + ^x 6 .yn -i(0) Iterating »(«) 1 2 + ^x 4 + \x e 1 — j/„_!(t)(ii + yn -i(t))dt j*(t ysix) yn (x) for =1- = oo. Iterating .. formula j/„_i(t). oo. = find yi{x) /((.x . n = = -> - . j/o 0. = 0. . yo) in the region x 2. is {*) dt 1 - . Identify to for 6. l. . W (a:) y2 (x) As n —> 7. -i -x* 4 2 = . .\x 2 . — tans. y\{x) 8.. ir/2.0 did i+ -x 3 + —x 5 + -tt^x 1 for \x\ H k then the iterants are k times the iterants given in Problem (a) = .. i . 4 8 7 2.Picard's formula < . Picard's formula 1 + -x3 yn{x) .4c8 7 2 3 x 3 + x 5 + Ts mx 16q Chapter 2 Review Exercises 1. For f(x. = Identify xq = we Iterating . Picard's y3 (x) - = >/3(z) x + ya ) (b) - — y„(x) is find y\-\{t)dt for n -\- = x = 0*0 3/2 I yo 0. yn (x) —* 0. 2. 24 o I .^x 3 1 + x + \x 2 + \x 3 + ^-x4 . = x we is find 12 1 + -x 3 + ~x 5 + —x 7 we use separation of variables to obtain y 1 2 17 O i. . 3. (a) Identify xo 3/ n (x) — x = 2e = 1 FVom dy (c) The Maclaurin = (l 2e* = 2e'-l-£ y n - - = 2e* + x + \x 2 y 4 (x) = 1 1 = t/a(x} If yo (b) If yo(x) = oo. y 2 +y < 2 = 48 (25 — x2 — 25 or x 2 y 2 so there will be a unique ^ +y > 2 25. and 1. x then m(x) = x + 2 kW = x + . j/„_i(()) -J/n-i(0. 1. 2.jf As n —* 0. 3. = and 0..-e = and yo 0. /(i. 3.8 As n -> 5. = = = x and y(0) = y2(z) series for tana. 2. . we Iterating .y) — (25 — x2 — y 2 ) we obtain fy (x.Exercises 2.. yn (x) forn = 3. is =1+ n _j(()) j/ 1. = x yn (x) —> e oo.y) = 2y solution for any point (xq. . separable. obtain ylnydy = xe x dx If => l)+c (j/ . = - e 21n(y 2 + l) +c. I. so becomes ux (udx + xdu) = 2 . exact y. so let x = Then dx — udy + uy.^y 2 = 4' ij/ 2 hi a 2 8. is a homogeneous V \ J differential equation becomes y) dy ylnudu — —dy. linear in homogeneous (m) homogeneous y (n) separable (o) Clairaut (p) Ricatti 6. Bernoulli (j) x and (k) linear in (1) 2 (b) linear in (c) Clair aut (i) l) y.4e x — dx = \x\n V ydu and the ylnu(udy + ydu) — (uylnu — y 1 + c. j This dy. Separating variables we „ . Ricatti (h) homogeneous x (g) linear in Bernoulli homogeneous. The solution 2y is 2 ln |y| Write the differential equation in the form y equation.= -ln|y| + *ln \ y let 2 (3u x 2 — In y) y = +x ux.. 2 + x2 = 9x 6 . since 4. since f(x. Then dy = udx + xdu and the dx ot uxdu = c y ^2u 2 + differential equation Separating variables we l) dx. or Separating variables we obtain In ji du =— uln |u| — u = — In \y + c y x(ln x 9.In 4 2 1 /j .y2 - In xe* 4xe x . exact. 2 y V + xdx — 1 we obtain 1 1 —x + — dy . False. exact. obtain u dx j\n(2u 2 + l) =lnx + + 1 = Cl x 4 2^ x i If y{— 1) = 2 then c\ = c 2y 2 => 2 2u + x2 = 9 and the solution of the initial-value problem +l= Cl x is cix 4 6 2y . is = (y — l) 3 — and fy {x.y) — 3(y x are continuous everywhere in the plane. The equation is homogeneous. 2x + sin2j. = 2i sin -1. y(l) = c= 1. homogeneous. 2 J — x = — y In |y| + cy. True. exact. (d) Bernoulli in x (e) separable (f) separable. -1/4.y) 5- (a) linear in a solution. .Chapter 2 Review Exercises = y 3. Separating variables cos 7. . The . /dw — \dy „ equation becomes e" differential 2 = I2y dy i/ is + dx (u x I 12w + xe" = --2 \ . Let + 6a: * . 50 . . £ -e _I . Write the equation in the form y y c is is (ar \i + 41 . The [ * e -2 "x1 = ye" differential equation dw 4 dx x w = — x 1 An . — dx If [x -4 wl ' <- y(l) = 1 2 " => e-*i = ~»c-^2 1 = is w= Bernoulli..2x) dx + 2x) e 1 => 4 . mtegrating factor =t.1 Chapter 2 Review Exercises 10... -^r + r^-T => ™ = -9* a du — =x+ dy that « 6x is is 1 differential equation u = xv so 11. I J +y- = 2 fx 3 4j/ + c = of the initial-value 2 x . integrating factor a An . f . Using r .-x 4 lnx + cx 4 v 4 — x4 lnx) ' .. . 3 equation becomes = = u dy 8x — + —-y = ±[(x*+4) y\=2x(x* + 4f => then c = e" or ^+w=x 2 => + + 2x. u 3 we obtain also exact. n / (aT dx = integrating factor . The differential that du du - An 1 w = Using Separating variables we obtain e"du If 3/(0} + Bernoulli. u- x then c V 1 . = —1 = then c u = xy so 12.lnx + e -4 _1 y = 1 and y = (x 4 - =t- x 4 lnx) 2 " + c => 4 x = -\y 2 we obtain -xy 2 \ 4 ^+ 4y = =t. .y — (cx + ce 2". + 1 -£-[(fa {Note: 9l2 w= + 6x 2 = —~~—ry y 1 and y = 2x „ .-320 e I e*" = 4y problem 3 is + c. Let 5 is integrating factor dx y-r- „.. x 4 y 2 or so . so = \ + c (x 2 + 4) e~ 2y so .x~ 4 io is x = . tii dw — dx 6 H + The or e" — dy = lM = 12y 2 . e*.(0) so 1. dy dy 15. l)y 3 = -3x 3 + c. e xy = 4y + 5.. . Write the equation in the form = -1 + ==* (6x dy and the solution -T-le^u] If 3. .320 14. The ^ + 6x + differential equation ^ „ . + c An = + [x> 2 i e* + ^+ +4 = An integrating factor 2x y. so 13.) = xdy + y dx. .. y 18. (()) yn(^) for 20. — = \y — and dx 2 dx y +x — ^ dx sin u — -Lz let +. it is = y(0) a Clairaut equation then c = -1 and the —x. 2. [e so that (4 — . 3.Chapter 2 Review Exercises 16. Picard's formula is so that yi{x) = 3~2x y2(x) = 3-2x + 2x 2 y3(x) =3- 2x + 2x 2 - yi (x) =3- 2x + 2x 2 - 4 3 4 3 2 4 2! and j/n (z) -* 2 - find yi(x) 2y) dx and y(0) 1. n= n = . Write the equation in the form -i [cos £ !r \ y } dx the differential equation becomes + {y' + l) 2 we = a: and • Then — - u=x sin —^2 = x A we obtain Separating variables 0. Let is e — dx u— y 1 x = ti' The equation becomes ?/'. = =t- + e -21 as n — > oo. family of solutions solution of the initial-value problem is y = is = xy' y = i ^+ ] — ^ (cos u) 2 cos u du = 2x dx cx + + (c I) 2 see that If . .. From dy = for = x — x - \ + c\e u' u ov v! +u= x. Writing the differential equation in the form y = with /(f) (t + l) A 2 . 3! 4! X + C2- . Identify xq xe z — e i- 0. 1.. 3. 17. t/o = x u — xe x - 1. An integrating factor so . Iterating we = i l + 1 i 2 = y' + y 2 -i(t). and e x + c\ yn _i /((.c. 1 ^] = 19. ^ 3 x = l + x + -x s y 2 (x) = 1 + x + x2 + — y= Picard's formula -x 2 -x-cie is + _(%n-l(()^ 3 we obtain + ^x 4 + ?/ = 2 ~x 5 + ^x 7 + e _2a\ . 2. obtain 2 = = 2x —x - - = —xdx y' 2x so that the differential equation of the orthogonal family and y 2 we obtain y = — y = and / y = y' and 2y 2 2 y 1 we obtain y = 1 Then ydy .3 Applications of First-Order Equations Differential Exercises 1. From v 9.1 — = o^x. = so that the differential equation of the orthogonal family x Then ydy 2. = = -dx + x2 = C2- 2^/y so that the differential equation of the orthogonal family and \--y\dy = xdx Then U 3 2 4j/ / + 3i = c2 is . = ax we obtain y' — — From y Prom 3i y 3. so that the differential equation of the orthogonal family is + ciso that the differential equation of the orthogonal family 2 and 2y \ay — y 2 = —2x 2 + ci. 1 so that the differential equation of the orthogonal family ia we obtain y '— 1 cf x 11—dx = 2x 1 dx and y we obtain y' Then y\ny dy 3 cix we = x2 + 2 In \y[ 2 j/ + c2 . is . 3.4j/ and y 2 1 ia o FVom y y' 5. Prom y y' 4. = From 2i 2 y = y y + x2 = we obtain cj Then Zy . 4. so that the differential equation of the orthogonal family Then 3ydy = -2xdx and Zy 2 is + 2x 2 = 3y 52 ca. cji = is y' 8. — y 2 } —x = y' xy + y2 = V — 2x = From y «' c2 3 = -- y' = -xdx Then 2^/ydy -y Prom y = 1 = C2- we obtain — c\ (x —^j2 =- 2 Then 2ydy . -— — . so that the differential equation of the orthogonal family is so that the differential equation of the orthogonal family is c\e~ e ClX .4x =_ 1 7. = From ax 2 +y 2 y 6. — —x dx ' = — = — . 4 -. we obtain Then -dy . ylny = 2. . + ax = 2 2 is C2- by 11. This is ~2 ] = i2 c2 + y 2 = c 2 y. 2xy r x* + y2 --.y1 x- is + y 2 «i|y| = y' 2 2 = y— -x — ^. = ux so that . 2 we obtain + CjX 1 In |x| cry + c =^ ^ — n If — n x \) = ' I mar \ +c 2 . we obtain - = x family y' — y2 (1 dy } c?.. 3y2 _ +r c . 2u J._ i_i In |x| 111 j.Exercises 10.a x y = c\x x2 so that the differential equation of the orthogonal 2xy 53 ..21 — = u — 3/' = .. 1 From y 2 — x 2 = c\x 3 we obtain _ . = From y 1 is y' 13. j. From y family 3 is + 3x 2 y = y' = x2 ci we obtain = + —+ .. x Let = uy so that dx dv — = u+w— dv dy Then y 15. ^2 du g = ^£ — z2 From x 2 +y 2 — 2c\x we obtain y' 2xy = „1 This a . - 2 _ a? J- /\1 f i y~ \ W I = ci => . 2xy = ux so that y1 = u+xu 1 . — = y' u3 is 2x Then - + y2 = y' so that the differential equation of the orthogonal family c\x From 2x 2 is = y' Then y 2 dy = —x 2 dx and x 3 + . _ = — ^ — _ r -In — ill I-. 7j rfu wu. >- —f .-- j/ 1 [ so that the differential equation of the orthogonal 5 £ y a homogeneous differential equation Let y .2 — u -2 — In |u| = => is = = u + xv! Then 2 14. _a — . Then - 16. so that the differential equation of the orthogonal family 2xy a homogeneous differential equation . so that the differential equation of the orthogonal family — — 3£ Then bydy = —axdx and by .2x 2 2 so that the differential equation of the orthogonal 2xy 2xy_ = 2x z This z y is a homogeneous differential equation Let y . From y a = C\X b we obtain y' ^ = y' 3. = Prom y y' — x2 = r £ 12. so that the differential equation of the orthogonal family — 2xdx = 4cix we obtain y' —s—-—* y — 3x 2 — and 3y y 3 = C2- . is y' = family 1 y —— 2xy t-j Zy x2 —u + xu'. From y family 23. 1 1 + 6xV+V) . Then = An x. — —y 2 obtain y' 1 In \x\ + -x 2 + c = 2 2y 2 hi 2 Ixl + x 2 + ci. + tan x) y so that the x x dx and 2y 3 we obtain cii we obtain .i d + | In \u equation 3y x y =a + 3. 1 = . a homogeneous ^=^(1 + 6^ + 9^) 1 + 17. .1 Exercises 3. .„-4 = ---x +cx differential . V 22. ==> + „ . 2xy From + x2 - = 2 dx ==> y x —+—x we = i/ l = 2ydy 18. ~ y x so that the integrating factor -ye y + ev + c l_i. so that the differential equation of the orthogonal family is c\ 1 y 19. From 4y + x 2 + 1 + c\e 2v = 4 orthogonal family H is —x — — [xy\=x rf I" 20. Then 2 dx — I l-3u —sdu= 3 ti + 3u . y (x — From y + 1 > y ^—x xz we obtain 2 + 3y 2 ) = y' —x dx 2 (x ci = l differential .2. = 3x 3 = e~ y sec x so that the differential equation of the orthogonal e~ v dy = differential 2 = —cos 2 xdx and 4e~ y = 2x + sin2x + C2- taimy so that the differential equation of the orthogonal family x = — xdx and 2 In j coshyl 54 + x2 = ci. so that the differential equation of the 1 x cix ln(ci 1 y x\ or = + xi — ey is =s> x . du= u + 3u3 l =^ This . ~ —x — FVom y = — dx + 1 dy family is 21. 21n(l I 2 2 - 3u I Let y . = Then y 2 dy — dx and y s = 3x + . equation of the orthogonal so = -y + 1 + ce 3 Then = [e 1 = dx +x = ?/ Ay -x~5 we obtain ^ In y y 1 = From y x 1 d — dy f cie 3 j/ we obtain c. 3 = {x 2 = + 3u 2 = ) c 1 (x'' + 3y 2 ) ux so that In \x \+c / ^ 2 =* ==> y ) . y' Then tanhudw y' equation of the orthogonal family is + c. integrating factor ^4^1 x y=-x 4_1 --x 6 +c +c we obtain = —e v cos 2 x. is . 2 From y' = is y sinh y „ tanhy = ^' — +x = — dy = -ye v 2 y dy = + y . = so that the differential equation of the orthogonal family is 1 + x- 2 l + x = l+SL Then . evx ==s- y' = is x4 so . An y. is 2 + 9u 2 — -12u . . — x. r cos cisin20 we obtain ^ = . Then dy = cos0 we obtain dd — = tan dr 6. If = y(0) 10 then c = 10 and y = lOe1 . so that the differential equation of the orthogonal Then cot 0d9 = dr — =i- In lsin#| = ln|r| +c =4- = r cisin#. so that the differential equation of the orthogonal ' 2. + y = — is y' family l ~a 28.„ sin 31. family x we obtain ci sin 3. = tan20 so that the differential equation of the orthogonal Then -tan20(f0=— r . = u(0) dy y we obtain ci Cie H 1 2'3 Then dx If y cot x so that the we obtain cj Then y . = —_ so that the differential equation of the orthogonal Zx*y y. is + y' = .cot dr sintf r r i dr sinfl <tf . 3x 2 ydx and r ^ dr = —cot j/ = ce 1 . so = 2-x + ce^ 1 . y . From = = r family = y' is is r 2ci y' and y = —^ 4~[ex y] ~ 2 = x | x + c2 is . . so that the differential equation of the orthogonal + y-1 -xe 1 =^ 3cix we obtain y' 2 In cosx\ orthogonal family so that the differential equation of the orthogonal family 1 = 3andy = 2- + 3x — Then y + y — e differential equation of the 5 i 1 "" dx and y' ^ 2 x 2/ 3 dx and y x y. .1 Exercises 24. ( —= -ln|cos20| 2 55 = Inr +c r 2 = C! cos20. . v' 25. 2 = dy - = y' we obtain —x— 5 then c 3xt/ a From family 29. = From y —- From i 1 ^ 3 26. . An x e y = 2e 1 + 3e~ x x2 = 'a + c2 ." j/ ^= x5 / 3 + c2 . integrating factor x ->xe* + c =» y is e x . From family 2 is = r sin 6 so that the differential equation of the orthogonal Ifien -z l + cos0 dr 1 - r— dr 20. .tanxdx Then ydy . From r ci(l+cos#) we obtain r family = — dr is r l+cos0 . = x = t/ = From x — 1 ^3 From x" + ya — is y' 27. r = 30. . r 34. the slope of the tangent Depending on how the angle a is we chosen. and ip the angle of inclination of the tangent to a trajectory. of the given family. This is + u2 ) = = In \x\ +c ±2 tan ~ . Then dr dr — -cot8~ => — In = sin SI I In Irl +c — r c\c&c8. r . de — =— dr r + coa 9 1 . From r r^- = = ci sec sinfl dS l+cos0 — — — — — so that the sm8 dr - differential equation of the orthogonal ° : — dr sinfl => — s<w = 1 cosf? we obtain 6 r Then . = — . y) ± tan a IT f[x.y) . In tan/3. Then t/ = u + xv! and 2 . y) tan a r isogonal family a trajectory +a= have will either = — <j> tantp. + cos0 l dr — r = d(? sinf? 33. using the . Since the differential equation of the original family is /(x. " dti = ~ dx =* ±tan u.^ 1 1 ±2tan _1 |-ln^l + |^ = 21n|x|+ci 37. — tt.cosfl] = +c lnr 1 = ==> r ci: r 1 - cos#" so that the differential equation of the orthogonal family is -tanfl. x homogeneous so dy/dx — = Troi0 — a f(x. From r = C[e fl we obtain = r ^j- 1 dr r = _l. ci — From r 1 . ^ dr so that the differential equation of the orthogonal family is Then — = -dB -9 = +c ln\r\ = r c\e~ e . See the figures for this problem in the answer section in the text. is Now. the differential equation of the let y = ux. i TV This is (j> + is fact that f{x. Let measured from the positive x— tangent line to a axis.In y = - . of the member be the angle of inclination.— \/3 — = lTV3si/i y i In (l is f{x y) y \/3 x y=- x^VSy 56 . r 35.. xu — ±1 ± u — => l^u ± Itu + v? J. we obtain + cosS = cot ^ dr r - In |1 . Since the differential equation of the original family the isogonal family is y' = yjx ±=.r— periodicity of the tangent function and the dy — — tan^J = dx tan{/3 tan/J±tana ± a) = -— 1 T tan p tan a 36. At the point where the curves intersect. (x 2 + y 2) = a..Exercises 32. family is 3. y) is y' —= = 1 T y/x . Then . the angle between the tangents a. line to any event. the differential equation of X homogeneous so let y = ux. Vt Self-orthogonality follows from the fact that the product of these derivatives 40. - 1 ci(2x + ci) we obtain c\ -1 == ±2\/3 tan =*• ±2^3 = -x± u. the differential equation of the t and y = coe _t sin i.y) is -lT«/v^. 2 + y 2) = = - f(x.1 Exercises y' = u + xu' 3. we obtain This dy — dx is satisfied x+y x by x —y = . the differential equation of homogeneous so let y = ux. Cl . . Then and xu — 39. yy* i ci results in exactly the 41. This shows that the family and y is -r- ax = = * cie sint — w +x . 1 is r c\G cos orthogonal family t same equation. Since the differential equation of the original family 1 the isogonal family y' = u + xu' y is V^ X ^ }^^. and = xu ~ 1 .I -1 tan" 1 yjx 2 2 In — da: + v?) = (1 ^ - In ^ - In (x ^1 + 2 In = + y2 ) = Ixl +c 21n|x| + ci ci- + y 2 and x -\ + y \yj \ y i. 1 \/3 T + u2 ±-~ tan ± 2 dx j x 1 1 u. is Then C2e -i cos self-orthogonal. +U ±y/3 tan = This . r ~— aii = => ±/ ±vJ— 1 from y 2 is In (x .= V ^ 3 v^j ^ l^y/VSx x^fy/ V3 = ±l/</3±u?/V3 — —= . Ftom + ci + l y = — FYom jt — we obtain ci = - —1.^ + u2 ) = In (l !/ tan-^-lnfl + = In +c \x\ 21n +ci |x| "V3 =* ±-j= tan -1 38. When 90% In 1/2. = 01 find 97 we find = 50 in fc = = fe ^ From dN/dt — t. 0. —— 2) t / 5 and Pq the f. we obtain =» = Po 61n2 10. From dN( dt 400 = N 3k e or e* = JV{0) = = kN we (400//V )V3 . k= 575 we find = = kt « 6597. ho.1 = 500 we obtain 500e 3taL15 r* 760<9^p> number. Setting JV(f) Let JV 1 In 2 ^ N(t) be the number of bacteria at time et (V3. Setting JV(i) e 2000 we then have Using JV(3. Let JV Let i P(f) be the population at time 500e From 5. = 4P kt .2 1. 15.000 and = e = f = 3. = Using P(10) .97.42.! From dN/dt Then JV(24) ( = = kt = » grams and N(Q) = iooetV«)("»ojr)M 10. .B< (ln2)t W 1 Setting P(i) . = fl. £ In 0. P= fct Problem = taa = obtain JV = Ne JV(10) = JV = will N = e fci . 100 = we obtain 100 ( . Let P= P{i) be the population at time P= obtain Poe kt ..96 hours.1 ^ * t. = 3Po 3 — t=^£« 7.3)ln(i/2) 6. We - have tb< ifo =? so that = tani/ji + tan (^2 2 coti/^ — — tan ^2 2/ \ Exercises 3. Let P= 4.!.000 Then P(10) = — 111 J ln3 = fc initial have 4 2. H = lnl — t 58 = « 136. t = 10 years. * 5 =j. Exercises 3.5 hours.** 1 we have be the amount at time JV(t) 7. Poe 21n2 (ln2)t/5 3 in 10. = 2PQ we Using P(5) find we have 3 = .3) = P=P (ln2)t/5 5 In ^ =i^ n4 N(t) be the amount of lead at time Using JV(6) and = kP we From dP/dt population.5. t and Wo the = initial 2000 we find = P = and P{0) ^lnl. Using JV(3) remain. «26.9 7 )4 Problem 6 we obtain go. kN and we obtain of the lead has decayed.000e-°' Prom dP/dt t.4P we Setting P(t) n e ' P Setting =10.390. N = 100e w „ ggg mg . Then P(30) 400 and iV(10) = we 1/2 = 0.9 years. 25.Exercises 3.2 8. g(. 50 so that = = 20° and T(\) = i § T(i) If +ce~ sm . E(t). jfc{T-5) so that Assume that dT/dt = k(T - then c rt and 1(3) fc/ 12 years. *(t. = As and i = 60 = A(t) 10 + T = = If then . C= -6 5 x 10 [f *(0) = -4 .-1/100 then c 1/1000 [e*/ So- 3/5. 20 the differential equation = 6e" 10 T= .75% then 5(5) 59. = Let I . = and £(t) 100 so that q = 1/100 + ce -60 2001 . k / 10. .25/ $6665. 5.000 then (c) 5 $5000 and r A= Assume that 12. = = Assume Ldi/dt = + Ri = 10) so that = = io L = i(t) = g = and 19. If Sa = (b) If S(t) =$10. = T{0) 15° then = t ce*'. -3/5 and limt—oo Since i(0) 18. 51*. 13.1472 amps. and £(t) then c = = 200 so that -1/500.-i 2 )=ln£ ^2 = £ A Solving . 1000. ~ (tl -(ln2)/fc. If T(t) 100) so that £(f). E(t) we obtain Assume Rdg/dt + Assume then .1 seconds.00098/ / $6651.45.06 minutes. and IO -4 . + it 2 C= 200. 36.4611 so that T(0) Assume and k - =S . fa- /t = -200cet If i^sinwt. intensity. Assume that dT/dt and c 14.00012378. It follows that At (2) ~ ( t2 = t - In(Ai/A2} 9. 22° then 98° then If t{0) = -» co = i(0) iQ so that -r^^^rcoswf + ce + & Rt/L . = fc 21n(2/3) so that T{1) = c = -80 and t = 145.67°. If = t wl5. have g -> 1/1000. . we is + cje"*/ 10 *. R= E(t). i = = A(ti) Aoe*' 1 A2 = . = Aoe^ A(ti) and => ife^-J—to^. The = kA solution of dA/dt =A A(t) is 2 =A = Then Ai . For j t = fttfg/df + + ce" i(. = i io + = = and £{(). and 1(0) t = I e kt 11. = T = 5 + ce*'. = where 5(0) -.82 =s = = and /(15) fa) = the thickness. so .0O5) = . = If J"(0) = 82. Aoe*' and fc 16.4611°. 20 di/dt If *{0) - + 1i = then 120.7 seconds. ln(39/40) so that T(t) Assume Ldi/dt + Ri = If ?(0) = 64. L 2 uj 2 + R 2 17. 90° implies sinwt- Z'w' = 30° then k = -|ln2 ( = and £(t) 70° and T(l/2) = 50° then-c = 60 3.1. = 55° and T(5) = + 100 50. 15.005) = <t< 1D i] (l/c)g c = = (l/c)g 2001 and .145-4 T(\) i ce**. = be the I(t) £ln FVom dS/dt = rS we obtain S .003 coulombs.40/2 kt e e*' for we obtain t. .600 years. a= An integrating factor -60 and i = 60 - 10 is e*/ 60e-'/ 10 . . = t dT/dt = k(T - that = e = If (*//<** - . 60e~ 3 60 i( * ) Ea . 60 - t) + c(400 - f) 4 . = 26. t>20.000 and 4(30) tank = . If in 100 minutes. If 4(0) of alcohol after 60 minutes is 4. + - empty 24 4)r (o = .60e-'/ 10 j 60 \60( e 2 -l) e and - 60 (e 2 = i -'/l0. Thus l).06(400 The percentage 400 minutes. Separating variables dq 20di/dt . From dA/dt = c . we obtain = 4(0) A = 30 and = 1000 SOe'^ 50 A= then c - 1ft = .1 gallons. If 1000 4(0) + ^ = ce"*'' 30 then c 100 If .38 pounds. From dA/dt = 23.18 = -12/400 and empty after + ce"'/ 50 ce"-'/ A = -10 - _ 5)t The tank . -CIn fc2 t we ijo = + we obtain 20.q/C is find c2 — = ~ fjii — 1 80 / 21. = 10 then = 12 then f 64. 4 = . 10 . Prom dA/dt = - 4 4 = 200-170e~ 22. 1000- 4(0) 25. 44/(400 4(60) is —— _ = 3-~+— 44 = 3- 4 is obtain ^ - -100. -1000 and + c(100 If 4(0) - t) 2 .A/100 lOOOe '/ 100 dt c we obtain = then c dA From -4/50 / 50 . c2 e~^ < t 20. = 10 5Q0 —— 200 A= we obtain 100 0. The . if A(0) = 30 then c i/ck2 = -170 and = oil 50 .1%.4 \ .4/50 we ^= From t ki - t) we obtain 14.2 For At > t * = 20 the differentia! equation 20 we want c2 e -2 = Setting g{0) fci = - dt - + = 2i = so that c 2 . < .Exercises 3. we obtain 4= 50 + ( + c(50 + £)" 2 . From dA/dt 4= 24. m . Owe obtain x . If &i < = fcsinl + c — — P= - . cie frsin K ci -P . If &2 t -(*i+t>)t.A/B . P= vq e( (l .{AfB)e~ Bt = A/2B then T = (In 2)1 B. As ( — * oo.Co we C) and C(0) k2 )P we obtain oo as t m m 9 I — * oo. Write the differential equation in the form dA/dt Then an .a). J If l ~l s Pee*™'- ' integrating factor . i If — gm/A. m\ C = Cs + (C .Exercises 3.( V fic so that x ~> A/B 33.kAi * v obtain = If £= S 32. From 31. P{0) ffcet*'-*")' Po From r 2 In [P| i Po then dO for every t. oo. + Using P— then /> 34.2 27. From d£/<i( = -E/RC m gm\ ( V dC/df = kA{Ce - 30. k cos = and . 61 e -(*i+fc2)i). (a) From mdv/dt = mg — kv we obtain v = gm/k + ce the solution of the initial-value problem — (b) As (c) From ds/dt ( oo the limiting velocity * = = v and s(0) 28.) m k\M. x. + = we find c = --™^-' and A = . = v(Q) we obtain so am *"/*". (l_ the materia! will never be completely memorized. d( = ^ 2 foJA = l i 10 —(6 . is = A-Bx&nA X(0) . = p( 0)> where Pq P= then as t -» oo.(L/m) dt we v ' and P= ^ — — — —— — — obtain A= * \ I 2 r d$ = * 1 e'* 1+ 2 ' ^ 2 2/9i is + — / mJa + (ki 4 = _^_+ ce k2 ki ^2 > 0.4(0) > 2 .Ct )e. A kiM .n and . Jf(T) vq then c is FVom dXjdt If 29. Separating variables we obtain = = gm/k. (a) From dP/df (b) If > fci as t — * = (*i - P— &2 then = and £(ti) » _kt/ Eq we obtain . . * ~+— w 150 150At -* co so that the amount here 1 of A— • 70minutes. = _7 f ^ ' = P(t) = obtain E where ~ bc > t =P P(0) From Problem 5 then c = we have ^= 2 Setting d?P/dt X{5) as 8. is P(a . 52.2X)(150 - .0005C / MOfc+ 5000 we obtain 500. be the amount of 10 then —= = oo.0005C) and C(0) 1 we obtain \G d£ C = ^°^Q5 1 e 2. f VfezM ) d P = dt a-bP + P .™"".c / f \ df 7.3 grams. so that " so that and 150 .000 then .000095238. dt* and using co. t From 1834 supermarkets.3 1. and 180A:t amount Then X(20) P= e^e . • as t and where k . and X(Q) = X = - 10 then 62 » 30 as we obtain -» 150 as 75 then dt B— i = ( t — X = X -» 60 oo.bin P) we obtain If .0001259.000 as ^ = P{a ^ + — P= — + From ^ = - c atfe*" 1 bEe\ a - 5. rrr-7 10 fcP) dt = JV(1) -t oo.00014.3 Exercises 3.7033.X) 2 X(0) = B— = = a/b.Exercises 3. and X(5) 33. = .w .lnP we fee"" v (1/6) ln(a/f> - - l) In Po) . = t -» 2000 as an arbitrary constant. ~= Prom = C(l . ^ = N(a - From Um - t = Then °( 1Q ) ' + = t • . = Let A" t — From ^ A.3 grams — = oo. .ce W e and iV Since - = 5 °° P= - ^'°°IL _ so that W . so that the and the amount 9. Now A"). Va*0.000. Xp^ Xj^ (iMz^L + \ct — = ~ — ) dX = kdt so that .9months.™ - " Q^. = = ' X (t) 6ce | - = P = e^e . 3. N= 1 (ltT 1. —^ln|a.cP"*) we 1 1 = 1000 we have a P) and P(0) If N= 500 we obtain C dC = — 1 -6 (a) (b) 6.000 and ^=P From P 4. and bN) and N(0) ^ =50. If and X(() fc(a-X)(/3-X). ' and P= e^ 6-1 If X(0) 29. ^^-^ In Pq. b and . and X(20) > and the amount of 0.150e X=— Q^ fc(150 dt .bin P\ = + Q t ^ c — ^ ^-^^ = ^= of b 2 obtain t C ^= &(120 at time of A— . p = = ^x 2 — ^ dx ln|x| ^ we obtain p — cosh ^-x + = and . In a) = 32 ft/s and #= - /?)(7 + -/?)'" Then . + — — we obtain 1 2[vi/v 2 1 + - T + T„ -2tan. 5291 mi/hr ft/s as v 12. If ' V2 then VlV2 + l/M In 25 l-vi/v2\ l + p2 = . \dx } then y = ^ and . k (t 4 From . ^ and . 50/ If h(t) - then * = 50^20 seconds. V^gli.T^) we ^(1) = From ^ot and w| ' 1/(47%) 1/(27%) T-Tm T + Tm T 2 + 7"2 ft(0) e wx ^ so that p=-2 (x n/v * .X\ .7 )0? .X)(0 . In \0 oo we assume that v ~> * ff = v (a ' «o (c) -1 X„> + a- and «o = \a\f~X\ = fci + c. . p 13.2 ft/s «w mi/hr.3 X= 10. .X)(j . .ae^-ftto -^dX = kdt - then .— Exercises 3. From y(l) dx l = + 1 FVom u(0) = and vi = i>2 . 14. 4000(5280) we ft 7760 ss -1 „.X)* (a kt+c obtain at .„ . (a) As . „ —= dX From k(a a= If . = y'(l) — If ^j.(/?-a)( 7 1 a-X + -Q) 0-X + (a-0)( 1 -0) l-X (a. ^=^ From = sinh ^-x.X) we X = a- and .165)(32)(1080) . .x~ v ^). a)( 7 ?/ — (b) Using - .1 20 we obtain dT = kdt ~m = 4T^H + c- -i ft = ["^20 \ - . = y'(0) follows that y 1 it = —\l+(^r) V2 \ p ./l - ' v\ obtain so that 15. find » 25067 36765.7 )(/3-7) dX = kdt so that -1 - {{0 11.7) (a ' = 2gR Uq = v^2)(4000X5280} /2(0. — } Exercises 3. a Clairaut dx +y y H dy .^cos20^ 64 = fcsin d0. t = . = Jtyt + c.v ^ that dw — — = dy -c 2 which 2x »t 21. -r 3/3 yfZjidb so that and A(0) = /2 10 - In (fiXJTr + we obtain 10 /(cscfl ^) = . 20. is . 2 = ±dy. ( j " 1 . If h = . dx ' + 2fcr 2 ) -9 - From If > /i W then find ^ 17. + c.. = dv + tjk/mgv 1 v — yjmgjk 19. 2k If y . and t/ > \ +a we obtain dz = 2fcsin0cos0 d$. 2 1 /'di/A . dx = -^L=dy. Using x > kd . From a = lOsintf. + y2 yjx 2 +c I _|_ / 1 » dx dw — — and w = y —4 \dy dy dy x \/k 2 dy dx dy — =— — we obtain dx dt — ^mg/k to see that v and the quadratic formula we obtain _-y± ijx 2 + y 2 ±J/ t 4 dy ^ ^i — yjmg/k or Then dt a family of parabolas.y _ + From m dv ^ = - fa> 2 and w(0) = -\/l00-y 2 = wo we obtain 1 — yjk/mgv v + ^gjk so that 1/2? + = vo + ^gjk vq — <Jmg/k e2 Divide this equation by e 2 \Zs k /mt and multiply by v di (g) From x — (dx\ \dyj I . From y [l + dy (y*) = 2 ] = as t —» oo. f 1/ 18. 2 then and x = we have . Using 2 + y2 = / dt —7 In y + 5y = a In x — 22.3 16.dx where y ^100 Then 10 J100 . 2 v 10 In k /3x $y\ dy = —— /& [ - fix | dx. i and dy The equation.smO) dB = + 2/ir = dr dt.1/2 = then r 1 ! 2 we table of integrals (2/xr 1= -.^sin20 + c. _ + 2y— = solution 2 x is dx + 2y- x 2 = = dy dx 2 u-i'so cy + . Let * :r. A — and = -4(0) 16ft 3 we obtain 4 CtZ the concentration As 0.8 + -» 4. z(0) = then -.017%.„ -r- v cos = . I r = rffl 0. dt / or Jo / fl <w ° / V .16x = ^ dy 3 c%.2e"*/ 4 . Then Ay . -j differential = -Ay and x + 64 4 equation of the orthogonal + qje 16* If . .-x2 + c 2 . l dy — = K2xy we obtain . is t -> 00 we have or 0. + Cij = From y(x 3 family — —x 2 y 2 y' so that the differential equation of the orthogonal 1 — is y' we obtain 3 . —x = 3 we obtain y — Ay — I6x so that the = 4x+ 1 + cie 41 is y From y — family 2 = a(x — 1 is y = —.„ . at lnM^lnll+d^'l+c or y = c 2 (l +c 1 e ak > t h/k ' . = when x then c 2^-^dt = From (a) = 9 0.4 7. . ) + cie"* 1' . dA — = 1. Since A(10) = 5.cos (?o vcos 9 JO . 2 = P we 0. ^ -^f-smOdS and (It (it = t ^ dt J = ] (b) Solving ( ^ ^ (cos = J \dt J .cos Po = 0(0) we O obtain ~ cos^o).2 - A = . Chapter 3 Review Exercises 1. 2(!/ 4. dt 6.Chapter 3 Review Exercises If 23. ^p(cos0 and separating variables we obtain cos #o) for ~jt dt i V 2 ff •'Co — l H~ r° / I etc T/i /2T ™ T = *2.. 8./— .7ft 3 . From . The orthogonal . Let A — A(t) be the volume of CO2 at time From t. If we obtain P= P(T) 4e = 2P 0i8t then T= so that P(45) ]- = ln2.. trajectories are arjr 2. From = ^f- 2 we obtain y' = ^ — -— x The —^-r 2) and P(0) jfcP ^= I) orthogonal trajectories are (y - 2) 2 = x . . From 5. .99 billion.8 ft From — 3 11. ^ .018P and P{0) so that the differential equation of the orthogonal 1 = P= obtain 4 billion Poe*'. = — kjx(a x) we obtain | — \ x dt 1 \dx a .1 = 64 3. From y family Y' = - 16x C2 - - . .06%.xJ = k\ dt so that x = r-. 4. = . -f- — and so +B + +B +B 1 w= and lim Tit) ^+ BJ + 2 . . f 66 > 10 = = get 10 the equation for the current becomes 0. 20. 3 + C3 e fc < 1+B >'. — t\ +c V40 - or Solving for i we 4 *"*' 2 °£ < ( ' 10 ..2i = 4. 10 Then t 10 -iln|40-2i| = — In Since i(0) < t < = 10. ~ 2ky/m 7 the square root of the result obtained in Problem 27 of Exercises 3. + 7b)]. at and v(0) . ^ = k[T-T - (c) becomes mg + kvn dv = mg — kv «(0) = 0. and y(0) = we From mv dy mg 2 t*y/ m — oo we see that the terminal velocity is v = ^ Letting y In e differentia! (b) Since v i mg "| - v2 follows that it _ f\ (b) Setting y 9. (a) Let . — ^-t= 1 we must have lim e ^ t-00 — is V{ dT -. Separating variables we obtain fc e dv — = — mg — kv dy * 2 Since trra — .2i it. I 20. B(Ti . . dv dv dv — = v — so that m~— = —mg — kv dy . . 1 40 w If - . Thus 20 . + B)T - = ^f^/' Separating variables we obtain 20 10/ dt .T)} MtJ first ^ Using y{0) •> impact at 1 We - . is (1 10.2. Chapter 3 Review Exercises . 8. = is ft „ This 2 dt we must have For r > ci = |10 2/"/l0 .*)-/ 2i = ci(10 At 4 or — i - gt 2 = t). Then 2) and = T(t) ^"J ^ + 0.v height v% The (a) is m — 2k ~~ = h we see that the velocity equation ^-=]n\(l+B)T-{BT T 1 + D 1 Since T{0) = 7i < 0.r2 t-00 T). ' - ' 1+B ^ rs = T2 + B(7i - solve f 1 \ 10 2i = then the - —— ^ = +T2 = )\ C3 7i = maximum find that ^Jmgjk k[(l kt —=-r =k r + [BTi +c + B)T - (BT\ dt. 1 ^T2 + BT^B UmT. From y = = that ci 2.C2 = 0. From y are y 8. Solutions are y (b) (c) (d) = e x cos x . y'(0) = c\ + C2 = so that ci = 1 and c2 = -1. Then is y = Then j/(0) Theorem 0. y{n) = -c\e" = so that ci = and c 2 is arbitrary. We have y(Q) = ci = 1.C2e x sinx we find y' = cie x (— sinx + cosx) + 026^(003 x + sinx).l). 6. j/(tt/2) = c2 e"'' 2 = 1 so that ci = 1 and c2 = e^ 2 The solution is x v 2 x y = e cosx + e~ / e sinx.4 Linear Differential Equations of Higher Order __ 1. 2cj • = ci 4. y'(0) -1 so that + c2 e .1 + C2 cos x + C3 sin x we find y' = —C2 sin x + C3 cos x and y" = — 2 cos x — C3 sin x. for any real numbers c2 . t/{0) = = = c\ = 2c2 3. We have y(0) = ci = 0. Then j/(tt) = ci . (a) We have y(0) = c\ = 1. and cz = -2. j/(jt) = — cie* = —1. . The solution + c2 x 2 we find = x is y{0) = a 2 (x} we have In this case ci - l) and + cj = 1. y'(l) = a + C2 = -1 so 3x — 4x In x. Since 7. We have y(0) = ci = 1. so c\ — and c2 is arbitrary.1 is = = 0. y'{\) — 2c2 = 6 so that cj = 1 and c 2 = 3. is 4cie 4E y _1 y(0) = ^e 1 — ^e = + C2 = c\ e/ (e 2 . + C2e~ x we 1/2 and C2 solution From y Exercises 1 . D x and y = 2x 2 1 y ci = 2c2X. .cae -1 Then y(0) = 1 . c 2 = -1. y = c^sinx. we have y(0) = c\ — 1. at x= = 0. The solution is y = —1 — cosx — 2 sinx. y'(0) = and y'{0) = 1 is not not violated. Two solutions 2 . j/'fjr) = -C3 = 2. The solution From y = ax + Qxhix that c\ = 3 and C2 From = ci jf = = we find y' = Ci —4. We have The 3. c\e x i/(0) is 4 eye * = so that ci = = ci = = find y' cie x — czeT x e (e + c^e -1 0. C2 = -e/(e 2 - 1/(0} = l). — e x sin x. 5.C2 = 4ci 2 2 -e~*. which is not possible.1 does not apply because y and y' are evaluated at different points. solution is y 3 41 = -e + = cj — = so 1 - a= = 0. + C2(l + lnx). c1e 2 Then '. + y From y = cie^cosz +. 4. The solution is In this case = 9. The solution is 2 3:r Theorem 4. —1/2. - e~ x we find x = 3/5 and c 2 = rffO) ^ j (e 1 y = The 2/5. = = ci+c2 = possible. j/"(tt) = c 2 = -1 so that ci = -1. y 4. for < x< £1 68 00. is = Ax 15. + (l)cos x = have nontrivial solutions when c2 n a nonzero rnr/b for 2 will y{0) x < C2 sin A7r have nontrivial solutions when 02 A A = -f will = c\ (If = (— 4)x — y(0) arbitrary and and c2 — tt/2 < x < set of functions containing f(x) 19. see that the functions are linearly independent since The = Thus we they can not be multiples of each other. the problem has a unique solution for + = = y(2) 3.2+x — W (l + x. x2 ) = +x 1 x x2 1 2x 2 + so that c\ y = is 0. ci + c 2 + 3 = 0. have y(l) solution ci = ci We have y(0) = ci-0 + c2-0 + 3 = 3. y(7r) = + ci cos Att = 0. 11. x. The Thus we require + c 2 sin 5A = / 1. 17. 0. y(5) or The problem 0. Since (l)cos2x 15 so that the problem has a unique solution for 0. which not possible. the family of solutions Ax + 01 sin Ax we have = ci = 0. 21. is = 16. (3)x + (0)x + (l)(4x 1 (OJe = - 3x 2 ) + 2 + (1)1 (3)(x — (1) sin 2 + 2 (-2)cos x 1) the graphs of /i(x) c\ cos 5A integer. Since a2{x) 14. the functions are linearly dependent. 4. (c) 13. Since (-l/2)e* 0. tt/2. = the functions are linearly dependent. + 3) = +x 0. (a) We have y(-l) (b) We have y(0) (d) We y is —x— 12. Since (—1/5)5 +3 = is — = + 16c 2 so that ci and xq c 2 sin A7T + + = 3 — oo < C2 sin c\ cos 4ci not possible. Since (-4)i we — y(l) 2 and xq tana: ci cos that shiAtt 20. Since a = -1 or A be a nonzero integer. 5A 18. + c% • = c\ + c2 + 3 = = —x 2 + x 4 + 3. v (l/2)e c\ the functions are linearly dependent. (If = A = 0.) argument shows that -H-+- 22. Solutions axe y = c\x — {c\ + 3)x + 3. linearly dependent. 3* + so that = 0. .) |x| 1 _I 0. y(l) — c\ + c2 + 2 4 C2 = — 3 — c\. .Exercises 4. Since (1)0 any Ax we have The problem From 0. W(x l '\x 2 ) = x1/2 x2 \x-W 2x (l)sinhx = -x 3/2 = require that ci j^t. 2. t —— t = 2^0. + {l)(x = 2 will = — = be A similar and /2 (x) =2+ functions &re linearly independent since 23. Since oi(x) = from 1/ = = and From y and — C2 sin sin5A +3 = + c2 + 3 = c\ which 1. — x ^ the family of solutions the functions are linearly dependent. y is = the functions are linearly dependent. = = 4- the functions are linearly dependent. 10. or -1. a. +1. > oo. (a) (b) If If y = y - 1/x then y = -1/x 2 c/x then y" -2^ = 1 = 32. IV (a:. 5- Hence. tV(sina:. For x < 0. . 2x 3 - = 2x 3 0. + 21nx and f2 are and f2 are of fi r* - + 2x In £ = 3 i 29. cosx 26. (a) + C2f{x) — The graphs for a// values of -co < x < for is on — oo < x < oo. a constant multiple of the other neither function (b) For x 7r/2. W^. so that c = for c 2 ^ or 1. i 2 lnx) = 4e . this does not imply that fi <x< x 2 lna. No.Exercises 24. 31. /i and f2 are linearly 4-H- independent on (—00. 1 + {y') 2 = zl —y1 + ^ = x - = 2/x 3 so that y" 3 then y" ± % x' ~ 0.x 3 ) = +x x3 sin = 25. dependent on any interval containing x Obviously.fi) = so = -2x 3 + 2x 3 = x2 x2 = „ 2x We 0. 4* xlnx 1 oo. x = 0. 0.-30e 4x # 4ar x — need c\f[x) 30../2 = +lnx = -J + In and y" implies that c a. 2x conclude that W(fi.e-V 4 - *) e x esc 2 e 1 -e" 28. If ji = cijfi + c2 jft = ct + C2lnx then . x -oo < x < x e 1 for = -2cotx ^ cot x esc tanx = x esc a.00). in as shown. <x< for linearly oo. <x< for = -2secxcscx ^ . + In x 1 for jr.cotx) = x 2 (3 + 2x) ^ 3x 2 1 4.f2 ) = 2 f2 = -x and for all real values of x. (a) Clearly y\ (b) If J/" y = + yi + 2 (j/) 1 1/2 and = 1 . 69 2y 3 0. satisfy jj" + -c= 2 (j/) = 0. cscx) - sec 2 cot — x e 27. In a. x 2 and x1 -x< 2x -2x 3^ W (fi. x(2 + lnx) ^ = /2 0. 3 * We the interval. 1 W(l+x. W(tanx. x" 2 In x) = 9x" 6 ^ is = c\x + C2X -2 70 + C3X -2 lnx. The (x. + + c2 xe x ' 2 W(cos(Lnx). The general 1 (e solution oo. 4. — oo < x < The for oo. The general solution = cie functions satisfy the differential equation <x< oo. The general solution The 3x . and are functions satisfy the differential equation cie~ 3l + cae and are functions satisfy the differential equation 4:t . The . = sin 2x) is w(x for = y is x functions satisfy the differential equation and are linearly independent since -co < x < The + C2 sinh 2x.e 4x linearly independent since )= 7e*/0 is = y 34.1 Exercises 33. = 1/x ? is = ci cos(ln x) + C2 sin(ln x) functions satisfy the differential equation and are linearly independent since W for 2x functions satisfy the differential equation and are linearly independent since y 39. general solution y x~ 2 . The general solution / pje 1 sin 2x.sinh2x) — oo < for x < oo. The general solution is = ci y 35. The 36. independent since linearly 6 ^0 = ci# 3 + C2X 4 . 2 functions satisfy the differential equation and are linearly independent since W for = < x < oo. The general solution The 37 cie cos 2x 3 x/2 The < x < oo. e y 37. linearly independent since Whoosh 2x. and are .x 4 = ) x . . W(e—oo < x < for oo.sin(lnx)) for 2e is y 38. cosh 2x cos 2x. sinx) for — oo < x < The oo.Exercises 40. = 1 and y'(xo) 4. a: By Theorem = xq. 44.4 since 02(2 ) independent since 3/1 y-l*l a 3* = x 3 solves x 2 y" — Axy' + &y = 0. . general solution = y 41. the general solution since we form a general solution on an interval for which for every satisfies x y(xo) in the interval. 42. 43. let j/2 = -x that It y-i shown easily It is form a fundamental = W and 3/2 (1/1. The and are functions satisfy the differential equation linearly W(X. To show 3 3 that ya = \x\ is a solution let 3/2 — x for x > and 3 for x < 0. The equation. set of solutions of the homogeneous a particular solution of the nonhomogeneous equation. 0. c\ = 4.1/2) = on /. 1 equation. and yp functions y\ equation. H-H (b) If x j/i > then 3/2 = x3 and W(y\. l From the graphs — x 3 and of yi — ys |x| 3 we see 3 that the functions are linearly independent since they cannot be multiples of each other. ze 2* form a fundamental = 3/2 +x— 2 2x 2 is the homogeneous sinx form a fundamental set of solutions of the homogeneous xsinz+fcosa. They are linearly 00. set of solutions satisfies y(a. = x~ f 2 and 3/2 — z -1 form a fundamental set of solutions of the homogeneous — j$x 2 — g£ is a particular solution of the nonhomogeneous equation.x 2 ^ a: satisfies y(0) 3/2 0. If x < then 3/2 = —x 3 and = and y'(0) — =s x2 of x y" — 4xj/ x < is zero at x 2 = 0.y2) x" = 3x (c) (d) x3 -x* 3x 2 -3x 2 = x" 2 3x 2 The functions Yi (e) The (f ) Neither function y a 2(a0 46. The = functions yi and yp The functions The (a) = functions y\ and yp = coax and yi = e i 2x and e is a particular solution of the nonhomogeneous equation.) = 3/1 and yp equation.cosx. Thus. x. Part (b) does not violate Theorem 4.o) yi(x)i/ 2 (x) - = and y'(xo) j/i(a:)yj(x) = 1 = at 1. Assume y\ ^ is = = a: 3 W [x x3 = x are solutions = 4 ^ for —00 < 2 and Y2 3 . 45.2 they are linearly independent since and = 0. independent since = e 2* and y% = e 51 form a fundamental set of solutions of — Ge x is a particular solution of the nonhomogeneous equation. is + C2X + C3 cos x + C4 sin x. + 63/ = 0.) bi(cosa. 1 = ^(^5. In Problems 11-30 the text.2 In Problems 1-10 (4) 1. —2x. W^ 48. function. W If (3/1.1 Exercises 4. e 5 / = e 5x . (a) Assume y\ and y 2 are solutions of a2y" 0. from we use reduction Define y = u(x) - Now we use formula 1 so y' If v> of order to find a secoond solution. = [e 5l w] — gives 72 u" + 5m' = 0. obtainitlg c . Then + aij/i + ao) + a response of the system to the input E\ + (U2J/2 + a if2 + «o) = E\ + &2 Ei- Exercises 4. ki k3 k2 ki lni| e 50. and ai(x) = 1. 2 1 a constant. y" = u".]nx + iDX <> )e^M = (*1*4~W*0 x E\ and 03^' +V2)' + <H> = +a = + (<*2I/i E%.fa = -ffl(0)-B2(0) The equation (b) is x (c) Let (d) Prom We + a iv\ + a oyi) °- The first-order linear.order equation ti/+5u. + a-ilf + avy = 47. in part (a) ( 1/1 + i/2 (ft!** - + an/j + oq = + y2)" + fll(yi is k2 k3 *o = (fok4 -k2 k 3 )e. 1 49. which has the integrating factor e 5l tu = c. = u'. Then from Abel's formula Then from If W(x ) ^ then in Problem 47 we have - xa the alternative form of Abel's formula in Problem we have = We '. . and y" + By' = — u' we obtain the first. is a 3j// solution is W = ce^-H" '"^" ^''' = W = ce" O ai(l)/as(()idI x in We we part (c) for see that if W(xq) = every x in / since =1- identify 02(3) identify 02(1) 47(c) is an exponential = x 2 and oj(i) W then = have a 2 y" 02(2/1 and where c = for every z in J.1/2) = yu/2 — y[y2 then dW to = yi (021/2 + a u/2 + ao^) . a. sin4x. Define y = v! = c/x 2 and u = ci/x. gives l e second solution = ui = is j/2 = c -1 . Now 3. (2 and + If if = we obtain v! 2j/' + = j/ e _a: (a:«" the first-order equation + w 2u') = u" or + -u' = 0. u(x) cos 4x so y' = — 4w sin 4x + u' cos Ax. Define = t/ = u' u(x) ce 5x and u = = second solution and u". y" — u" sin 3x + 61/' cos 3x — 9u sin 3x.2 w= Therefore 2. = Therefore u" = = u' 5*.Exercises 4. is j/2 = — xe _I = e -2 . = Deiing y u(x)xe 1 so l/ = (1 - ije-'u + ie -I u'.4y' + Ay = and u. x —w= 2 which has the integrating factor H a: e*ff/' = x1 Now . . e which has the integrating factor e~/ dx ax w= — is t/2 = u"-u' = i/'-i/ = = -r1. which has the integrating factor Now ^ Therefore w' first-order equation = 2 [(cos csec 4x and u 2 Ax)w] = c\ = tan4x. (cos4x)u" -8(sin4x)u' 6. c. e E .x)e~ x u' - . so + u'e 2*.' If u) — = and u V+ V+ a + cix 4e 2 Taking C2.[e y' 4. £IX w= Therefore 5. we obtain the first-order equation u/— to ti' Therefore 3/2 = y" «'. — [x 2 w] = A second solution gives x2 w — c. and — C2 we see that 4e 2l u" = 0. x = j/" xe~ u" + 2(1 . — Define y = u{x)e 2tie xe 2* 2x ce 1 it = ce x y" = e and A . A . is 3/2 = tan4xcos4x = u(x) sin3x so y' = 3u cos 3x + u' sin 3x. a second solution is 2*. y" = u" cos 4x — 8u' sin 4x — 16u cos Ax and y" If e w= -8 / u' + = 16y we obtain the tan4tiit = 2 cos 4x.x)e _ *u. gives A (cos 2 4x)w second solution — c. 0. 4e = ci 2l 1 . c\e 5x so 1 j. Define w= y = u' = — u" or 8(tan 4x)tu - = 8(tan4x)u' = 0. = dx = = lna 7 a:) lnx x(lnx) 1. (x + dx - iy dx . t/" - e*/ o V 4- ^WV + Je^u o y and 6j/" If ui = u' e (S/6)M we obtain the = + y'-?/ = e l/3 w= = w' 11.x2 l) [-7TT- 2 a dx 2 dx = (x = -2 - + 1 x" 1) - = (!+!)/ / x. |x|. (In A = 6 2/x we have is j/2 = 5x/6 fe -f-(7Mdx fl — dx — x dx — x § j J % ^W7h^r 14. Identifying P{x) 0. 1) - 2x (x + + l) . we have = is = w= 1/x we have A second solution is 15. x 2 x"' / j x _6 dx = -7^ 3 -3 . is 3/2 = e -W 6 e s/3 — e . Identifying P(x) ce -51 ^ 6 second solution 12.2 10.2x . = x = x j^—dx = 2 . Identifying P(x) §u' which has the integrating factor -/(2/. A e In — second solution 3/2 second solution W= = = c. Identifying P(x) + have — = y2 A u" or Now e 5*/6. ~-[e**/* VJ] Therefore + 5u')=0 (6u" 2(1 (* x 1 ?2 + x)/ + !) + (l . Identifying P(x) and u = — 7/x we V2 A + w' first-order equation = |iu is j/2 = second solution = = gives C\e~^ x ^.1 Exercises 4. Define y — u(x)e x / 3 so \e x !\ 1 y + e*/V.I 2 ) / <x 1 ^+ (l . .x 2 ) we have e -/2(l+x)dx/(l^.)<i« 13.T / 2 In \x\. -dx j- x 2 cos(ln x) tan(ln is t/2 + 1-x 1 fa-li 11 we have X C03 ~ ( xcosflnx). = | ] -^ = fa -j. dx ~ = d: we have 2x) . we have fe -S-d*/* m = xj - ^ dx 76 = — fdx = xj xln|xj. dx ^ = x] ^ / e -« x u— . — x) dX = x3 ^^iiWH f X x) = e -i* — j e~ 2x [i e 2r + xe 2x - ±e . Identifying P{x) = 4x/(l y2 A second solution 20./ ~3dx/x x^o^x) J x a x^snrfmx) . Identifying P(x) e = = — 3/x / j i"1 ^ A second solution 19. fe~ = x fe~ -~y dx-\-x j j = = snv^ln x) = e -2xy = | . dx.x 2 ) we (l have B = / c -/-^Mi--*}^ = | e -Ki-A second solution 17.Exercises 4. = e-^ J(l+ 2x)e 2*dx = 1/2 — x. x ^—-e e X) -g-JxAj/d+a) zj—g—dx- x a! i 2 sin(lnx). Identifying P{x) is J yi / J . e -*+ln(H-z) - X] 2 .-dx/x - [xsinflna:)] [— cot(Inx)] is j/2 ( we have = xsai^ax) y2 ) — x)\. = A second solution 18.2 A second solution 16. =— x - + e - . dv = -^dx. Identifying P(x) V2 = = is ^ + + A second solution 21. Identifying P(x) is j/2 = j = — 1/x — 1 . + In (1 a x . dx^xj ( du——e x l + a: ) e -» — X. Identifying P(x) is y% = = -2x/ + x 4- x2 2. x xsin(lnx) = — 2cos(lnx). . v x* - e dx we have — dxj + x j — dx = -e = -1/x 1 x 2 sin(lnx). dx e x . . . we have -5 — /(lnx) — fx— 5 (ir/l x A second solution 24.2 A second solution 22. 3 . e ~ J ~ ldx/x xi . Identifying P(x) In sin(lnx). Identifying P{x) = is j/2 = xln|x]. J cos-'flnx) A i we have - dx — e x j gi+Ini — dx = 77 e 1 f j xe x dx = e?{~xe 1 - e x ) = —x - 1. = + 3x cos(lnx) tanflnx) = sin(lnx).:. l)/x . = is . Identifying P(x) ?/2 = = cos(ln x) 6 = x3 y% is dx r^- st. . dx r —(3.)- in . . I _7 X in / / 1 _n\\ dx = J*f 1 i 1 we have 1) ^ = e 3* | (3x = + g3x ^_ xe -3. 3/2 = e" J 3x + I 28. Identifying P(x) / -stt 6 2 J x (lnx) dx I — r s-t. = —7/x we second solution — f have 2 A x \ -^•^(. Identifying P(x) — - . * . _ | efa e _3^ _ = |_ + 2. Identifying P(x) = ^2 is x5 = — 5/x .Exercises 4./2 = dx cos(ln z) / J = is 3/2 x3 = is j/a x A e 3x — — . 2 = In second solution 25.f -{x+l)dx/x 5^ = cos-'flnx) -d» = e*/ second solution = ) Lux/ . Identifying P(x) e e31 dx ^ is 3/2 = -a. we have B -J0<fa A second solution 23. = — 4/x we second solution I x 1/x we have r A = . = -(9x + 6)/(3x + ( \ 2 m-^j = 1 have „-/ J -7ds/x 27.^-(^^ 26. = — 3tana: we y2 A + — sec 3tanxdx x tan x + ij/z + x tan a. . Identifying P(x) = is 1/2 x second solution = is j/2 = -(2 have = j e' i = 30. sec + In - In sec x j sec x | 3lnstKX ^J dx dx = Jsec 3 xdx + tanx|.- e . y" = \<? fe x - 3*. general solution is . We . Thus y 34. see by observation that a particular is x = a + cieT x + x. The 3 (SAe *} = e + 246 = -4 e _a! -. 33. Identifying P(x) 1. 31 = 5e 3 *. — 9Ae 3x and The . yp = y% is = x. 1 We we have The see by observation that a particular solution = y2 = —3 QAe 31 - x . Identifying P(x) = e^ 1 . is yp — —1/2. Identifying P(x) + c2 e'!I - we have a particular solution we try yp 3 !I " cie = j e'f^dx = general solution y2 find 2) e we have y To -2x + is 32. Exercises 4. ^ j dx = e = Ae 3x Then . Identifying /"(i) A = second solution general solution is 1/2 = — + 2a. + tan x\ we have W = | e" J "P^W 1 ^ = | e 21nl+ *di = 1 1 V dx = A second solution 31.2 A second solution 29. Identifying P(x) is 2 2) e*. y solution (x = A= c1 e I / -4 dx y' J = 5/2 and yp + c 2 e 2x + we have - x r e V2 78 Ax e x dx = e 2x 3 3/le *. 3— ^ then and = Z£ e .3 m= 11. From 4m 2 + we obtain m= and ^ . 4. Then 3a = 1 and -4a + 36 = yp = ga: + | and the general solution is = so a = j/J. . + -./ ^_4 e -/ + v\ y\ 4 y{ e -/ + y« / J so that y'i + Py'2 + Qv2 = (y'{ + Py[ + —~ £^ re -fPdx Qyi) I dx vt dx = °- Exercises 4. m = -2 ± %/5 so that y = qe'" 24 ^) 1 + C2^~ 2 ~^ x From 12m 2 .8 = we obtain m . 2.1 = we obtain . 10.Exercises 4. o. ^ . To find a particular solution we try yp If j/2 = J/i / ax + b.—1/2 so that y = c\e x i A + cae~ x l 2 m 2 + 4m . 3. 14. 7.2\/2 and m . 3:r. ' 2 . Then = 3ax-4o + 36 = x. 6. j/£ 1/3 and 6 = = 0. / . . m = -1/4 so that y = c\ + cit~ x 4 From 2m 2 — 5m — we obtain m = and m = 5/2 so that y = Ci + oie" x 2 Prom m 2 — 36 = we obtain m = 6 and m = — 6 so that y = cie 61 + C2e~ 6x From ro .2 = we obtain m = -1/4 and m = 2/3 so that y = cie -1 4 + c2 e 2* /3 From 8m 2 + 2m — 1 = we obtain m — 1/4 and m . ~ m — 6 = we obtain m = 3 and m = —2 so that y = cie 31 + C2e~ 2x From From m 2 — 3m + 2 = we obtain m = 1 and ro = 2 so that y = cie 1 + cae a *. 5.3 A second solution is e and 0-4a + 3(ax + b) A particular solution 3x . . From 1. is !/ 35. = = cl e I + C2e JI + -a.5m . 2 1/ . From m + 8m + 16 = we obtain m = — 4 and m = — 4 so that — c\eT Ax + C2xe~ ix From m 2 — 10m + 25 = we obtain m = 5 and m = 5 so that y — cie 5x + C2ie 5x From m 2 + 3m-5 = we obtain m = -3/2± %/3§/2 so that y = Cie (-3+^>/2 + C2e (-3-VS)*/2_ 12.cie 2 ^ 1 + c2 e _2v 21 From m 2 + 9 = we obtain m = 3i and m = — 3i so that y = ci cos 3x + ci sin From 3m 2 + 1 = we obtain m = and m = — i/y/Z so that y = ci cosx/y/3 + 02 sini/v'S. 4/9.-2^ so that y . 13. . 9. 8. . 3x and e~ so that -5 *. e so that i m = 3/4 ± a/23 i/ 4 so that we obtain y 17. + C2sin:r/2). V23 x/4) c2 sin + ci&~ x ' 2 + csxe~ x / 2 m= and j. and c3 x 2 e~ x . ± -1/3 m= we obtain y ?/ V^3 x/4 + (ci cos m= we obtain y 18. and m = —3 so that . From m3 — m 2 — 4 = we obtain j/ 27. 24. From 1 /3 = m3 — 4m 2 — 5m = e ( Cl From 4m3 + 4m 2 21. m = m 3 + 3m 2 — 4m — 12 = + C2sina:). From in 2 16. = QeT* + c 2 e 3x + c 3 ie 31 = —1 + c3 sini/3a. cos x i .3 — 4m + 5 = 15./2) + C23: + cje c\ . m= 1 c\e — 1/2 m = 3. and 5./2) m= —1. m— so that _3. m= cje~ 2x m— —5 —2. . so that + c2 e 2x + c3 e. Prom = m 3 + 3m + 3m + 1 = 2 lx c\e c2 sin 5j: + c3 e ± . From m 3 + 5m 2 = = m= c^e* we obtain 1 = = From m 3 — 5m 2 + 3m + 9 = From + e~ x/2 = we obtain 25. From m3 + m2 — 2 = we obtain y 26. Exercises 4./2 m— cie" (c2 . m = we obtain +m = = \/2i/3 so that cos y/2 z/3 _I ^ 2 y 20. m— + c 2 xe _:c + 80 so that + C3sin\/7a. m = —1/2 ± \/7 i/2 (c 2 cos\/73. From 2m 2 . and m = 3 so that —1. m = we obtain j/ ± i/2 -1/2 m = 0. and and 0. m = —1 ± = x + x m=2 and + e~ x ^ we obtain = 1 2.. ^/Si/2 so that m= ?/ ^2 x/3) m= —1/2. —1/2 m— we obtain y c2 e + (c2Cos v^a:/2 m — 0. From m3 — 1 = we obtain ?/ 22.3m + 4 = = Prom 3m 2 + 2m + = 1 From 2m 2 + 2m 3l/4 +1= = e" 19. . —1 so that . so that + C3 sin x). m= 2± we obtain —1. e^fci cosx so that cos:r/2 (ci m = 0. + ci ci y 23. 3x we obtain y 35. = = m= 3. m = 1.16m = 6 c\e we obtain y 33. and y = 2cos4i - 2 then 39. +C2e 2x + + 12m3 + 8m 2 = y then -JZi/2 so that so that From 2m 5 . m4 + m 3 + m 2 = From = 30. bo that y and y -1 and 5c 2 4:c 2x . m = we obtain = e = c\e x + C2e~ x .8 = = y 29. and m= 1. 40. so that . m 5 + 5m 4 . x + C2ze~ I + c3 e + axe* + c5 e~ 5x we obtain 3 then c\ 1. m = ±V5i so that —3.2m 3 . If y(0) = 2 and j/(0) = —2 ^sin4x. m = 0.3 28. and c$x cos m = 2. and y e 4l — If . we obtain y m = 2. -1/2. 2 2iC m — ±4i so that y = c\ cos we obtain + m = 0. c 2 that y c2 = e 4a - — ae~ x + C2e -5x -3/4. m = 0. so ci — 8m + 17 = we obtain m = 4 ± i so = —1 then c\ = 4. 4q + c 2 = —1. If y(0) = 1 y'{0) = If j/(0) — and 1 1 -5 . + c 2 sin 4i. 2. + ca^re 2* + c$x 2 e 2x (c3 cos \/3 a.Exerci99S 4. and y (ci = \e~ x cosx + C2sinx). m = 0. m = 0. we obtain + 24m 2 + 9 = 21 + c 2 + e" 1 ' 2 ci y 31. we obtain =. m = = c\+&ix + c3 e~ x ^ + e we obtain 2 3/(0) From j. - fe y{0) -5 *. m = 0. m— and 2. c\ cos . — 17sinx). m = 0. V3x/2 + CiXs'mV^x/2. we obtain c\ - C2 — m + 6m + 5 = m = cie" 1 m= m= -1. C2 c\ 4. c\ -1/2 + C2e~ 3x + C3 cos \f2 x + asm V2x. m= ci m= 3. x m= and m — 1. so ci = 4. = (4cosi —17. so that y 3/4. c\e x/2 .7m 4 From ± + c4 sin y/lxfty —1.6m 2 + 12m . m 4 — 7m From = — 2 = m From = 18 = we obtain = m 5 — 2m 4 + l7ro 3 = From = 38. 4 and . -1. 2 y 37. m 2 + 16 = c. ca m= - = —1 0. and (c4 cos 1 and = 0. From m 3 .10m + m + 5 = From m = —5 36. and m= 1 ± so that 4i + C2X + C3X 2 + e x (ci cos 4x + C5 sin 4x). FVom m 4 — 2m 2 + 1 = From 16m4 y 32. cie m = 0. and m = 2 ± 2i 80 that + c5 sin 2z). and + c 4 cos2a: + Cssin2x.'(0) = m —1= From and c3 e = -1/2. m = ±y/3i/2 and m = ±\Z%if2 so that m= we obtain m = 2 so that + C2xe x + c$e~ x + Cixe~ x + y 34. -d m= 1. c2 m= so ci = m= 1.C2 From = m= —1 . and _2a: m = ±2i so that -2. —3 so that + 9c3 = 1. \c\ + ^c2 = 0. -£ci + §c 2 = 5. + 11/4. y(0) +c 2 e 3x/2 andy = 12C3 Jaw" + c3 e" 3c3 — = --e. c\e Fromm 2 +m+2 = Owe obtain m = -l/2±V7t/2 so that = e . = 3/ . c 2 x + 5xe*. . 6 *.c + 2c2 - = cie 2iC xl 2 c\e~ = cie -2 + cae + — e~ 10 + ci 6 e lx 15 21 + e _I = 1/6.^ 2 If y(0) = and y'(0) = then ci = and c 2 = so that y = 0. then 0. and y"(0) y + c2 e cos\/7x/2 1 e If . c 2 = 1 then = Cl e -ci 1/10.8 = = we obtain y{0) = and y'(0) C! so ci = -1/6. . and y — (ain^x — cos jx). — — \/3. 4. m = y + c 2 + C3 = 0. _. so that 4ci = Ify(0) . (ci and y 1. - 36C2 1. 0. -7. From If = - 5/36. 1/6.1 ± \/3 y If = = C2xe x andm = —6 so that y = —6. = -e +e If y(ir/3) = and 2/(jt/3) = and y .^=sin\/3x^ 82 y(l) = and 21-2 2 c\+C2e~ 6x +C3xe~ 6x 3x - If . m = If y(0) . 2c2 e = 2 so that y 1/ ^/3 = + C]e m= and 1 2 m = ±i so that = ci cos we obtain ^^3 m= we obtain 2 3/2 so that y -7/4. 37-1 3a. 42. - i .|e" 61 + and y we obtain v V From 3.2V3c3 = = -^e 21 +e~ x ^cosV3a. m = 2 and m = .0. so cj = —1. c 2 = 1. and c2 cos \/3 2 + \/3c3 = -1. c3 = 0. 01 m From i/fl) From = m 2 1 then Cie + c2 e = 2 + 1 then -c\ 47. = = -fe^+^e 2*. -6C2 0. = — 3m + 2 + 1 = 0. soc! = 46.-1/2V3. and y = 5. a. ci so ci + c2 = = m 3 + 2m 2 — 5m — 6 = y(0) = -1/6. c2 -5/36.Exercises 4. c3 . . From 45. * + C3 sin V3x^ 2c 2 . 5 and + c 2 shi\/7x/2). = 4c2 1 31 ' 2 = — V3~ cos +sinx. —z-C2 ——c\ + -c 0. FVom2m 2 -2m+l = Owe obtain to = l/2±i/2 so that y = e^ 2 (ci cosa:/2+C2 sinx/2). ci + 01 = = = 10 so c\ m and 1 = 1 so that = 5. y'(O) = 0.c 2 sin 2. 1 = 2 = so ci 48. 5e* 1/ 4m2 -4m. c2 = = m3 . C3 = -e~ so cj Fromm 3 +12m 2 +36m = we obtain m = 0. Ifv(O) = -1 and j/(0) = then c\ — — 1. . and y"{0) 1/15.x + + c2 = 2ci = and -1. m= and 2. ^(0) = 1.3 = Owe obtain m = -1/2 andm = andi/{0) = 5thenci+C2 = 1. and y"(0) = -7 then ci 49. c2 m = —1. 0. . = C3 = c2 . m 2 — 2m + = 10 then = 1 = c\ m we obtain 5. 44.3 41. FVom i/(0) 43. + 1 = & . so c\ 2. --sini. = C2 —2. and y - ci cos2i 1/ + = cie 5x + C2xe 5x — e 5^ — xe hx If . is 0. 1 1 = -e*--e 5 so ci = 1. and 1.3 50.m = From y — x x + c 2 e + C2xe + C4X c\ + C2 = 0. If y'(0) = and 1 = — 2cosx. = 2. a. -l = 0we obtain m = 1 and m = — 1 so that = cie 1 + c 2 e~ z or y = C3 cosh x + cy sinh x. Now. cie 5 - 0. £/"(0) + C2 + C3 = Cl so ci m= we obtain 1/4. 3. m= ±2i so that m= ±i so that y 2 m=5 and . y(0) = 1 and . y (m-4){m+5) 2 = 58. = 1 is a root of the auxiliary equation. y'(0) C2 0. and Cl 0. o m= 1. 4 ao that y C2 = —1. we obtain = and y = cosi ci + C2sinx. Since = — cosh x cosh 1 From the — 6m + solution yi = dividing the polynomial 1 — loj = m 3 — ^-m 2 + 7m + 5 e x 2 coshfx cosh + is y'" the differential equation 5 !/ = 1) . and _ 4 m= 0. = §z: 0.3m 3 + 3m 2 . ci sinh 1 + C2 cosh 1 = 0. m4 . 6q = 4. C2 = — sinh 1/ cosh 1 and 2 From 1/ If y{0) . C4 + C2e — y"{Q) = 4. m4 = Owe obtain y = Cl + c2 x + c3 x + ax 2 From then = ci = 2. smb. and 5. 2 we obtain x . by m — 1 gives m 2 — 8m + 17. C3 2 = e If . and y 52. C2 = 2c3 3. y'(0) 2. we obtain we obtain 2 then 56. = 1 and j/'(l) = then ci = 1. Cl = = 1 then + C4 = 0. 1 Prom = 3: y 53. Cl = so ci 2. 1. C2sin2x. C2 = = = -1/4. c 2 1. so ci = 1. Cl y"'(0) 5 = + C2 - C3 = 0. then 2 = + 3x + 2x 2 + 0. m = — 1. andy"'(0) = 5 m = 0. m 2 — 10m + 25 = = m From then 55. y"(0) = + 2C3 + 2C4 = 1. If j/{0) = y(it] = y'{ir/2) = and c% sin 1x. y(0) + C3 = C2 a— Prom m* — 1 If y(0) = = = - 0. y'(0) 0. m = 3 . j/ 51. eg 1. = y(l) 54.Exercises 4. Since fm + Q (m sinhl 2 . Cl — C2 — = C4 1. 1 15?/' — IOO3/ = 0. and m = ±i so that y = ciex +C2e~ x +c$ cosx+tysinx. Therefore m = 4±i are we conclude that mi m — 9m + 25m— 17 3 73/' ——— sinh x sinh 1 -r~cosh 1 m 3 +6m 2 -15m-100 the differential equation y'"-y!/" + 59. = Hy{0) —1/2. cosh 57. 3 .coshx . = 2-2e I + 2xe -^rV. and C2 m= so that 1 ?/"(0) = 1 then + 3C3 + 6C4 = 1. -1/2. ci + = ci — 4 and y m +1= From ci = —2. 0. e'^cosx we x 4- e ta (c2Cosx m 3 + 6m 2 + m-34 by [m.(-4 + i)][m.3 the remaining roots of the auxiliary equation. A. . is + y" 16y = a differential equation . a v^m + l) (m + \f2rn. we find 3 Then yc mi = m2 into the differential fx + f . Substituting into the differential equation y 2.2 — e" x ^ ^3 cos + C4 sin ^= x 3. 9. Exercises 4. From yp = m2 -+- 3m + 2 = we find mi — —1 and A. .C3 sinx). J- = + c 2 xe 51 30 and and we assume — 10-4 + 255 = 3. we obtain ro = -l/\/2 ± i/v^ so that ^ _ € i/v^ f Cl coa _L 1 + ca sin x^ + m. = 0.2. 3) a m 7) = m — 7m (m — 3)(m + 3) = m 2 — 63. = cie (m — 6)(m + 2 = -4l and the general solution of the the third root of the auxiliary equation. yp = | and 5 - 5. + c 2 xe 0. = -1 Then yc —2. and the general solution of the differential equation is y 60. Substituting into the differential equation we obtain y 3. Since FVom _4x m 2 + 16 = a differential equation (m — 3 m +1= (m 2 4 Now sinx. Prom 64. y 3 ci cos -a: — 2A — + c2 sin -x + = cie 51 84 and we assume yp = 3 and c2 sin Then j4 §x and we assume = jj . From yp = 4m 2 + 9 = we find mj = — § i and From cie we obtain + c2 e~ m2 = 1 1. = i)] e m2 = and i Therefore m3 = 2 differential equation is + c2 e _4l sinx + c^e1*. Substituting p Then A = § B yp = j/ .Exercises 4. From the solution y\ = c\e . 6.(-4 = y 62. -4 . c\ cos 9A = Then yc = equation we obtain = c\& x + 3. cie 1 Sl 25-4 and = + c 2 e 2x Then A = 3. Since 65. Since mi = -4 + conclude that Hence another solution must be y% auxiliary equation. is — y" 7y" — 9y = 0. a differential equation is y" — 3y' — ISy = 0. 21 = m 2 — 10m + 25 = = Ax + B. cosx m 2 — 3m — 18.4 1. 61. is 4.t + gx + |x + 15.J are roots of the dividing the polynomial m 2 + 8m + 17 gives m. + / / l/v 2±i/v 2 l) and y"' is equation differential 2 m= 0. Substituting into the differential A = -3 B = yv = £x . = we find mi = 2 + 4i and ni2 = 2 . + C2sin4x) 3x — (Ax + Bx + <7)e + 12C = 0. obtain 24 . = (-4x 2 + 4x . + C2e~ 1: '' 2 and we assume yp = 4coe2x + 5sin2x.Then 2 and we aaaume yp .§) e 3x and 24 + 65 yp i/ 8. y 7. Then yc — e 2l (ci cos 4i + ci sin 4x) = Ax 2 + Bx + C + (Dx + E)^ Substituting into the differential equation we 20 and we assume yp . j/ 6.^ and yp . and ^ y = cie 3l/2 + c 2 e" l/2 - ^ cos 2x - ^ sin 2x. From 4m 2 . Substituting into the differential equation we obtain 4 = 1. y 5. C = ^ D = . 13D m. Then A = 1. Then yc — cie _2a: + C2xe~ 2x and we assume 2 ?/p = Ax +JBx + C.4 4. + 5x 2 + 4x + ^ + f-2x . . 124 + 125 . C = \ yp = x 2 . B = -4.^] yc = e and 1 .0. yp = .]§ -2. 3x Then yc — c\e equation we obtain —64 2. and \A + B + C = 0.cos2i . Then .Jgsm2x. B = 4. + 4x - 2 C2Sin-\/3x+ (|-4x and 1 {§) e + 4x + ft + (-2x - mi — V3i and mi = — v^3i. Prom m2 + m — 6 = we mi — —3 and mi = find = Ax + B.. 5x 2 ci cos \/3x find mi = + | . + ci& 2x and we assume = 2 and A — 65 = 0. Exercises 4. = + cje 2 * + 3 cie" * ^ - . Substituting into the differential equation we obtain —19 — 8B = 84-195 = 0. \m 2 + m + 1 = we find mi = ma = 0.2 = -5 Then • yc = cie e B = 4.4m -3 = = we = -26 20A = yp = 100. Then 4B= . From m 2 . From m2 + 3 = = e we 2l (cicos4x find £ = . . 24 + 5 = -2.8m + = cie -21 + c2 xe~ 2x + x 2 - 4x + ^ . 1 and . cicos\/3x + Substituting into the differential equation and 124 = Then A = -4.4i.Jg. -48.85 + 2QC = -6D + 13E = -164 + 205 = Then 4= 5.4x + | and From . C2sim/33: we obtain C = -f 3x 3l '' a . we find mi = + Bx sin 2a:. = Ax. yp m2 - — Axe ix . Prom m2 — m = we — mi find and m2 - 1 we obtain Substituting into the differential equation y 10. and cos 2x + C2 sin 2x — —x cos 2x.Exercises 4. = ci cos2x -f C2sin2:e and we assume m\ = yp = = c\ 4B = and -|a:cos2:z. and ~2C = -1.4l + ^ = — 2i. = we find m\ = Substituti:ig into the differential + ±xV 2 . y 14. y 11.E = -3 -8B + 6Z> = -12A86 1. Then yc = cie _2x + c2 and we assume 2x 2 Substituting into the differential equation we obtain 2A + IB = 5. + C2 Then A —3. + 2m = we find mj = —2 and m2 = 0. Then and mi e yc 41 . . B = £ Prom . 16 = m. 2x — m + \ = we find mi = 2 x/2 = A + Bx e Substituting into the yp = 12 + j^V 2 and A = 12. Then = ciW 2 + c 2 xe^ 2 + y From + C2 + . . + C2e . . . yp + 3x. Then yc — c\e x ^ 2 + C2xe x / 2 and we assume equation we obtain \A = 3 and 2B = 1. 2i Cl e 3 we 2 find 2i and 3 we obtain 25 + AF = + 8£ = 12D = -4C + 2. Then yc = c\ cos 2x + C2 sin 2x and we assume = (4i + Bi + Cx) cos 2x + (Dx + Ex 2 + Fx) sin 2x. Then 4 = § B = 2. Then yc = cie 4x + C2e~ ix and we assume equation we obtain 8. . = C]e x m + C<2 = —A = c\e x Then yc 0. Then A = ~\ . and = y 13. From yp = -4. Substituting into the differential equation From yp m2 + 4 = 4l Substituting into the differential equation we obtain B = 0. = cie~ m2 .2 ]-x \ + 2x + ^-xe~ 2x 2 - differential .4 = 2. Then A = | yp — jxe^ mi = 4 and 12 -4.4 m2 + 4 = Accos2:r = 3. C = \ yp = \x 2 + 2x + |ie _2x and Prom 2 . m2 = — 2i. yp = Ax + Bx + Cxe~ 4A = 2. = 3x and and we assume yp = 3. 12.4 9. Substituting into the differential equation we obtain 4C = 0.4 = 16. 1 2 1 + C2Sinx — -x cos x + -ism x.C = 12 — sin2x.4 Then A= -fa.B=-£.yp = ^cosx. and 2C-5D = 6. B= 0. Then A = 0. — A cosx + Bsinx + Ccos2x + Dsin2x. From m2 — 5m = we ci . and -AC . 62 lo n xJ + ) J — — r«2 p j/ Then yc i. C=§ = ci cos = .3D = 0. Then A = y p = —j cosx — ^cos2x + ^ sin 2a.Exercises 4. Substituting into the differential equation we obtain we 2x e e we -3. mi — -6 and m2 = 4. From m2 + 2m + 1 = = (ci cos2x + C2sin2x) + ^xe I sin2x.2 — y 16. + —x and m2 697 .4=|. Then . cosx mi = find . and yp y = c\e~ + C2xe -T — x 19— . £= F= fa. * = "A* 4 + fr 3 + = v 17. Then y c = c\er x —1. SB*. . 12A-15B = -4. From m2 — 2m + 2 = A + 2B = 1 ?/ 19. and x e (c\ cos differential 2x + ca sin 2x) and equation we obtain . C = 0. x H x. From m 2 — 2m + 5 = we ^ . = y 18.8 = 0. . . and 2 yp 2 — . £> = £ yp = — ja co8X + gxsinx.cosx 20.4 = -\ B = 0. = c\ + cosx c^sinx and we assume = (Ax + Bx) cosx + (Cx + Dx)s'mx. 5 and m2 = Then yc = 0. and y 15. 0. Then A = -fa S = $ C = D = -|g 4 yp 2 3 . Then x cosx (ci find .^sinx + C2 sin x) + mi = TO2 = 7 1 5 5 -e 2l cosx — -e^sinx. c\e 5x + and we assume ca = Ax + Bx + Cx + Dx. B = \ yp = ^xe 1 sin 2r. — cos2x Jo 2. From m2 + 1 = we „ + + C2Sin2x 2a. find = Ae and -2A + B = and we assume yp 2 = 1 — i. 65— IOC = -1. -4A = 2. Substituting into the differential equation we obtain — 20A = 2. -3C + 4£> = 3. Then yc = we assume yp = Axe? cos 2x + Sate* sin 2a:. = 1 — 2i. + + C2xe~ x and we assume differential equation we obtain D= B = Q. Substituting into the 4S = 1 and -4A = 0. 2A + 2D = 0. 0. Then y c = e*(ci cosx + C2sinx) mi = 1 + i and 2x cosx + Be sin x. 25 \ In — x cos2x + — i' sm2x. . mi — find and i f 1 \ 12 = (-^x 3 + fx) cos 2x + ^x 2 sin2x. -2A= 1. and -2B + 2C^0. From m 2 + 2m - assume yp 24 — we find = A + (Bx 2 + Cx)e 4x . Substituting into the 2. 2o Then y c — cie" Substituting into the differential equation 87 and 61 + x and we C2t* we obtain —24.. -»d cie find kt 1 + C2 mi = x + 2i 1 4 53 14 . Substituting into the differential equation 4^4 + C = 1.£ cosx + ^ sinx. mi — m.^x + m3 = — rri2 . and 014 = —1. 6j?-C = -1. = 24. m4 — m 2 = = we 1 -+- 2 sin x. and and we assume yp 25. andB + 6C = 0. Then yc = cie I + c2e 2a: +C3e -2x + c2 e 2x + C3e _2lc + 7 4 + ^xe* 3 + 7 xe 5. = Ax and = we B + Cx 3 e x 4- 6C = Then -4. m 4 + 2m 2 + = C3X cos x m = 2 and m3 = -2. y = x 2 . Then . £7=^.2x . and = x 6 2x Substituting into the differential equation . and we we obtain C4X sin x and A= 1. Then -4 = |. . / = Then yc 2 2x cie * +C2xe +C3e~ 2x Substituting into the differential equation .yp = J y From j / = 4+Bxe I +Cxe 2x Substituting into the differential equation we obtain AA = AC = 1. . + caxe 21 4- mi = find (±x cae" 21 2 ) e Then yc 1.[\x 2 + y = ci §x) e~ x . and p find mi = 013 we assume yp = Ax 2 -2. 10C . B = -2. + */„ . and we assume and = ?/ —6 cos x + — 1 1 + C2X + C3e Br . C = -§ + c 2 xe + c3 xV . Then A = 1.12A Substituting into the differential equation A = -\ B = Then .^x 3 68 ^x 2 + ^x^ e~ x . and -4C = 2. -IB = 0. + cix + cge 8 * c\ = 3. and A--\.4 cie 2 find = B= -1.4 2S+10C = = 20S -2.B = -^.~x 2 - Ci 4 22. Then A^\. = cie x c%xe x + C3X 2 e* and we obtain —A = 1. . we obtain and ^x 3 .^ = 5 + ^xe 1 + Jaw 2 *. 4 = i and TO2 = ran = —i. Then yc = c\ + C2X + c$e x + c\e~ x and we assume yp = Ax + Bx 2 + (Cx 2 + Dx)e~ x Substituting into the differential equation we obtain -6A .-x - . = 7712 and — 013 Then yc — 6. From m 3 — m 3 — 4m+4 = -3B = -1.C=-^. From m 3 — Qm 2 = we mi = find 2 yp = Ax + B cos x + C sin x.2D = 0. Then and e 21.-x 3 e x .fl = J. From m3 -2m 2 -4m + 8 = we find 2 = y 23.Exercises 4. ro3 = 1. 3 . C=^ .Then yc — ci cosx + C2sinx + + Bx + C. C = -3.yp= 3 -\x . D--\. . we obtain .x i: mi = 1. and + c2 x + cge 1 + c 4 e _I . 21 m 3 — 3m 2 + 3m — 1 we assume yp 3-4 cje mi = 2x and we assume yp = (^4x + Sx )e and 6A + 8B = 0. and 3 . -1.4 = -| B = 0.yv = -§-(^x 2 + fax) e 4*. From -B= 0. C = -\ From we find 3 . yp .B 3 o ci cos x 4- C2 sin x 4- C31 cos 1 + C4X sin x + x2 — 2x — 3.2 = 0.^x 2 ) 3 . = -±x 2 . and m3 — —2.4. B= y 26. ma = 2.3. -3. 24^4 = c\e .fxV. = -^. 21 . and = 4+ | cos2x. so y 34. S = § C = 0. . Thus y = ae~ 2x + cze*/ 2 . into the differential equation we obtain -8A 0. B -7. . We = have yc cie -2* A= equation we find From the initial + cie x l 2 and we assume yp = Ax 2 + Bx + C. . = 4 — 4cos2x. 9. .4 27. and -3C = 0.7x 2 . We have yc = c\ cos2x + C2Sin2x and we assume yp = A.\ — i and rug — — i. We have yc = Cie equation we find 1 + C2e _I A= a | = e~ (c\ cos x + c<i . Thus y = c\e~ x ^ + C2 . Substituting c = C] cos u + C2 sin i and we assume yp = A cos 3x + B sin 3x + Cx cos = = = \. From m 2 + 1 = we find mi = i and m 2 = — i. c2 xe~ Substituting into the + ( B x 3 + §i 2 ) e -2*. -3B = -4.\xe~ x = cje 1 + C2e _3: + ^ sinhx. Exercises 4. j/p a. and p = -^sin3i + \xoasx.2D 0.. = Ae~ 4x Substituting into the differential sinx) + 7e~ 4x From the initial conditions we and we assume yp 2x = e ~ 2x (. so _ 19x _ 3? 5 We have yc = c\e~ x ^ +C2 and we assume yp — Ax 2 +Bx. 2x so Qx 3 + \x 2 ) 33. x + c2 e _I + \xe x . We write sinxcos2x = ^sin3x — ^sinx. We have yc = e~ 2x (c\ cosx equation wj find obtain ci + c2 smx) A = ^ Thus y = = -10 and C2 = 9. e ci We = have yc differential From the c\e -2x + equation we initial 200' I/5 . Substituting into the differential equation we find A = —\. conditions we obtain y 31. — ci + 9xe~ 2x + + 30x. 30.-200. = cie' 2* + .10 cos x and we assume yp + 9 sin x + 7e~ 4x = Axe x + Bxe~~ x . .37.Then A = 4. From rr? + 1 = We write 8 sin we find m.200 . 28. -SB B= C = \ £> = 0. Then yc = c\ cos x + C2 sin x and we assume yp — A + B cos 1x + C sin 2x. Substituting into the differential equation we obtain A = 4. .3x 2 = C2xe~ 2x and we assume yp find conditions A = ^ and we obtain y = c\ 2e~ 2x B = = 2 § and — (Ax 3 + Bx 2 )e~ 2x Thus y . Prom the initial conditions we obtain c\ = 200 and ci .\ a: j/ . = A -2z 5 = + and oi 186 gl/2 _ = Hp 7x 2 . Then + Dx sin x. ) Substituting into the differential and B = — \ Thus cie . 29. Thus y = c\ cos2x + C2sin2x — \ Prom the initial conditions we obtain c\ = and C2 — V2 so y = V2 sin 2x . j/ .19x . so y 32. . Substituting into the differential = -19. Substituting into the differential equation we find A — -3 and B = 30. and -2C = Then A = 0. and C = -37.3x 2 + 30x. Exercises 4. Substituting into the = — fo/2w and B = 0. ^3-65 COs2l -65 S1Il2x A= ci x sin ~ cos 2x 00 i - we assume yp B= 3: and j From so .72 + C2 sinwf + coswf = ci fb/(w itt 15 7 = 7—= (w 2 - 7 ) cos 7*. 7 1 . and D = 5 Thus . (_F and we assume xp Fo/(w z ~~ We + - 2 x— 37. Thusx = c\ Coswt+C2sinuJi— (Fo/2u>)t coswt. x 5e~ (fb/2<J )sinurf we obtain conditions initial - C2 sin uit find x the — . = c\cosx + ci C2sina: . = A + Bcos2x + C sin 21.l2a. j/ cie~ x + C2e 3x - . + 7—. B = 0- and ) /2w)fcoswi.4 Prom 35.C = 0. ^ 2 the initial 38. conditions we obtain c\ = and C2 — Fq/2w 2 so initial have i e —5. I 1 H 2 A = — 5 B = —^ = = —\ C2 4 sin O and sin a: and we assume find if c\ 6 3x so 0. 2a:. = 7 and C2 cosw( c\ differential equation Prom we obtain Thus . We have yc = ci + c-ie? differential equation we + we obtain -* 1 = e F0 + y = 3 55 3. . 00 n 4 . B — ^. the conditions initial y 39. Substituting into the so . A = 0.cos y = and we assume yp we obtain conditions ) C2 it From Substituting into the Thus . w 7 COSwt J + C2 sin x + x ci cos 1 = — . We the conditions initial have x c = the ci — y — 7e We = Q cos uit + we A= have yc = ~ sinhx. differential equation Prom x + C2 smart and we assume x p = Atcoswt + Btsinuit.7 2 and 2 Substituting into the differential equation we find y = A cos 7* + B sin 7*. G 52 csxe x and find c\ and C2 — sm „ 2x. — sr. We have j/c = cie -1 differential equation + C2e we ( = —\ t J = Ax coax + Bxsinx + C7cos2i + £>sin2:r. ^-x sin + x ~ x — . Substituting into the .V + ^e 5 *. H 3 —^ C- =^ 2. p . so we find A x 36. „ 90 - = Ax + Bx 2 e x + Ce 5x C= 2a: j . = + C2C 1 + c 3 xe x + —12. and Thus sin 2x.COS 7*. We have yc — conditions 2 = ~ i c\ and t/(l) ' 2x + eX ("S c\ — — cos C2 . and we assume yv = Ax + B. jt/2]. We have yc = e*{c\ cosx equation we find A = 1 + C2sinx) and B= 0. and C=— 1. Substituting into the differential Thus y = ex {a cosa:+C2sin:r)+:E.4 FVom the initial conditions we obtain y 40. solution of the boundary-value problem is y 42. and C3 . oo) — c%e x sinx + x. + c 3 sin\/3x) + -x - y so 9. The = — 0. — —H ^ ^x = Ax + B + Cxe -21 C = | Thus 5 t and + 2 -xe~ 7a = C3 + sin - .4 we have y = = c3 cos 2x and B— = j . FVom + Bsuix = y(0) c\ cos = 1 on and Substituting [0. Substituting into the differential Thus = cj cos x + C2 sin x + x 2 — 1. + e^ci = any 7T. C2 = —11. real number. 6(cot l)s'mx + x2 ~ 1. From y(0) = and y{ir) ?r --- we obtain — 7T Solving this system we find c\ problem = and C2 is We have yc = C\ cos 2a. 1 FVom 5x and we assume yp \/Zx = qe" 21 + e I (c 2 cos \/3a. the initial » 41. = is y 43. r . .Ue x + 9xe x + 2x- x e (c2 coa + = ci + cssim/Zx) equation we find A = I2x 2 x e + ^e \ B — —| .\ + h'*- = A 2 + Bx + C. so . ci we find . j/ we obtain ci (cos l)ci Solving this system we find ci = - 1 = 5 + sin(l)c2 = 6 and C2 = —6 cot 1. oleosa: Thus y + c 4 sin 2x. . and . We = have yc cie -2x = 11 Substituting into the differential = 11. 1. Prom 3/(0) = 5 cosz + C2sin:E and we assume yp A= equation we find = e we obtain . 6 cos a. The solution of the boundary-value On (7r/2. + 2x ca sin 2x y'(0) = 2 + we 3 sin x on obtain .Exercises 4. C2sin2i and we assume yp into the differential equation [0. B = 0.ir/2]. (D 3 . D 3 + 4Z> = D{D 2 + 4) 13.8D 2 + D(60) Ae x l 2 - D4 because of x 3 1)(D = + 20e 2x = (D .2x) = D 3 (30x . t — = . D 4 + 8D = D{D + 2){D 2 -2D + 4) D 4 y = D 4 (10:e 3 .2)(D + 5)4e 2 * — (D - 20. (D 3 .5sin8x) 2)(8e 21 ) 22.4D 2 7.Exercises 4. (D .4frc / 2 = = . D 3 +2D 2 -13D+10 = 14. D 2 - 5 a: e" 3 * + e 2* = (d .2) 2 {D + 4) 2 = Ae*l 2 = . 2 \)y a £> a (603:) = %De x l 2 . (4D 4.40cos8x) + 64(2cos8z .cos7r + .6){D + 2) + 8)y = +3 2. (D + 64)(2cos8i . (2D - = {2D - l)4e x i 2 19. 2D 2 . sin 7T H 6 —1 + j = — C3. 92 D5 2) + 3) = (D . 16 - 128 cos 8a: because of x 4 320 sin 8x = [C .2D 2 + 7D-G)y = l.2)28e 2 * = 56e 2 * .4 = Solving this system we find c\ Now x= continuity of y at 1 and — 5 H 7T — 1 . {D + 3.S&e 2* = = D(-16sin8i . is tt/2 tt/2 Exercises 4. (3D 2 - 5.5sin8x) 15.2)(3D - 12 + 2) = (D . C3 = | .V5) (D + y/E) 10.sinx 6.3D - 2 = {2D + 1){D 11. D4 (D-l)(D-2)(D+5) 17. D 3 + 4D 2 + ZD = D(D + 16. 9D 2 - 9.2D 2 + D) = 8. sin 3 Continuity of 5 —| — 2C4. + + <x< ^ sin x > |sin2x.128 cos 8a: + 320 sin 8x+ 21.2) = 18. Hence | — Thus y . D 2 - 5)y + 5£> 4 l)y = + 5D) = = 4£> 9sinx e x 4x (3D .cos — = — ZC3 shitt + or + cos 2x 7r/2 implies cos or = c-i cost 2c4 | and the solution of the boundary-value problem !cos2x |cos2x g sin 2a.5 = 1. Then 04 = | sin 2x + | sin x on C3 cos tt + C4 sin 7r 2 j/ 1 at x = implies 7T — 2smx + . D 3 + 10D 2 + 25D = D{D + 5) 2 12. {D 4 . D 3 (D 2 + 16) because D 2 (D 2 + l)(D 2 + 25) 29. e~ 3x 37. 24. . and 1. D 2 + AD = D(D + 4). e 5x . e (3 2 5) .2D + 5) because of 1){D l) - 2 1 because of (D . 1. D 40. 2 + l 2 ). x. e . D 2 -6Z> + 10 = 39. e 12x . 1. —9A = 54 or .6) 2 26. Applying D to the differential equation we obtain D(D 2 - Q)y = 0. 28. D 2 (D . Then y and j/p = solution = + c 2 e~ 3x + c3 A. and and e I e 2x cos2x e~ x 5} because of sin x and e 2x cos x e" 41 ex e' Sx / 2 . e 1 x .9D - 36 = (O - 12)(D + 3). e 3l sin:r 5x 7x Exercises 4. (D + 1)(D - 30. Substituting yp into the differential equation yields is y 2. because of e _I and x 2 e x 3 32. The genera! . D(D - 2) 25.6 1. 35. 3 . 36. D(D 2 + 1) because of x and xe6* because of 1 and sinx sin 4a: because of 2 3 2x x 2 and because of 2) (D + 2D + a: of e and co95x x. we obtain D(2D 2 -7D + 5)y = 0. sinx. cos\/5x. 1.Exercises 4. 3x c-je Applying D to the differential equation = cie ix + c 2 e- Sx - 6. sin \/5x 38.6 23.4 = —6.D 2 -2(3)D + . £> 2 .10D + 25D = D(D 2 x.4D + 2) 33. e *. D(D 2 . e xe 3l cos:r. 5 1. x2 x4 34. D2 + 4 because of cos 2x 27. D(D - 31. 6 Then y and yp = 5A = —29 A. and the general solution y = cie~ 2x 2 6.^. The general solution is + 3x. D 2 to the differential equation we obtain Applying D 2 (D 2 + 4D + 4)y = D 2 (D + 2) 2 y = 0. Then y and j/p = j4i. Substituting yp into the differential y 4.Exercises 4. B— 1. is + c 2 xe~ 2x + ~x + 94 1. Substituting yp ci + c%e~ x + case"* 4- C4X into the differential equation yields y 5. Substituting yp into the differential equation yields general solution D Applying or A= —29/5. 4Ax + (4A + 4B) = 2x + 6. we obtain D(D 3 + 2D 2 + D)y = D 2 {D + l) 2 = y 0. = ae 5x/2 + c 2 e* + C3 = to the differential equation cie hxl2 + C2e x . we obtain D(D 2 + D)y= D 2 (D + l)y = 0. D Applying to the differential equation = +c 2 e ci A= equation yields _31 3. Equating Substituting yp into the differential equation yields coefficients gives 4A = 4A + 4B = Then A= 1/2. = — c\ + C2e~ x + czxe~ x + A— 10. Then y = cie' 21 + c 2 xe~ 2x +c3 + c4 x He and yp = Ax + B. . Then y = ci + c2 e~ x + C3X Vc and yv ~ Ax. The is y 3. The general solution is lOx. obtain. Then y and yp = Ax 2 + Bx. and the general solution 8 2 ci is + c 2 + c 3 e _I + -x 4 . 2 Substituting yp into the differential equation yields 6 Ax 6/1 Then c3 x 4- = cl to the differential equation + c2 e -1* ** is 2 19 .-x 3 + 8x 2 :k D 4 to the differential equation we . + -x'. B= -8/3. —19/9. .4 A= 2/3. Equating = c\ + cae" 3 ^ + ct x + (2A + 3B) = 4x- coefficients gives 2. and the general solution y 7.Exercises 4 a 6.—x. we obtain D 3 (D 3 + D 2 )y = D 5 {D + l)y = 0. Applying 8. D3 Applying to the differential equation we obtain D 2 (D 2 + ZD)y = D 3 (D + 3)y = 0. D 4 {D 2 -2D + l)y = D 4 (D ~ ify = 0. Then y — c\e x + C2xe x + c^x 3 + C41 2 + c^x + c$ yields . Then = y + C2X + c3 e~ x + c 4 x 4 + c 5 x 3 + c%x 2 ci _ = Ax 4 + Bx 3 + Cx 2 Substituting yp into the differential equation UAx 2 + (24A + GB)x + {6B + 2C) = St 2 Equating coefficients gives and yp . B= = D3 Applying 4 + 35 = -5. 124 = 24A + 6B = 6B + 2C = Then A= 2/3. C= y 8. = 8 0. 2C + Then A= 1. we obtain (D-6)(D 2 + 2D + 2)y = 0.3)y = D{D . Applying D—6 = Cl e 4x to the differential equation + c^e~ 3x + equation yields TAe41 ^xe 4x = e**. The general solution is and yp y 10. Substituting yp into the differential coefficients gives A — 1/7. Equating .3)y = 0.6A = 6.1)(D + 1)(D . Then y = cie 3x + c2 e~ x + c$e x + c4 _ 96 = 5c 61 . £> = 32 C\e 0. Applying = 1) general solution — e~ x (c\ cos a. Substituting yp into the differential equation yields Ax + {B . Then y = cie 4x + c 2 e~ 3x + c 3 xe ix _ Vc = Are 4 *. we obtain to the differential equation -D.6A)x 2 + (6A .4S + C = 2B . Then y and yp = j4e 6a .Exercises 4.6 = Ax 3 Bi 2 + Cx + D. to the differential equation D{D - 1)(£> 2 50Ae 61 is + ci sinx) + e6T - ^j we obtain -2D. C = y 9. Equating . and the general solution . The y 11. coefficients gives D{D — e _I (ci cosx + C2sinx) +C3e 6s: Substituting yp into the differential equation yields A— 1/10.12)y = (D . Equating coefficients and j/ p 4- 3 A= gives 1 B .4B + C)x + {2B ~2C + D)=x 3 + Ax.4 .Af(D + 3)y = (D . x D= 4 x 3 2 + C2xe + x + 6x + is 22x + 32. B = 6.4)(D 2 0. Applying D—4 = 22. The general solution ^ sinx. 21 yields solution is + c 2 e.~. Equating Substituting coefficients gives . and the general . Substituting yp into the differential equation + 8Bx + (6S + 8C) = 3e~ 2x + 2x. Then y =d cqs2t + C2Sin 2x + C3 cos x + a sin x + c5 l/e is + . Applying B= and c\e 3x The 3. B= 1/4.ix + \xe~ 2x + ]x . Then y = cie~ + c2 e~ ix + c$xe~ 2x + c*x + c 5 2x Vc and yp 2Ae~ 2x = Axe~ 2x + Bx + C.4 y 14. we obtain to the differential equation D 2 (D + 2)(D 2 + 6D + &)y = D 2 (D + 2) 2 (D + 4)y = 0.6 and j/p = Ae x + Equating B. we obtain \){D 2 + 4)y = 0. is + c2 e~ x -e x + 3. Substituting yv into the differential equation yields coefficients gives A = —1 = y D 2 {D + 2) 12.Exercises 4. 2 to the differential equation (D 2 + 4 lb we obtain lXD 2 + 25)!/ = 0. Applying D(D 2 + 1) p into the differential equation yields 24 A cos x i/ — ci cos 5a. Applying D +1 2 -3/16 = cie" 0. Equating coefficients gives 2A = 3 85 = 2 6B + 8C = Then A = 3/2. Then y = pi cos 5x + C2 sin 5x + C3 cos x + a sin x yc and j/p = Acosx + Ssinx. general solution — 4Ae x — 3B = 1 4c — 9. Cy 13. + C2 to the differential equation D{D 2 + = and sin5x B= + 1/4. 24B sin x = Gsiax. Then y = cie~ + C2Je~ 3j: 3l: + c3 xe 4:l + c4 e 4T — Axe ix + Be^. and the general solution 1 ~ 5 xe 98 _ 3 1 x e 36 6 10 . 100 is . Then y = cie 2x + eye' 5 * + c3 xe x + c\t x + c 5 x + eg = Axe x + Be x + Cx + D. Substituting yp into the -6Axe x + {5A .4 144 Then A= B= -1/49.6B = -10C = 3C differential equation yields - 1 0. y y C= -1/10. Applying = -1 = Cl e- to + c2 xe~ 3x to the differential equation (D - 2 1} {D 2 + 3D - is . B- -5/36. and j/p -6-4= 5-4 Then A= -1/6. and the general solution y 16.Exercises 4. ^e 2 1) {D .4) 2 (£> + 3) 2 = j/ 0.6 = Acosx + Bs'mx + C. 2.cosx + sinx — = 4/3. -3/100. = ae Or + 1 c?e 2 10D -- D= —St Equating coefficients gives 1 . Substituting yp into the differential equation AQAxe + (UA + 49B)e 4x = -xe 4*. we obtain = D 2 (D - 10)?. we obtain to the differential equation (D - Equating 9)?/ = (D . and C — —2.IQCx + (3C .—xe 4* + 4x .2){D + S)y = 0. Substituting yp 3j4cosi + 3Bsinx + 4(7 = 4cosx + 3 sin x — 8. D 2 {D — D l) 2 2 + 495 = 0. The general solution is into the differential equation yields and yp = y 15. Applying (D — 2 A) ci + C2sin2x + cos2x 2 4) (D 2 + 6D + coefficients gives -4 4 .10D) = xe x + x. 2/343.6B)e x . B = 1. Equating coefficients gives and yp yields ix 49. Exercises 4.\f{D + l)y = 0. .4 + AB)xe x + (2B + 2C)e x .-x 2 e x + 6 to the differential equation (D + \f{D 2 4 -xe x - is 5. Li 19. Equating coefficients gives . B = 0. The general solution is y = cie~ x + c2 xe~ x + ]-x i e" x . -5.6 D(D — 17. The general and yp 3Ae x .D — x 2 e x + 5. Then y = pi^ + <%e~ x + c3 x 3 e x + CiX 2 e x + c 5 xex + ce d Vc — Ax 3 e x + Bx 2 e x + Cxe x + D. and C = 0. y 18. Then y — x e (c\ cos2x + C2sin2x) + e*(c3 cosx + C4sinx) yc = Ae x cos x + Be x sin x. Equating coefficients gives and yp &A= yields 1 6A + AB = 2B + 2C = -D = Then A = 1/6. Applying (D + l) 3 C= = 1/4. Substituting yp into the differential equation 6AcV + (6. Substituting yp into the differential equation yields we obtain cosx + 3Be x smx = e x s\ax. and the genera! solution + c^e~ x + -x 3 e x . 4 we obtain + 2D+ l)y =(D+ ify = 0. Then y = Cje~ x + C2xe~* + cjx 4 e~ x + c<ix 3 e~ x + c^x 2 e" x yc and yp = Ax i e~ x + Bx 3 e~ x + Cx 2 e~ x Substituting yp into the differential equation yields l2Ax 2 e' x + &Bxe~ x + 2Ce~ x = x 2 eT x Equating coefficients gives A = 1/12.D(D . Applying I} 3 we obtain to the differential equation D(D - 3 1) (D 2 - l)y . . Applying D 2 — 2D + 2 to the differential equation {D 2 we obtain -2D + 2){D 2 -2D + 5)y = 0. c\e x D= 5. B= -1/4.4 = and B = 1/3. 4a. 5i — 2x cos The 10Bcos5x - general solution is 5x. Substituting yp into the differential equation yields {95 I x I 27v4/4)e cos3i — (9^4 + 27i?/4)e sin3x = — e^cosSi + e sin3a. Then y and yp = . Applying x e {ci cos differential equation we obtain -2D+ 10) [p 1 + D + = (D 2 .-^e 1 cos 3a: . 4 Then jl = -4/225. .^-ex sulZx. Substituting yp into the differential equation yields 2Bcosx 2 A sin x = 4cosx — sinx. Equating coefficients gives A = 1/2 and B = 2..6 solution is = y (D 2 2x + C2 sin2x) + — e x sin a. The general solution is and yp y = c\ cosx + C2sinx + 100 ^3: cos a: — 2xsinx. and the general solution = cie~ x is ^ + C2xe~*ft . Applying + C2 sin 5x + C3X cos 5x + c^x cos 5a: D2 + 1 = ci cos 5x to the differential equation (Z> 2 + 1)(D 2 4- into the differential equation yields A — —2 C2 sin and B = 0. o D 2 — 2D + 10 to the 20.2D + y 10) (*> + 5) If = 0.. Then y — ci cos + C2 sin x + C3X cos x + C4X cos x a: Vc = . we obtain + 1) = (£> 2 + 2 I) = 0.-1 4 27 -&4 S- 1. Applying -28/225. Equating coefficients gives y 22.4xcosx + Bxsinx. Equating coefficients gives and i/p - -— A + 9B . Substituting yp 20sin5x. Then y = cie'^ 2 + C2xe~ a ^ 2 +036^ cos3x + c 4 e sm3x a: yc = Ae x cos3x + Be x sin3x.Exercises 4. 10-4sin5x = cos 5x + = c\ cos 5x Bxsin5x. B= y 21. D 2 + 25 to the differential equation we obtain {D 2 + 25)(D 2 + 25) = (D 2 + 25) 2 = 0. B = 1/8. + C5 + Bx sin 2x + C.2C + D = -C= Then j4 = 2. and the general solution V3 — C2 sin and applying 2: D(D 2 + 4) + 2 cos x is + sin x — x cos x. to the differential equation D{D 2 + 4)(D S + 4) = D(D 2 + 2 4) = we obtain 0.Cxsinx = xsinx. and ci 24. — v/3 cos 2: + 1.r + C2 sin --- + C3 cos + C4 sin + c$x cos x + c%x sin x a: a.4 cos x c\ cos —. Equating coefficients gives A = 0. Substituting yp into the differential equation yields 5 + 5 cos2x. Applying + l) 2 to the differential equation (D 2 2 + 1) we obtain (D 2 + D + 1) = 0. Then y and j/p = -4i cos 2x -4Aain2x and C= = ci C32 cos 2x + C4X sin 2x 1/8. coefficients gives B + C + 2D = O= -A .Exercises 4. a: Vc and i/p = (B Equating + Bs'mx + Cxcos x + Dxs'mx. Writing cos 2 x = h(1 + cos2:r) D = 0. C= -1.6 (D 2 23. 5= 1. 2 ) Then y ~ c\ -\- cix + C3e 81 + c4 x 2 + c 5 x 3 + c$x A . Then y = e -/a . Applying + C2 sin 2z + cos 2z = ci is cos2i + C2 sm2x + ^1 sin 2a: + D 3 to the differential equation we 8 obtain D S (D A + 8D = D b (D + 8) = 0. + 4£?cos2i + 4C = The general solution j/ 25. Substituting yp into the differential equation yields + C + 2D)cosx + Dxcosx + {-A -2C + D)s'mx . Then A = 11/256.Exercises 4. . Equating coefficients gives 4£> = 1 2C + 4D = -4B = -1 Then A= -7.^xe x + ^ 1) = D 2 (D - l) 4 = Then y = cie x 2 I e we obtain . . C= -1/2. and = cie* + C2Cosi + cgsinx - y 27. D(D — 26. to the differential equation D 2 (D - 1)(Z> 3 and the general solution 7 + ^e" 1 .3D 2 + 3D - + C2xe x + c$x 2 e x + c4 + c 5 x + cex 3 e x 102 is 0. Applying D 2 (D — 1) £> = 1/4. B= 1/4.4Be~ x -A = xe*~ e~ x + 7. . + 6B = 2 + 24C = 9 16A 485 96C = -6. 2 1) C= -1/16. Applying B= 2 {D + 1} D(D - l) and 7/32. and the general solution we obtain to the differential equation (D + 1)(D 3 - D +D2 1) is = D(D - 3 1) (D + 1)(Z> 2 + 1) = Then = de + C2 cos x + C3sinx + a + c$e~ x + <%xe x + 31 j/ 2 x c?x e Vc and yp = A + Be~ x + Cxe x + Dx 2 e x .6x 2 Equating coefficients gives and yp . Substituting yp into the differential equatioi Wxe x + {2C + AD)e x .6 = Ax 2 + Bx 3 + Cx 4 Substituting yp into the differential equation yields 16A + &B + (48B + 24C)x + 96Cx 2 = 2 + 9x . and p yp .3D + 2) = D(D - 2)(2D 3 2 + 2){D + c 2 e 2x + c3 e xf2 +c4 + c$xe 2x + cee' 2) {D + 1)(2D - 1} = 0. Applying . The general solution + C3e x + e^are* + ^x 2 4. 2 (D + 2) = 0. + c2 e 2x + cze x/2 + 1 + 4 1)(D - B 1/9. y 28. Applying = = 1/2 and ci + c2 x B— 1/2. c%xe x + c3 x 2 e x - and applying 13 is + x + -x 3 e x . . 2A + 2Be x = 1 + 3 e .2QCe~ 2x = 2 + and yp and C= 2x e Substituting yp into the differential equation yields 2x _2x Equating coefficients gives = = e + = D(D — 1) to cie~ x A . j/ into the differential equation yields Equating coefficients gives -A + 3B = 16 ~B = -1 6C = Then 4 = -13. ^xe 2x - ^ e ~ 2 "- we obtain 2D 3 + D 2 = D 3 (D .3D 2 . B— C= and 1.2} 2 ) .2){D A . + c3£ x + c^e 1 + + 2 C5X + C6i 2 e I Substituting yp into the differential equation yields A= y 30. is the differential equation D(D - 1. and the general solution y 29.6 = A + Bx + Cx ex Substituting (—A + 3S) — Bx + 6Ce x = 16 — x + e*.Exercises 4. c\e x + + e" 1 )' = 2 + e 2x + e~ 2x 1. Writing (e* = and the general solution 1/6.iUV is 1 to the differential equation we obtain D 3 {D . 6 D(D-2)(D + 2) to the differential equation we obtain D(D - 2)(D + . ) Then y and yp = Ax 2 + Bx 2 e x Equating .if = 0.AD = D h (D . Then = y x cit~ 2* Vc 21 = A + Bxe + Ce~ 2A + 9Be 2x . coefficients gives D 3 (D — 2) ci . —1/20. C = = S 31.. and yp = A + Bxe x + Cxe~ x Substituting yp into the differential equation yields AA .1)(4D 2 + = 1) 0. The general solution is y . and the general solution + c2 x + c$e 2x + ae. and C = 1/6. = cie 1 + c2 e~ x Equating . '.\xe x + \xe~ x o o . y 32.e 2x 2 and yp 3 into the differential equation yields . 0. Exercises 4. Writing 2cosh x = D(D - e x = The c\e + e~ x 1){D + general solution x l2 + x 2 c2 e~ ^ 1)(D 4 - + C3 cos \-x + c\ sin \x + \xe x ^ 2 2 D(D — 2 = cie x + + 5D 2 + 4) = D(D - 1) c2 e~ x 1) to the differential 2 (D + 2 1) (D . Equating coefficients gives -8A + 24C = -24B = -48C - 160= Then A= -5/16. Then y = c\z x t2 + c2 e~ x ^ 2 + C3 cos -x + c. and O= to the differential equation (2D . B = 0. B = —1/6.2)(D equation + 2) = + cse 2x + c4 e~ 2j + c5 + c$xe x + <^xe~ x yc . -1/16.6Be x + 6Ce~ x = -6 + e x + e~ x Equating coefficients gives A = -3/2.6 Then = y + c2 x + ae 2* + f^e"^ + c 5 x 2 + cex 3 + ci c7 x 4 + cgxe 2* Vc 4 21 = Ax + Bx + Cx + Die Substituting yp 2 (-SA + 24C) 24Bx 4SCx + IQDe 2* = 5z 2 . we obtain . Substituting yp into the differential equation yields A~ 1/8.1)(16D 4 - 1) -1.j sin -x + Csxe x^2 Vc and yv — Axe x^2 coefficients gives . . Applying (2D - 1) -5/48. a Then y x 2 e ^ is 1)(D and applying 8Ae*^ — + c%<?* + c 4 e~ 2x - ^ I 104 .2* - ci 5 ^x 2 - j^x 4 - is ^xe 2* we obtain = (2D - 2 1) (2D . 1 + x. 35. and j. Thus we find A — —1/4 and yp = A. we obtain — 64. j/ p . and The complementary function is yc — ci+c2 e bx . Using y' initial 2B) + 2Bx — x. 8c 2 =2- e _I . The complementary function is = yc c\e Sx + initial Using .T 9 conditions imply d+co — 1 1 2 j/ p . 8 8 Substituting yp into the differential equation B= = 5/8 and y 34.4 Substituting yp into the differential equation The Sx c2 e ax y = Cl e y' = Scje + c2 e~ 8x - 81 - 8c 2 e~ D to annihilate 16 = 16. Thus A = 9/25 and £ = -1/10. conditions imply = 2 and C2 = — 1. .x + -x 2 . \ 4 8x conditions imply ci+c 2 = - 8ci Thus ci = = C2 The complementary 1/2. function is yc = c\ + x 4 c2 e . Using ^9 . + = Ax + Bx 2 Thus A — —1 and x we find 1 . . The initial . = l 1.6 33.x + -x 2 = -c2 e~ x .4+25) — lOBx = —2+x. D 2 to annihilate x — 2 we find = Ax + Bx 2 Substituting yp into the differential equation we obtain (— 5.Exercises 4.ci+c 2 e -C2 ci to annihilate we obtain (A c\+c 2 = Thus D 2 and y The 0. function find yp -6ca = --. -f- 2B cos x — 3C cos 2x — 2A sin x — 3Z) sin 2x = and D = 0. -11/3. conditions imply 3 d Thus 37. <5 The initial conditions imply - 7T sin x -Cl Thus ci = -7r and c2 = The complementary xe x +5 we find yp 0.l) 2 to annihilate + Cx 3 e x Substituting yp into the difFerential equation . Thus we find A= 5/4 and The initial y = cie y' = cie x + c 2 e~ 6x + ^e 2* 1 - 6c2 e- 61 + \z 2x . + 4) to annihilate Substituting yp into the 8 cos 2x — 4 sin i. . The complementary yp = Ae 21 . — — cos 2i. - and 5/28. .Exercises 4.cos 2x o = — ci sin x + C2 cos x + 2 cos x — 2x sin x + — 1 sin 2x. ci = -3/7 and c2 The complementary — 8cos2x 4 sin x we difFerential A= 2. D— Using . C = -8/3. Thus 8 . function 41 = u y 41 e 1 125 is = yc 9 ^ + —x c\e x x 25 c^e~ 6x + 1 ox 125 . and 36. 106 . is i/c = c\cosx + C2sinx. 10 2 to annihilate lOe 21 we obtain SAe 21 Substituting yp into the difFerential equation = lOe 2^. and equation we obtain B = 0. and y 38.6 Thus ci = -41/125 and c2 = 41/125. y y' = cos x ci + C2 sin x + 2x cos — a. 3 3 ^ = C\ + c^e x + csxe x Using D(D . = Ax cos x + Bxsinx Using (D 2 + 1){D 2 Ccos2i + Dsin2x. = = — function —— 11 jt cos x + 8 2x cos x is = Ax + Bx 2 yc x e. Using D i to annihilate x 3 we .JiV + ^zV. and y 40. The initial y = e j/ = e 2l 21 (ci + cos2x C2sin2x) 3 + J-x + -^r + ^x 8' 16 32 2 . and find 2 yp 3 . C2 = The complementary x + e x we 3 3 o2 o 2 . Thus A= 5.12D)x + 8Dx = x Thus A = 0. = ci 8.xe + ±x e x ) x + caixt? + 2e x ) - e x - xe x LV.6 we obtain A + (2B The initial + QC)e x + 6Cxe x = y = ci y' = c2 e y" = c2 e + c2 e x xe x + 5. and — = 6e e x + 3xe x + 5x. B= -1/2. B = 3/32. -3/64. and C3 — 3. . and + c3 xe x + 5x~ \x 2 e x + \x 3 e x x 3 x + c3 (xe x + e x + 5 .2sin2x) + [ci{2cos2a: 02(2 cos2x 2ci = ci 2. and C= 1/6. + \x 2 e* + conditions imply d +c =2 2 + C3 + 5 = C2 + 2c3-l = c2 Thus 39. + C2sin2x). C .A + Bx + Cx + Dx Substituting yp into the differentia! equation we obtain 2 3 3 {8A -4B + 2C) + (85 . A s in2x) + ax + + + ^3 |^ + C3X 2 + c^e 1 .Exercises 4. 8 find yp + 2c 2 + |. 3 — + ^x + -x conditions imply Q= Thus + 2 sin 2x)] + = e 2l function = Ax 3 + - (2cos2x is Bx* yc = + Cxe ci x . Using D 2 {D - 1) to annihilate Substituting yp into the differential equation we obtain . = 2 4.SC + GD)x + (8C . 3l (ci cos 2a.3/16. y = 8 function is yc The complementary 2 -l. and D = 1/8. C2 = —6. + C4 = + C4 + 1 = 2c 3 + c4 + 2 = C4 = + c4 = is yc = 0.1 -4x = and (D + 4)(iZ) - l)x = (D + 4)(xDx - x) = (D + 4)(x-l-x) = 108 0.. and 24 c\e x + cie~ x . . is yc = c\+ C2t~ x .6 (-6A + 24B) .Dx-4x = x(0) + 4x(l) . — Using {D l)(D 2 — 2D + 5) 2 to an + Bxe x cos 2x + Cxe x sin 2x + De x cos 2x + Ee x sin 2a: The complementary function 9- + xe x x we obtain Up 42.^x 3 + c2 2 ci o 1. Exercises 4.^x 3 2 C= -1/24.\x 2 -x+ B= -1/6.l)(D + 4)i find = {xD 2 + 4xD -D-4)x = xD 2 x + 4xDx .24Bx + Ce x = x + y The initial = x e A= Thus . C2 = 1. and conditions imply Cl Thus x ]-x x . + c 2 x + c3 x'2 + ci c4 e + 2c3 x + Cie x - y' = c2 3/" = 2c 3 y'" = c4 e + eye 1 x - 1 Ze x = + 2. 2e and 24 . -1 2 /fi sinx. + xe x . Applying the operators to the function x we {xD . The complementary e*{2 + 3a. x e + xe x 6 41.cos2a:) C3 = 0. Using D(D + 1)(D 2 + I) 3 to annihilate x 2 sin x we obtain = Ax + Bxe~ x + Ccosx + D sin x + Ex cos 1 + Fisini + Gi 2 cosx + 43. and function = Axe x e~ x + yp + xe x . —2. «2 = — cosx. Thus. cos x secx we obtain sin = ua x sec x cos x sec i — — tanx = 1. auxiliary equation is m + = 0. the operators are not the same. Since y^ (x) = n > for 1. m 2 + = 0.Exercises 4. Exercises 4. n) 44. The = + «iJ/l U2i^. 1 Then ui = U2 In| cosx|. so yc — c\ cosx + C2 sinx and 1 cos W= Identifying f(x) validity of a particular x sin — sin x x = 1. = y for -5r/2 1.sin x sin x - 1. where trigonometric functions are involved. + tanx|. and + C2 sinx + cosx (sinx — In | secx + tanx|) — cosx sinx . in the following problems can take on a variety of forms. yW + a n_iy( n -V + an • • + aiy'p + a = yp + + + + a —= it. cos x tanx we obtain = — sin 2 tan a: = cos x — 1 = cos x — sec x cosx iiq Then uj = sinx — y for -tt/2 < x< hi sec = | c\ tt/2. yp especially checked by substituting 1. = cosx c\ + C2sina: + cos x In | cos x\ + x s'mx it/2.7 The particular solution. The back into the it auxiliary equation is = form can best be differential equation. so yc = ci cos x + ci sin x and 2 1 W= Identifying f(x) = cos x . x cosx — sin x. The < x< and x. auxiliary equation Then — -2 cos x — <x< sin + C2 sin x x x = 1. cos x sinx we obtain — — sin 2 x u'j_ — u2 cos x sin x. «2 y for -tt/2 The co. auxiliary equation Identifying f(x) x sin x - 1. tt/2. cos x we obtain u[ = -sini(secxtanx) = .cos — sin 2x — — x — - 1 — ~x £ cos x sin x. auxiliary equation m is + = 1 0. The „ 1 1 1 u\ = U2 = — .7 Exercises The 4. so yc = — Identifying /(x) =x— = sec x tan x tanx. | c< . x — -x cos x ci cos cos x u\ - 00. and = is 2 tanx. so yc = ci cos cos x cos 2 x x sin x — sin x in sinx In cosx| | + C2 sin x and sin — sinx = 3 n cosx|. cosx we obtain til — — sin 1 cos u'2 = 3 cos x = x cos x (l 110 — sin 2 xJ . and y for —00 < x < The — c\ cos x + C2 sin x + = ci cos x + C2 sin - sin 1 cos 2 x ro is 2 + 1 = 0. = — x sin x sin + cos x c\ cos W= Identifying f{x) — so yc x and C2 sin x = 1.tan 2 x = u'2 — cosx(secxtanx) — —— and ' - 1 sec I ci cos x + C2 sin x + x cos x — = c\ cos 1 + C3sinx + xcosx — m2 + 1 = 0. cos 2 x a.(cos 2 x — c\ cos x + C2 sin x + . ] secx tanx| + tanx| 4.7 Exercises Then u\ = 3 cos x. — ci cos x + C2 sin x + ~ cos 4 x + sin 3 x — ^ sin 4 3. so yc — cie x W= Identifying /(x) + sin x In = coshx = \(e x + e x ) we obtain + C2e~ x = and -2. 00. = -- «2 — secx.+ .sin x 1 C2 sin a. —00 < x < The Ci cos x + 2 x — sin 2 x~j + sin 2 x 2 — sm 2 x + o •J = ( cos 1-2 + . y 112 = sin x — and 5 sin x. cos x x sec^x we obtain sinx Then ui - -- 1 — — sec x COS 2 U2 = In I sec x + tanxl and y for -ir/2 7. = so ye cos x C] cos x Identifying f(x) = + sin — sin u. 1 1 + sin 2 x) for 6. auxiliary equation is m +1= 2 0. . C2 sin x = 1 and 1. auxiliary equation is rn 2 — 1 = 0. The < x < = c\ cosx + C2 sinx — = ci cosx + C2sinx — 1 + sin x In cos a: sec x | secx + tt/2. = c\ cos a: + ci sin + . le-.-e x - x - x e ) oo. — e Identifying /(re) and sinh2x we obtain „-3x 1 4 4 4 4 Then 1 T . auxiliary equation is m 2 .Exercises 4. —oo < x < The = cje = cae = c%e . 1 + c4 e x + ae x - i _. —oo < i < The oo. Then 1 -2* * + -1 x u^--e and y for 8. auxiliary equation m 2 — 1 = 0. i i _ + -ze 1 . 12 and 4 = cic* = cie z + c a «-* + ^(e 2l -e+ C2e I + !l11 ) -sinh2x o for 9. -e 1 + -x(e + -x sinh x . IV so y c = cie 2x + cieT 2* „2z „-2i 2e^ -2e -2x = 112 and .4 = 0. so yc = cie x + C2e~ x is W= - = -2. 1 x x + c2 e _. 3" 00.\xe~ 3x 3x ?xV 3 * 4 1 „„-33x (l-3x) -±xe- „-3i 3x + 4 c3 e-r 1. y for 12. auxiliary equation is m 2 + 3m + 2 = (m + l)(m -1 W= Identifying f{x) = 1/(1 + e T ) e -e" 1 + 2) = so yc 0. > x The = cie 21 + cae" 2* + 2 (e * In ^ - \x\ e~ 2x dtj >0 x . auxiliary equation is m 2 . —00 < x < The e^ + c 2 e~ 3x . auxiliary equation is m z . 0.2i =- Hi - ~ 1 Then Ul = ln(l + ex ).I )e" 1 ln(l + e 1 ) 00. + e* 1 cie -* + C2 e- = cae~* + c 2 e. = ~\ x Then u'2 - -61 1 = "24 e„Hte and -3x = —J* «2 and y = Cl 24 = tic Ci e - for 11. -00 < x < The u2 = ln(l + e 1 ) 2x - e 1 . 3x . = cie~ x + c 2 e~ 2x and -2i = _ e -s« -2e" 2a: we obtain + e1 1 . and + e~ x ln(l + = e + e1 e*) + e~ 2x ln(l + e x .^e~ 5x .e" 1 ) + e. so yc = cie 1 + C2e 2x and 4.7 Exercises — e^/x we Identifying f(x) = obtain u\ ui 1 = = —e 4x /4x.2x + (1 x . 1 r "' x I t j* ^ --j and y for 10.3m + 2 = (m . . so yc — cie 3x + C2e" 31 e W= Identifying f(x) = = Six/e^ we obtain u\ 3i e 61 |a:e 1 111 = and -6.9 = 0. Then l/4x and u'2 u2 = -ln|x|.l)(m - e x 2e 2x 2) = e = 0. = c\e = eie T + c2 e 21 - x + c2 e 2x -e x e cose x 21 sine +e x cose x -e ^sine* 1 oo. so y 2 W= e e Identifying /(x) = e 1 tan 1 1 xe 1 1 xe c — c\e x 1 + e r x we obtain ul ti 2 = ^ = e^tan c 2i : = — xtan x = 114 tan 1 x. y for 14. auxiliary equation is m 2 — 2m + 1 = (m — I) = 0. auxiliary equation is m 2 + 3m + 2 = (m + l)(m + 2) = 0. so yc = c\eT x + c 2 e~ 2x Identifying f(x) — sine 1 e~ 2x e" W= -e~ x .-3z Then in = — cose 1 . Identifying f(x) = e 3l + /(l e 1 ) we obtain 1x e.7 Exercises 4. -co < x < The = «2 e x . + c 2 e 2x + e x ln(l + e x ) + c3 e 2x + x + e x )e (1 ln(l e + e 2x ln(l + e x 2x + ex ) ) oo. t + 1 4 Then t*i = H + ex ~ 1 e ) y for 13.-3x -2e~ 2x we obtain ^ sine 1 e „-3x = -e 2x sme x _f. —oo < The a: < x u2 . = = Cl e = cie* x ln(l e = e* + ex l - e' x +e x 1 + e1 and ). cosx — sine 1 and . 1 x + c 2 xe x and and . ^ In (l «2 = = — e c\e x + C2ie x 2x = tan cie 1 -1 x.7 e Exercises 4. y 16.2i x ainx + C2e I cosx + xe x sinx + e T cosxln I cost! cje x sin x + cje 1 cos x . so yc — and e^cosx e^sinx e Identifying f(x) = e I x cos x + e 1 sin x — e x sin x + e x cos x seci we obtain I'm TP 1 — 4 u\ = x. 2 *(l + x2 1 ) + x2 1+x 2 ) ' and + c2ie T - ^e 1 In (l + x 2 ) + xe x tan 1 x is m 2 — 2m + 2 — [m — (1 + j)][m — (1 — i}} = 0. + ex 1+x 2 for + x-\n(l + x 2 1 (m — W= u\ 1 oo. auxiliary equation Then + x 2 )^ xe" we obtain u2 —00 < x < ^x tan"" 1 x so y c 0. and 2:1. c\e (e <ip*- . Then ui = U2 = tan — i x — - x tan x In x +— + 1 ( x' and y cie = cie —oo < x < for 15. auxiliary equation Then e 112 = In cos | y = 35 T = sinx)(e x secx) -e 1x 1 — — tanx. The . and 00. = The 1 1 + c2 xe x + 1 tan ^e x [(x 2 - + cgxe 1 + + |j a: l) tan" m 2 — 2m + 1 = is e = Identifying f(x) 1 e / (l + x2) x l) 2 xe x = = e = — 5 In + a: 2 J. xe -5x e -lQx = e -5* -lta e e e ~bx -S* u'2 Then u\ .llnl . 1 2 --2 x r ^-x T e x ^2„-n + x*e \nx-x 2 e i x x lnx-^x 2 e 0. > x The = cje = ct e -x X X + C2xe -ix + C2xe x „2„-a: x 1 . + x irx*e Vax l + .cf* t anc . = —-x tai + -i 112 = x In x —£ and y for 18. Then u\ 1 1 2 2.. "2 x xe Ul = x e ~ In x = 21 In a. Jin > I xi) > t and 2/ = „-5x JJ cie - 31 + c2 a:e -5x e -Si 31 /" 5f X e j dt / Jzo 116 t 01 + xe -5i [ X C / Jxa 51 .z e -5i =— —— I Jin U2 = xq dt. x e \nx = . —5.dt. < x< tt/2. auxiliary equation m 2 + 2m + 1 = is IV Identifying /(x) = e~ x In l) x e = + (m = 2 so yc 0. — -Sxe-^ + e. The 4.5* 51 „-l(te e fx we obtain e u. auxiliary equation is m 2 + 10m + W= Identifying f{x) = = (m + 5) 2 = 25 so y c = c\e~ hx + oixe" hx „-5i „-5x xe e -5c- 0. cje _T = + C2xe~ x and xe —e~ x —xe~ x + e~ x i we obtain .-3» x e = . /* e~ 5t / —=.7 Exercises for -tt/2 17. 1 sin 3a: + 3a e = x sin 3e 2x . + tan3a| — ^e x sin3a.cos3x + C2eI sin3a: — — e x cos 3a: In | sec 3a: + tan3a| 7t/6. 3x tan 3a: we obtain (e 3a. so yc = cie* cos3x and W= Identifying /(#) = \e x T cos3a ~3e x sin x e x+ e e 1 3c cos cos 3jt x smZx)^ tanZx) 1 sin 9e 2x 2 9 cos (e^cosSxlfe^tanSa:) . auxiliary equation x C2e sin 3i is 3m 2 . The > 0. \xe*t 2 + x 2 e ' = (2m — W= Identifying f(x) — e^sinSxcosSa: + C2eI sin3i + xl2 I) so y c a 2 we obtain xe x / 2 e^ 2 VT^ 1 4c 1 4ex 4 Then 3/2 U2= |\A . The 4.x2 + l sin 1 x — c\e x ^2 + C2ie x / 2 and + . -(cos3x . — 1. | + tan 3a | 1 «2 = — ^= cos 3a and y = I cie cos3x — ~ for -7r/6 20.(1 . < x< ~e x cos3xln cie^cosSx | sec 3a.sec 3x) 9 3a: 1 Then ui = ^ sin 3a: — ^ In sec 3a.6m + 30 = 3[m - + 3i)][m .3i)] = (1 0. auxiliary equation is 4m 2 — 4m + 1 \e = je^Vl — 2 = 0.Exercises x for 19. —00 < x < The C4 + C2 cosx -f C3 | sec sinx — ln| cos. = -1 < x < The c^ 2 + c 2 xe x t 2 + ±-e x'2 fl 12 ^ . a: sin tan x 1 W 3 = = cos x — sin tanx a: x = — sin 1 cos x — x sin — sin x — cos x 1 = sin — tan a: we obtain cos x 1^ /2 1 sin x cos x cos 2 x = — sin x tan x — sin x — cos x — 1 = cos x — sec x. auxiliary equation m 3 + m = m(m + 1) = 0. so yc = ci -f C2 cos x + C3 sin x and 2 is cos 3. auxiliary equation is m 3 + 4m = m (m 2 + 4) = 0. so yc = ci + C2 cos 2x + C3 sin 2x and 1 W= cos 2x sin 2x -2sin2x 2cos2x —4 cos 2x —4 sin 2a: 118 . 1 W Identifying /(x) = = tan x sin x — cos x a. | x In sec x | + tanx| 00.7 Exercises 4.x2f + \x i fl 2 j\-x* + \xex ' 2 sin" x 1 o 8 1. - cos x — sin 1. and y for 21.r| x + cos 2 x + tanx| cosxj In — sin. cosx tan x Then u\ = — In 112 = cos x U3 = sin I cosx| 1 — In! secx + tanx! and y = + C2 cos x + C3 sin x — c\ + sin 2 x — sin 1 In = for 22. so yc = ae x +C2e 2x +C3e~ x and e e Identifying f(x) | — e 31 x 2e 2x 1 21 4e -e~ x e" = 6^. = Vi = sec 2z cos 2x sin 2x -2 sin 2a: —4 cos 1% 2 cos 2a: Sin 1 = . 4 —4 cos 2a: sec 2a: Then u\ = U2 = --x - In sec | 2x + tan 2i| 4 = «3 . + <x< In ^ | sec2z + tan2i:| . cos 2# 1 — — -2 sin 2x tan 2x. 1 we obtain e 21 1 6e 21 1 2e* Se 2* e e e 31 4c 21 x e~ _I -e e" 1 e~ x x -e" 1 3 e * e" 1 1 -3e 41 ~ 6e 2 * 2e 3 * 1 6e 21 3 £ 2 2a. .7Xcos2x 4.sec 2x 4 —4 sin 2x 2x 1 2cos2z "4 — 4sin2:r sec2a.In cos2x| | and y for -tt/4 23. The = ci + cscos2x + C3Sin2a. auxiliary equation is m 3 — 2m 2 — m+2 = (m — l)(m — 2)(m+l) = 0.Exercises Identifying f(x) — we obtain sec 2x «i 4.^ sin 2a: In 4 8 o cos2a:| 7r/4. U2 - je*. 1 a. y = = for 24. x 2x 2e' x ie 2x e ' 6e 2 * 3x 6 and + c2 e 2x + c 3 e~* + c2 e" + c 3 e - \<?" + ^ + i-e 3* 24 + -e J 1 o 00.—x — ~x 4 = ci+ c%x + c$e ^ + —x 120 1 2 x - ^ 1 - 1 55 and . auxiliary equation is 2m 3 — 6m 3 = 2m 2 (m — 3) x 1 W= e = 0. so yc = ci + c2 x + C3e 31 31 3e*x 1 ge 3* Identifying }{x) w'l = x 2 /2 we = — obtain e Wi 9e^ 2 1 3e = g e 3x W.- 7 Exercises 4. = = 9^ 9e 3x 3z 1 1 6" 9e 3s Oe 31 x 2 /2 1 U3 3* 9e 31 /2 e «2 3x a. „2x e e Then u\ = -^e 2". —00 < x < The and Cl e x cie-^ = 113 jje 41 . 1 9e 3 * 9e 3 * ~ 1 2„-3x 6 IS x 2 /2 Then 1 1 4 111 = 24* "Si 1 U2 =- 1 X 3 3 T8 "3 X 54 8 81 243 and y 1 1 1 4 1 . Then ui = -e-/ 2 u2 = = c1 e (^-2 -gie*. Then ui = z 2 /8 and Thus y = 1/2 J c. x^2 = /4 we obtain x/4 and u'2 = so y r _ = c\e x^ 2 + C2&~ x ^ and -1. is 2m 2 + m — = (2m — 1 z/2 fi ±e*/ = (x + l)/2 1 i= 0.Exercises 4.7 — oo < for 25.p c l e ->* ^ + -x'e xf2 c2 e + ftf -t- -J- and 2 The initial 16 conditions imply + <=3 1 1 --c 2 - ^ 3 Thus 26. and W= Identifying f[x) =1 C2 2 e l)(m + 1) = 0. —xe^/i. -x -e"* we obtain 4 = -3«*(* + l). The x < oo. The C3 = 3/4 and C2 = auxiliary equation 1/4. auxiliary equation 4m 2 — is 1 = (2m — l)(2m + — 1) 0. Thus !/ and l/a + C3e~ 3! -a:-2 so yc = c\e x/2 + c^e x and . - = Identifying f(x) u2 = -ze 1 /4 + e it xe /4. 27. and * = C 36 + g e 9 4 122 + C 9 - 21 + C2e 41 and .\e~ x .-6e -2i -4e. ci + c2 - — = 36 2ci - 4c 2 + —= 1 0. 7 Exercises 4. conditions imply 8/3 and C2 = 1/3. y' The initial Thus ci = = oCie l/2 x -c 2 e - 1. conditions imply 25/36 and r2 = 4/9.4:c e~ x we obtain — -e„-4x — -e -3i 1 = Ui 2 * 6 3 Then Thus 1 1 -2=: -x 18 12 4 6 18 9 and y' The initial Thus cj = = 2 Cl e 21 - 4c 2 e~ ix + \e~ 2x . — cje -4r . and = Tho auxiliary equation m 2 + 2m — 8 = is e 2at 2e Identifying /(i) = 2e _21 + 4) = {m — 2)(m e 2x so yc 0. ci - c2 - 2 = 1 -ci - C2 - 1 = 0. - cie 21 + c^xe 2* and „ D 2x „4x 2e Identifying f(x) 2 2) 2xe 2x 2x + e 2x 6x^ e 2x we obtain — u'j = 6x 2 . . The auxiliary equation ism 2 — 4m + 4 = (m — B 2x W— (l2x 2 = so yc 0.Exercises 28.) 2 so u\ = — -(lna. 4 -z 3 ) -2.i3 - 4 y — y yi —x and yi x 2 e * l) .2xe 2x + (i (z 2z + . . FVom .x3 and y' The initial = 2c ie 2 * + c2 + e 2*) + e 2 * (2xe 2x 2c\ ci = 1 and C2 = + 2e 2x - 3x 2 ) identify f(x) — + c2 = 1 = 0.Zx 4 ) ^de^ + axe^ + e 2 * (x 4 e 2x + (4x 3 - 3x 2 ) xe 2x . Thus y = Cl e 21 + tW + (2x 3 .) 3 and v. (x 4 . Then 4 2 .2 = — Ini x so U2 = 2(lni) 2 . and = e = e 2x 21 29. Write the equation in the form and 3 conditions imply ci Thus (4x (4hix)/x. + -*y = . 4.12x 3 u'2 = 12i ui = 2x 3 - 3x u2 = 4x 3 - 3x 2 - 6x.Inx x x — we compute xhix x 1 xlna: 1 + lnx Now u\ — (lna.x3) . . 1 — x4 u2 = so x5 1 T . 4x 4 Thus Vp ~ 4x 4 3x 3 ~ 12x and y = ye + yP = cix 2 + c2x3 + 31. From = ?/] a: 2 and yi = 2 3 x x a. U2 = sin x.7 Exercises 4. + x ^2 . x 3 /x 3 = "1 80 and x 2 /x 3 u2 — .6 . 3 X° we compute = 3x 2 2x Now I1 X -— = -? 3x 4 4 = x4 - 2x Ul = 3?' . ^cosx and l ^sinx we x cos x — jx x 3' 2 2 ' Now uj = sin i so ti^ = cos x so and Thus j/ = c\x = cix ^ 1 '/2 cos a: + C2i ^sinx + x cosx + C2X 1|/2 sinx 124 l /2 + x ^2 2 cos x . Thus yp = -^x(\ax) 3 + 2x(laxf = §x(lnz) 3 and — y c\x 30.zv + jf = 3 1 y and identify /(x) = 1/z 3 .w) = — x — yi 1 /2 i 1 /a x sinr l y2 — cosx x ^sinx ^ cosx — ^x 3 tii = cosx. Write the equation in the form and identify /(x) = Prom x W{y\.4. Write the equation in the form + C2X In i -x(lnx) 3 4- . . 7 32. = y\ cos(ln x) cosflnz) W= and 1/2 = sin(lni) we compute sin(lni) _1 sinfins) cos(lna. A= so that . secllnz} 1 . so «j = m jco5(lna. = ix 2 — I61 + 21. = B= cie -1 —16. (a) Write the equation in the form „ and identify f{x) = secflnxj/x 1 2 From . so U2 = inx.) x Now tan(lnx) . Substituting into the 4. + (In x) sin(ln 2) and — — %L x Thus. are equation we find 4A + (b) (lni)sin(lnar} . a particular solution yp (b) The general solution = y for is — cos(ln x) In (a) We ci cos(lna:) — n/2 < \nx < n/2 have yc differential = c\e~ \ + C2sin(lna. . and C= B= particular solutionis yp +C2xe _r and — = A 21.) + cos(lnx) In |cos(lni)| + or e -7r / 2 < x < x + C2xe~ x e"^ 2 . . is sec(ln x) in the differential equation 33. U2 = ie -- x e x a -2x I I e -2x e~ e" /x = jx = -1 1 " a. The bounds on and we assume yp A= have y c due to the presence of — Ax 2 + Bx + C. * xe -1 — ie _T e Identifying f{x) A + 2B + C = -3 2A We Ina. cos(ln x) | * + e _T x e~ /x we obtain u.e Exercises 4.jj. and C = — ^ Thus yPl —2x — jQe 3x Next we use variation of parameters to find a particular solution of y" + y = cot x. = -xe~ x + xe~ x In x. . is results of (a) we take yp = xe~ x kix. Assuming ypi — Ax + B + Ce 31 and substituting into the differential equation we find A = 0. We have yc = c\ cos x + C2 sin x. = 2 4x - 16a: + 21 + xe~ x lnx. — e 31 + cot x is . cos x W= Identifying f(x) = cot a. — sin x sin x = 1. — cot x|. We use undetermined coefficients to find a particular solution of y" + y = 2x — e 3*. cos x we obtain ui sin x cot x = — cos = - "2 = 2 cos X cos x cot X sin a: ~ cscx — smx. + sin x In | cscx — cot x\ + sin x cost = 2 sin x cos x + sin x In Using the superposition principle we have that a particular solution of y" yp = ypl +yP2 = 2x — — e + 2 sin x cos x + sin x In 126 | cscx +y = — | cscx 2x cotx|. U2 = lnx and yp Since (c) -xe -1 Adding the a solution of the homogeneous differential equation. We have . a: Then ui = sin U2 = In esc x I a: — cot x\ + cos x and j/pj = sin x cos a. and (b) we have yp 4. u\ = —x. B = 0.7 Exercises Then 4. (-oo.4D + 5) Identifying P(x) — we 2) 2 have = cos 2x .Chapter 4 Review Exercises Chapter 4 Review Exercises = 1. y2 15. true 51 .oo). y 2. x 9. Section 4. False. = The statement would be independent on an interval 2 -Sx + (-2) if dependent. 11. Section 4. From m 2 — 2m — 2 = 14.dx I 12. True 7. 5.0) 6.2/x we y2 13.2 In x — — dx = e* we obtain m 3 + 10m 2 + 25m = cos 2x sin 2x. Problem (l neither -x 4. A + Bxe x 10.1. 45. J y From 2x — cos' 2x J ± 1 e x and 2 j x dx = Tj^e 1 - \/3 so that m = -1/2 ± \/5/2 so that -W ' we obtain y = I ci m— ci ^ cos -^-x ^ + 0. consider Problem multiple of 0. if it a constant multiple of the other. From 2m 2 + 2m + 3 = = sec I 2 2x dx J ^ tan 2xj have / m= we obtain = e P dx = -2x . see 8. Identifying P(x) — cos = ^ = -2. (0. = = and fzix) 2 ) + linearly "Two read dependent even though x + x2) = 2 (D - 2 [p .r—. see 3. False." is (2 These are x. m= C2e" 5* 127 ^ + C2sm -^-a: —5. — —5 so that . False. and to + c3 ie~ is not a functions fi(x) and fo{x) are linearly 31.1.oo) or (-oo. iV. Then 23 + C5X + CfjX° C4I y= and = A + Bx + Cx + Dx 1 i/ p solution coefficients gives A = is 9 Substituting yp into the differential equation yields . Applying (D - 3 l) to the differential equation we obtain (D - 1) 3 Then y = cie 1 c^xe? 4- + 2 I c3 x e + 3 c4 x e x + cs^e 1 yc and j/ p = Ax 2 e x + Bx 3 x + Cx e A x e . 1/2. y D cie~^ /2 + 2m 2 + 6m . 35 + 2C) + (55 .3D vTT ^11 \ ^ —^ x + c 2 sin — — 1 + . and c\e x C— + c^e 1 + to the differential equation D{D 2 + l)(D 3 1/12. . From 3m 3 + 10m 2 + 15m + 4 = = y 18.5D 2 + 6D) = D 2 (D 2 + 1){D . The general ^^e 1 solution - we obtain .— Chapter 4 Review Exercises 16. Substituting yp into the differential equation yields \2Cx 2 e x + GBxe 1 + 2Ae x = Equating coefficients gives A— 0. 4 x ^ + e~ 3x/2 = e m= '. and _ j__ _ v^l x 222 \ 46 + 36 _ x D= 2 + (D . ± -3/2 -\/7/2 so that c3 sin m- -2. Applying D(D 2 + 1) = 0. x + C2Sm C= 46/125. ci cos J + C3 + 5) = 0.2D + x + 4/5.4 = From 2m 4 + 3m 3 19.9D)x 2 + 5Dx 3 = -2x + 4x 3 (bA Equating 3 = e a*/a I _ vTT /" ClCOS B= -222/625. 36/25. B— y 21. Applying = m= ( 3x/2 ' I D we obtain i (D 2 .6C + 6D)x + {5C . The 4 3 1) = (D - general ^ 20. and + c^xe~ x c2 e~* ^c2 cos we obtain to the differential equation y + m= we obtain Cl e~ —1. . l) 5 = 0. and m= ±-/2i so that = c\e^ 2 + C2e~ 2x + C3 cos \/2 x + q sin v^x. -1/3 and m- ^-x + m- m = —5/2 so that — 1. 2m 3 + 9m 2 + 12m + 5 = From we obtain y 17.3) = Then 1/ = ci + c2 e + c3 e is + c 4 x + c 5 cos x + eg sin x to 128 0.2){D . . Applying Substituting yp into the differential equation yields = B= 4/3. . Setting cosx. y gives -1/5.Chapter 4 Review Exercises and yp = Ax + B cosx + C sinx. 6. . and + 8 cse x m— and 1) Assuming j/p = .c cosx)(e taEx) _ .4x + 5 + Csinx and substituting S = 0.-e [m — e e Identifying f(x) — e x 1 x cosx sin x + x e x sinx. — and m? — 2m + 2 = is = —1 we j/(7r) auxiliary equation + C2X + c\ is obtain m2 — 1 = so that - = c\ e " —2A = equation yields differential . .D = D 3 (D . 1 cos x — e x sin x + tanx we obtain . is .4 + (5B + 5C) cos x + (-5B + 5C) sinx = A= Equating coefficients gives y 22. c\ + 1r c2 e = 24.\ sinx and = = y Thus. 1 - e sin solution of the initial-value problem -x - and W= „ 3. cje1 sin x + cze? cosx . The 1 + . so that 1/ e^fci — m e I-7r cosx + C2 sin x). 5 . so yc = = -e 2x cos x .cosx is Then + C2X + cie x + cix 2 ci = Ax 1 Substituting yp into the A = —3. C= 2 sinx. 2 and yp 23.-= c2 = — | The C2 e 2 0. The general solution is = 1 - solution } y The 4 It + c%e^ + The general 1/5. ±i and 1 = C2 (m — l)(m + — 3x 2 6. The auxiliary equation is 13 — m 2 — 2m + 2 = 3 x . . y Setting j/(0) = 2 and y'(0) = 3 = 1 -x--smx. auxiliary equation j/(tt/2) .1) = 0. +C2e c\e z we obtain + c2 = ci - ci Solving this system we find C\ = -^ and y= 25. and C = — \ Thus j/p = -x . — (1 — i)] — (1 (e I .sinx. = ±1 and 3/ into the differential equation = c\e x we find + C2e -X A = — 1. 0. D to the differentia! equation we obtain D(D 3 . Equating coefficients . we obtain + C2 — 4 = — 4 + 3c 2 + c 3 = a ~ -§ y= 28. The auxiliary equation cie x er particular solution is ex + e -i tan"" = -4. e = x . so yc = cos x c\ cos x — sin x 130 + C2 sin x and sin x cos x - 1. The auxihary equation is e 1 - m2 + 1 = 0. The x cie sini c\e x — am x — U2 s\nx c + C2ef In cosx e x and .-e x + + e 2x + c 2 e~ x + e 1 tan" y Setting y(0) „3x 3sc e Vc A -2. + e2x 1 2m 3 . ci Solving this system 1 and the general solution and y"(0) 0. y = = 26. \e *' e2 + e" 2 tan -1 1 = (2m - + c2 e 3!C + l)(m ca^e - 5 3) 31 - is 4. is + tan x\ — e x sin x cos x + e x sinxcosx — e'cosx In + C2e T cosa.Chapter 4 Review Exercises Then uj = — cos x. = so y c c\e + C2t~ x x W= Identifying f(x) 2e x /(e = x I +e ) = — Then Ui — tan -1 e 1 -e2 + - 112 . Thus *n-L 3x + xe 3x -4. = j/ 27. . yp j/'(0) 1 e i x = -4 = = = l/2 cie c\e x/i = find c\ = | . so that + tani| . — auxiliary equation x sec | .13m a + 24m - is and e 1 4 secx we obtain + ex | + tan:c|. and C3 — 1. 9 1 + c 2 e 3x + c3xe Sx - § ici we + and . cosx\n secx \ m 2 — 1 = 0. yp The initial = — C3 sin x + C2 cos x + 1 .sec x.sec x tan x. . Li and . C2 sin # 2 x sec 2 x 2 Thus y = c\ cos £ + = ci cos x + C2 sm x — -1 cos x sec > i + smi tan x 1 sec x 2 = + 1 — cos 2 x cosx 1 C3 cos x + C2 sin i + — sec i.cos x A + -sinz + £. conditions imply c3 + = C2= Thus c3 = c2 = l l 1 2- 1/2 and y = . u. z. 11—x— = — 1 2 cos U2 = = tan x. — Then ui = cob x sec x = sec i.— Chapter 4 Review Exercises = Identifying f(x) sec 3 i we obtain smx = — sin x sec i x = . . . 107 + it ft* 2. A Since sin x(t) 4. A upward released from a point 3 feet above the equilibrium is The velocity of 2 ft/s. a second quadrant angle and tan _1 (-2) + 7T = -1. A weight of 4 lb (1/8 slug).034). attached to a spring.-2. A = ^4+1 = 2V2 and = tan^> —= -1. weight of 2 lb (1/16 slug).(0) = ci = c\ cos \/2' + C2Sin -J2t and gives = -l and x'(0) 132 = \/2c2 = -2-/2. . and 2.034. c2 = = is c\ is 3 lb/ft.5 Applications of Second-Order Differential Equations: Vibrational Models Exercises 1. cos At + c 2 sin At and 1 = x'(0) and sin(4( initial Thus = — 4ci sin At + 4c2 cos At gives x'(t) 4c2 . 5. conditions to x(t) -\/2cisini/2t + V2c2 cosv^t a. = <j> < > and cos 2\/2 sin(5t- Applying the initial 0. > z(t) «s and cos$ < ^ Applying the 3/(() = — — ?r/4.-2. z(0) Then a fourth quadrant angle and 0. <t> is ci = = c2 1. 3. position with an initial 2.7 feet below the 4 lb/ft. conditions to x(() = = ci c\ = Thus 5. The equilibrium position. cos5f = -2 ci is spring constant + c2 sin5( and x'{0) and - — — 5cj sin 5t + 5c2 cos bt x'{t) = 5c2 gives 10. attached to a spring. + 2. <j> 4> = is ^ and tan <p = —^- . -1/2.1 Applying the initial spring constant conditions to x(t) = x(0) Then = c\ -2. and A = Jl + 1/4 = Since sin^ 4> jr/4). released at rest from a point 0. x(0) = A= \/16 ci cost + — — c\ sini + C2Cosf C2sini and ^(t) = -4 and x'(0) = c2 = gives 3. Thus 0. and A= < Since sin Thusi(t) 6. and 2\/5 = tan ^ = 2.471).107). 0.Exercises Then ci = = —1. = z'(0) 10e 2 = 1. A = Vl + Since sin0 = ci so that the period tt/4 = Prom mx" + 120x = we obtain ns/mft. c\ = 4. = quadrant angle and first tan -1 2 ss 1. tan~ 1 ^ + = tt + tt 0. VIM - 01 is and cos0 > < 5 shift From mx" initial and l and 1/10. and 0. = quadrant angle and tan -1 10 « 1. . = = ci cos 8t + C2 sin 8f and x'(t) c\ and 4 = i'(0) 8c 2 = = — 8ci sin 8t + 8C2 cos 8( gives 16. C2 = 2.605. a is 0.605).927). + - a first conditions to x(t) = c\ + a — 3. t 4 + C2 sin —. = z(f) 8.= f 1/4 slug.107. 2. Applying the initial = conditions to x(t) cj + C2 sinlOt cos lOi = — 10c sin lOt + IOC2COS lOi and gives = x(0) Then ci = 01 1. +4= y/l = conditions to x(t) = ^. C2 —2. C2sin y^30/mt so that the period . 16x Thus 1. and = tan y~ = 10. 7.Owe obtain x 10. and and cos0 > > —^ = = tan + 3. +9= -4 5 tan0=— and = a fourth quadrant angle and is ~~ tan -1 (— g) —0. and the weight ci cos 2^/30/m ( + is 8 lb. = and cos0 > > -4. = 4 ci m= = cos —7= . < and cos v/5sin(\/2i =s Applying the initial is 0.464 =s 3. « i/ToTsintlOf Applying the Then c\ — Since sin x(t) 9.471. Thus x(t) ss 2\/5sin(8t+ 1.927. A = V16 + 4 = Since sin0 and 1/5 a third quadrant angle and z(0) Then 5. 2 i(9tt/32) = v^/4. 18.588). The 20 kg mass has 20 kg: i'(tt/4) -5sin2i = x(0) (b) - = x(0) (a) (c) If 0. - obtain y/E/2 x m/s. and n= . x' If x = AcosSt = then i = (2n + 1)tt/16 for n = 0. . 1. and and x'(0) x'(0) = -10 sin ( + 16x = 0. -by/2 m/s. m.6). ar'(0) = x'(0) 2 = = we obtain x we obtain x = = -jcos4\/6f- sin4v/ 6*- we obtain x = £cos8t. and -10cos2f. masses at the equilibrium position. approximately 97. the 20 kg mass vibrations. is . 3. x{0) x' = -lOcosf. -\-12x 72a: and lb. From fx" + 15. x(0) = -1/4. it will make 8 complete . -5sin2i and x' = x(tt/12) 16.3 vibrations/second.1 Exercises 5. is feet. = -1/4. and = 0. m=30/ir 2 C2sin2( so that the frequency 0. x(tt/6) = -1/4. 10 m/s placing both is odd. 1. nir for = 1/2.. From 50x" + 200x = 17. 0. x'(jt/2) moving upward. x'(0) and is + 0. x(0) = 1/2. 19. In 4?r seconds x!> sin(2i 134 0. (a) and the weight slugs. 0. = = -10 sin t = 50 kg: x'(n/4) so that The 50 kg mass even and downward when n -1. x(n/8) = -1/2. x(0) - 0. From \x" 14.^sin4i = —^cos(4f - x The period = -10 we then 2 sin ((cos (-1) moving upward when n From x" = -10 we the larger amplitude. — — 4sin8i so that x'(3ir/l6) ~ 4 ft/s directed downward. . From 20a:" + 20a: = 0. + 40x = 0. From fx" (b) (c) = 2/2iv is = 13. 2. and x'(Q) .-2 we ( = w/2 seconds and the amplitude is = From \x" +x= 0. i(jt/8) = 1/2. x(0) x'(tt/2) = is is obtain 2 obtain x = m/s. 2 we first Let ki and weight. If x' = f cos(10t .927). 2. 1..814 ft/a 1(3) (f) ( g) (h) = a/'(3) 59.6a:" + 40a. so From these equations we w= + = 1- denote the mass in slugs of the Ak\k2/{k\ effective spring constant k 32m = &i(^ so = sin 3t for . the numerical value of the W — mg — 32m. = 4(40)(120)/160 Using x(0) = and x'{0) = = 120. From 1. 1. 2.721 s.927) = then 10* . .0. If x — and the weight is moving downward for the second time. Let — first = m = ^ k*i = 120 k/m = sin8\/3 we compute the 120(32)/20 = 2 —7= sin / y3 \ -7tt/18 + 37 . 2. we have from or k = 225. . x' = If x' = 3 then t fci Now. . 2. . is w = k/m 2 2-k/w — ir/15. and x'(0) ^ sin lOt ft + cos5t + 0. 1.927). 1. n = 0.702 ft/s 2 = thent = ^(0. 1. 23. 1 and 2. . (d) 5..927 + wr) for n = 0. — then lOf (2n + 0.Exercises 20.. weight is + 192 and x" 192x 0. — 4tt 3 + 2n7r/3 and t = -?r/2 + 2h7t/3 n= for 1.0. = -0. 5 we obtain 0. be the spring constants and fo) the effective spring constant of the system. = 0. . (j) If x = 5/12 thent = ^(tt/6 + 0..927 = or 2tt + 0.927 + 27iw) for n = 0. .927 = tt/2 + tot or t = = ji 0. 3 From k V3 — — — cos 3f + 47r/3). 30.597 ft X '{Z) = -5. and x'(Q) = ^- sin"(5t + I + 0. .927 (e) for 71 = 0. and z'(0) = —/3 we obtain (i) If x t 22. + 2n?r) 0. 2. and the period 2-k/IO is = n/5 cycles. = x(0) = x If 21. m = = 40 and 20/32 so obtain x{t) 24.927 + 2jijt) and i = ^(5tt/6 + 0. From x" + 9x = 0. Since the find 2fci mas = 3/t2- and The given period of an 8-pound weight 30 2 = 32m = k -71/4 = is Ak fc 2 ^ of the combined system 1/4 slug.927 + 5 (f = -1/3. = -f cos lOt + (a) x (b) The amplitude (c) 37r = 7rt/5 is = i(0) 5/6 and k — 15 and 2jitt) = §sin(10t 5/4 we obtain \ sin5t - = t = -2/3. x(0) = -1. .0927 l)ir/20 for . Now. . t = 0. 2V3cos(3t .. x = 5/24 then From 2x" * = + 200x = 0.927 + 2rwr) (k) If x = 5/12 and x' < then = ^(5tt/6 + 0.. and x'(t) = ±f ft/s. we obtain vq w sm uit .A.^sin5f] = 28. x'(0) — weight 46. 5 1125/8. = —c^/A 27. .4cos(uf 2\/2cos(5( + 5tt/4). or f = (n/2 — <j>+2nn)^ Thus. hi = 375/4 and = fc] Prom x = c\ coswt + C2smut.. the time interval between successive 2ir/w.. Thus a'(t) = when cos(wt + 0) = or am(ut + maximum acceleration occurs when = Thus. i = A cos(w( + = xo cos tot H (wq/w)^ first o wo = 22 «^ and xq. = — ui 2 Asin(ut + 4>).4>) we compute the acceleration a(t) = x"(t) 3 <j>) . the displacement is ±. Let /I amphtude = +4 is x(Q) A= = + yjxg cos0 = c\/A and sin0 = 2V2~[^cos5t. lZfcf = k2 first then minimum and maximum . x' = Auj cos{ut+<f>) = 0. 136 maxima . = is. A + uji velocities occur when x" = —Auft cos(wt+#) 0.4sin(wf 4. tiating we find a'(t) — — j4cos(wt + 0). Using T 2tt/w orw= 2jt/jT we see that the magnitude of maximum acceleration is w 2 A = (2it/T) 2 A = 4tt 2 A/T2 From x(t) . If x is zero. If x and is = ( Asin(uif-l-i^) the = (—n/2 extremes for x occur when — 4> + 2mr) for n = 0. Finally.2. . we obtain so that I If _ equation. A i = .Exercises We now 5. 26. Differen= ±1. 1. 30. that 29. .88 is lb. <j>) then the when i = =A ci -c 2 — coswf — sin . have the system of equations + ki fcl 2ki = 225 = 3fc 2 . Solving the second equation for ki and substituting in the 4(3fc2 /2)ft2 + 3fe/2 Thus. the value of the x so that the 12fca = hk2 ^ = 32m= ^ = 25. x(Q) = 1/2.. 153 second. + 4.232 meter. x = 1 - 9. resisting force numerically equal to the instantaneous velocity.e" 4' and 8 1/4 second.25 i(0) 0. and i'(O) = we obtain x = \e~ 2t - \e~ st . is The weight (b) heading upward (a) = from the starts 1 ft/s. equilibrium position with a damped with a ft/s.lx" + 0. and x'(0) 1 = =stn4( = 2tt. (a) From \x" is + 1&e . and a/(0) then f = 5 ^ we obtain x = 5te~ 2. and x'(0) then f = = we obtain x = 4te~ tt .2 1. the extreme displacement is x(0) = 1 meter.0. 16te" 4 '. x(0) = 1. = 0. * x' = y~ st (5 . (a) 5e~ 2v/3i (l 5y/2e~ l /4 FVom x" + e x ( (a) damped with a x(Q) x = = heading downward -1. x(0) = = If x' 2\/2(). + x' + 2x = 0. (h) (b) x 11. first time heading upward is f = 1. e .384 second and the extreme displacement is x = —0. (a) below (b) from rest 5. *. A 16-lbweight is downward velocity of above 4. From \x" .. (a) \e~ st - I0x' From i" + = is starts 2 feet above the feet. 12. resisting force numerically equal to 2 times the instantaneous velocity.Exercises 5. 3tt. A 2-lbweight attached to a spring whose constant is is The system 1 lb/ft.2i = s i n (4t so that the we obtain = e _2t lo) j-cos4i - = then £sin4(]. = x If -2 From \x" + -fix' + 2x x 10. and x'{0) = -12 then x = -fe" 2 + fe" 8 m (ie — l) is never zero.2 Exercises 5. 3. (a) below (b) heading upward 6.25).5 2. (a) above (b) 7. if x = |e" 81 [e 6t = g In 10 = 0. ^ . equailibrium position with an upward velocity of 1. and x'{0) = 1 we obtain x = e^ 21 Qcos4f + £sin4(). . i(0) = 1. x' The weight The system 2-lb/ft. 1 From O. 0. = -1. 8e" 4t - displacement 8. If x' = then t = 1/2 second and the extreme feet. attached to a spring whose constant is is + x' + 5x = 0. 4tt.4r' + 2x = (b) « = 2 e-»^ — —7=cos4t -/5 (c) x = If then 4t + 4.294 seconds. = V^/i t and second and the extreme displacement + 16x = 0.2e 6 ') = when * = $ In § « 0. J). m= —^±. see that the roots of the auxiliary equation are 2tt h^JlQO-0 2 so that 138 = 6.5 -.75x" > 5/2. . l . x(0) 2) so that + c 2 sinh-09 2 -18t c2 = x--^e. x 16..(-14-Tv^)( .3t + If 18* = -3/03 2 and x'{0) = 2 we and = -- v we obtain (t. 14 -2/3. and x'(O) v > 18. > and 3i/2 we find that the roots of the auxiliary equation are m =-f ±|03* -18 and cj If 15. then = 5/2. ~ 18. From x" + 6x' + 9 = 0. x(0) = 0. x = then t = 2/3(u - = = _^ (-H+7V2)t p 6 0..Exercises 5.2 -*^ (b) x = (c) x= If e then 4t+7r/4 n= for ^ + co B 4t = = ^""»in (« + gin*) it. so that the times heading downward are ( = (7+8*1)7^ 0. cosh -<J& 2 = x(0) 0. -2)(e- 3t . From 40i" + 1120x' + 3920x = T X 17.\\J 100 — . then < p< then + fix + 6x = 1 = — ^0±^402 — 5/2. and x'{0) 14 From x"+0x'+25x = we The = quasi-period is 7r/2 obtain = 2 we obtain + ^. a - 25 > If 40 25 = 2 If 4/3 25 < If 4/3 2 From 0.5 -1 13. 1. 2. 2 ft/s. 2jt. 3tt. x(0) = and Prom 40x" x'(0) = -2 then + 560x' + 3920x = Cl 0. Prom -^x"+0x'+5x = Owe find that the roots of the auxiliary equation are m (a) (b) (c) 14. -e " sin 7(. . From equation (17) (2n + l)7r/2 i/w which |ir.Xt sm interval \ju 2 - A2 f + 7] - A2 .2 underdamped motion 19.A2 (15) touches the graphs of - y — ±Ae~ xt tt </> 22. if x = then * = |tt.X2 1 = Ae~ u sin j^w 2 . For x for = e- M d cos \ju A some constants and (? - 2 A2 B and some constants the quasi-period of 20. . is ^?r. -A 2 (a) If S > (b) If x ~ -^e" 21 sin (4t + tt/4) is t and A2 f £ + -I- ^ is 2-k/^/lj 2 — A2 2ir/Vw 2 — A 2 that . is slow. = Ae~ Xi sin (Vw 2 — the ratio of the values at e 24. between successive intercepts vV-A The time fl] The difference in times between two successive maxima is 2wf \/u/2 (w+l)7r-^ 21. + = Be..Exercises 5. . The time interval 7. x'. .X2 ~ for which + 1)tt/2 (2n y/ui 2 ..X 2 1 + c2 sin \jio 2 and x' for t 2 between successive values of __ (2n — nir ~~ + 3)w/2 - yfui ( 2 7T <p . . and if x' is then ( = v'w 2 x of 2tt. The is . so that the ratio of consecutive is -A((+27r/v^^A !I) very small then x n is slightly larger then 5 = than x n +2 and the rate of damping 27rA/\/u> 2 - A2 = 4tt/4 = tt. extrema 23. quasi-period of x maxima = 2 The period half of the quasi-period. tt. so that the time interval between intercepts and tt/4. - nir 4> <f> -A = 2 _ tv/2 ~~ Vw 2 -A 2 e~' sin(( + tt/4) ' is 2?r. 37t.. x(0) 2 + + 4 -' . so that the equation of motion I =ie-'" x'(0) xp + 8x' + lfiz x As 5. . 6. t oo.Exercises 5. Since c2 *e + = cie _4i + c 2 te _4t and 7sin4().2. -!I we obtain x c = cicos4t + C2sin4t and cos4t-2e i. the amplitude approaches ^85/4 as the external force is F(t) — kh(t) so that 140 mi" + /3x' + kx — kh(t). = X = transient 3. ai'(0) - = we obtain x c -^-e-'(24cos4( o25 0. x(0) and 0. i = sin(4t By Hooke's law = 2t + . = so that the equation of motion / . f — > = ^e" 625 cc the displacement x From 2x" Xp = + 32x = — > 4t (24 5e" *cos4( — 2e~ sin4t so that x = --cos4t + 1 7.219) - and lOOt) te- 4t -4t and -icos4t. and 9 -sin4t 4 2. + T = 5 then x c ) -1. ' ! 'sm4(. 0.3 1.g5g e"' sin4t so that From x" we obtain x c = Cie~ At is 4 4. From x" + xp + 8x' = — jCOs4( I6x = = 8sin4t. and x'(0) e ^2 4 (a) x" + 2x' + 5x If = and zp = + C2 sin -^-f1 1 1 64 " * 2 x = 1 3/17 = 12cos2t + 3sin2i. .0. and x'(0) 3sin2t so that the equation of motion + Sm (C ° S = e _l {c! cos 2i+c 2 sin2i) is + 3 sin 2t.-^e~'cos4( . x{0) = xc — ™ (cos 3i + sin 3f) and x p -2. z(0) is \ SU1 2 10. l e~ cos 2t (c) steady-state <*») then \/47 ci cos I ^/if ~3 ° 0S [ 2.897).3 Exercises 5.e sin 4t. t- ^e _2t sin(4( . 68e~ 2! cos4t. If + Ix" \x' + 6x = 10cos3i. i(0) = 0. = x'(0) + 1 -e = >. 3 8. and x{ir/3 + mr/2) - -1/3/2 cm. (a) If g'(7) test (b) = then 7 (y 2 + 2A 2 . at the = + 32 — -sint. If x" + 1\x' + u^x = J*b sin xc 7* describes = e -At underdamped motion then ci cos \/w 2 - A2 t + ~ c2 sin \Jui 2 \2 t and F 2 4A 7 If sin^ cos# = = cj/^/c2 (w 2 +4 — 72 ) . Othent = w/3+nn/2 and many t = other values of n/G+nir/2 t for which 0. ) = -2A7 / v/(w 2 .V3/2 cm. z'(0) 13 . COS ft.2A 2 ) = = y/u 2 2 . 13 1600a: = 1600 sin St.7 2 2 + 4A 2 72 ) then the equation of motion . = x'(0) £ ra = 0..sin2I +Tr:cosf + 13 (a) we sin£ so that x 9. 2.Owe = £sm4t(2-2cos4i) 2 = = obtain x c c\ cos At + ci sin 4( then 1 - . (d) x(tt/6 + ror/2) . C2sin2t) x(0) = . A2 and . (e) 10. 1. /\j (w cos^ 2 (w 2 - 2 . = e 56 -2t( a [-— + From lOOx" = and xp .72 ) 2 -2A7F0 -7 + (w c2 /^ + — 7 2 ) 2 + 4A 2 7 2 2 2 4A 7 ) c 2 .A 2 ) 2 + 4A 2 72 or 7 = — 2A 2 F /2Xy/u 2 + 0)- Vw 2 . + Prom \x" - and x p 2x' + + jfcoat = ix f| 20cosf.A 56 „ cos2(-—.. 1. and 0. . x(0) = and 0. Note: There are = |cos4t-f cosgi = n = 0. ..Exercises 5. sin0 2 + (w 2 -7 2 . .. 13 (b) x1 e~ 2( (cicos2t -jsinSf so that x x= = obtain x c 72 . . first derivative .sin4£ - 3 o = htt/4 for sin St. 2. .2A 2 The .w 2 ) = shows that g has a maximum value The maximum value of g is 9 so that 7 at 7 = (V^ 3 . = If (c) Ifz' §(l-cos4t)(l+2coB4t) for extreme values. is F sin(7* \j{u 11. so that ) From x" xp < for various values of 0. sin 7* —Jo* — Jo = ^— tsinwt. The system 3 + 2x' + 6x = 40sin2i = Vu 2 . = \t sin 2t + |( cos 2t. and x= 2 cos 3f — + 5 sin 3( — 18 19. 142 u = 5 -t cos 3f 6 + v) = 3(7 — — sin u sin v we obtain = 5(7 + the result cosucosw w)t and w . (a) Prom x" xp = + uj 2 x = shown is = = J^cost*.u y . . x{0) = = ^ we obtain ic = .80/2.3 12. and x'(0) x(0) = — cos2t — = jt/4). this 1-J% then the roots of the auxiliary equation are the resonance curve and the family of graphs If < where we require that 5(7) 13. 2w = x'{0) we obtain x c = and _lim = — 1. = From x" + w 2 x ip 17.2-^12-4 = uj — < or 0. < we obtain x c = c\ cos3f. xp = + C2Sin2f. = 7 (b) yjZ When - Fa 2 = /2 3. -|(cos3(. Exercises 5. F 1 . z(0) = x'(0) (Fot/2ui) sinuii + 4x = — 5sin2t + 3cos2t. and cos(u Letting follows. 2V(cos7i — — - 7 2 + 5sin2t = 0. and From x" + 9x = g amplitude of xp 5\/2. when given by is -5cos2t x Jb . (a) From cos(u — sin w sin w = v) = j[cos(u cos u cos v — v) — + sin cos(u + sin v v)]. = cicos2t + C2sin3t.f and (sinwf. cos 7i. and 2 = hm focoswt. </6. \x" 7 15.sin2t and x'(0) = — = = — 2u f sinwt 1 we obtain i c 3 5 4 -tcos2f. 4 + -tsm2f + 8 5sin3t. x(0) 16. —27 and 0.0/2 > 3 in resonance is then 2 and g(f) (Focos7()/ (y 2 00 f-^w 'is 2 underdamped motion. ci coswi + (vjsinwi ci cosu* + and 2.2X2 = is = xp 14. . the cos iOt — 72 ' (Jot/2w) sin wt so that x 5v/2sin(2i - F = — coswi) x 18. C2Sinu. (a) x" If + 0X1 + 3x = and m = \ (~0± \J0 2 — 12^. damped.1071) time when 40* + 0. Solving the differential equation q" conditions g(0) initial = = <j'(0) g(t) = initial q(t) 2 3. Exercises 5. Solving the differential equation bq" The 4- imply is as (ci cos 40i + c 2 sin40t). underdamped. Since 4. — + c\ cos At C2sin4t 4.4 1. 6 Exercises 5. = 60 we obtain = At the circuit the circuit -201 e ij'(O) — = ^sm2V5t + — 4L/C = —20 < q(t) t as = conditions q(0) 16? —.cos 2. cj cos 2t/5t + ci sin2v^t + 1/5.5676 coulombs. 2q' + 5 and ^(0) = ^q" + and q(O. ' we obtain 5 x — — ttt(cos It — cos 5f) 12 sin6( sin t — cos6t cos ( — sin 6t sinil = 12 5 — sin 6t sin t. V5/50. The initial conditions Thus ^25 + 25/4 zero for the e~ 20t first sin(40i + 1. 0.. The Thus.4 (b) 5(7 lim (c) v t-0 2ef ' 20. ^5cos40t+ ^sin40t) 4. 27 27 ' = 10cos7i.4636 = jt or .15/4.0509 second. q(t) 5 and c 2 c\ \t is critically = = The charge e _20t 5/2. and = x'(0) = — — [cos6( cosf — 5 — — coset am ft (-. Solving ff g(0) = — L/C = 0. = If e — w) sin et sin ft w =s so that = hm x — (Fo/2e7)sineisin7f. Thus = ^cos2v5t + \. x(0) 0. 100g + —15/4 and ci + — 100g = 20t imply = imply we obtain ci is = and — c-i i(t) we obtain = and and 0. Since i? 2 5.0 + 25x = From x" then 7 am ft.Ol) « = 0. 15 sin q(t) = c% i(t) — q(t) — it. 150/13 and -50* B= + 2B) cost 100/13. Solving q" + initial + 20g' + 1000g = 100 sin f.4.' Exercises 5. z To use Example Then X = L 7 -1 ^(60) Z= yjx? + -5^" = /Y2 + 400 w and 144 13. The . Solving q" = g(0) + lQOtf and ^(0) q(t) Solving iff) = + 2500<j = 30 we obtain q{t) = cie -50t + c 2 te. Therefore the charge cos3( + c2 sin3t) '(ci is never + 10.50 ' . Thus 10e~ 3t (cos3t + sin3i) maximum see that the and i(t) = when charge occurs 60e 3t — t sin3t. Solving \q" and (f(0) = + 20tf imply = 300g = ci we obtain = 6 and oi = 7. .4) sin i steady-state charge = 150 - jg. sini+ rT T3 150 . = 1/35 and g(l/35) + sin%/3 1).012 we obtain yc = e~ l (3A A= l^te maximum see that the 9.4 + 6. Qp(t) and the steady-state current charge occurs (cos v^Sf z(t) when t = 26. and Substituting into the differential equation t. tt/3 and q(x/3) « 10.01871.2e- < e" 0. Thus q(t) Setting q = q(t) £ = q(t) 201 . + 0. 8. conditions q{0) = 4 -2. initial conditions -10. The = 2 imply c\ = -0. The 0. From and Z= \ZX*TW we see that the amplitude of ip(t) The Eo differential equation is = 100 and 7 = 60.432 coulombs. steady-state charge has the we find 50 cost. ^g" is z aV z* V 11.0370. 3 m .50t + 0. 24.70te _5Ot -I- (35 - 2. 3 in the text we identify . sin t is *j>( ( ) = 100 -— smi + —cost.012e. 10. Thus = we -0. The conditions as 0. is .012 and c2 = 1. Solving §g" + 10?' + 30g = q(0) = g*(0) = i(t) = imply c 2 q(t) Solving = find e 40* we = we Be" 1/3 which implies 300 we obtain = c2 = = 10- cie~ 2at + C2e~ m The initial .cos f + — 100 = .50 ' + 2q' + 4q = = A cos t + B sin form yp Thus.3333.012. 5880 and i Equating + = -XjZ and cos<£ = RjZ.1603(60* c2 sin 10*) + 3/2. Solving £7 — I/C7 = for 7 we obtain . q(t) -» 3/2. The initial = -i e -' ot (cos 10* + sin 10*) + \ -+ oo. = -^cos60* + ^sin60* + 100g and (400C - + = -—sin 60*- and the steady-state current i sin 60( 200 sin 60* qp (t) IJ ^ cos 40* - if = imply c\ ^ sin 40*.2400B) 13. Since Eq is constant the amplitude will be a maximum when Z is a minimum. Thus tan0 = -X/Rk Now equation (t)f« 4.1603(60* = - is a fourth quadrant 0.5880).6667 and find + As p (t) we obtain ^(f) {. Li = e.10t (ci cos 10* + = q = -1/2.4 . Solving L7 — I/C7 = for C we obtain C= l/i7 2 . 145 .Exercises 5. Thus 150 we obtain q(t) q{t) 14. 15. (a cos40t p (t) + conditions q(0) 10^ — 1 + {2400A + c 2 sin40i).4 sin = = 60* is 4. By Problem 10 the amplitude of the steady-state current is Eo/Z. « + B cos 60* + Csin40* + D we obtain coefficients state charge * -0. Substituting into the differential - 160OD) sin 40* 16005) cos 60f A= -1/26. OCi = + + 400 cos 40*. Z will be a minimum when X — 0. Since Eq is constant the amplitude will be a maximum when Z is a minimum. where Z = \/X 2 + R 2 and X ~ Ly — l/Cy. The steady-state charge cos 40*. <j> s=b 12. where Z = y/X 2 + R2 and X = L*i — I/C7.4 From Problem then 10. and sin 40* + 1/17- The steady- — cos 40* 1 1 I D= I is Id q'(0] (1600C C= -3/52.1600. R constant. Z will be a minimum when X = = \j-jLC The maximum amplitude will be Eo/R7 Since is 0. By Problem 10 the amplitude of the steady-state current is Eq/Z. Since R is constant. i p where sin$ angle. Solving \q" + 20^ + we 1000? = has the form qp (t) . Solving \q" 0) -0. 13 B= + 400Z?) cos 40* — cos60* + —4 1 4/17. EdjlkL. qa where find Eacosft. io i imply ~ ci we 3 sin?*. t At cos kt + Bt sin kt where k find ^cosfct. the EqC + cos 1 c2 VLC 1 H 1 charge is — EqC/{\ — LCy 2 ) and form of qp (t) coefficients we ' C2 = iij^/LC A— q{t) i(t) \ = = qo = and ci cos and ^(0) —ciks'mkt 5= fct = = \js/LC 71. = C2 1 — £7 5 ^ Equating ^ ^ 1 qp (t) and ) where we is i(t)= FVom Example ( differential equation -7 2 )Scos7t = and 72 + c2 sin lOt. qp {t) is i(f 1 ) 18.LCf + k 2 q = -2kA$mkt + 2kB cos kt = obtain conditions g(0) 1 qp (t) is we EoCt .(*) The B = 0. The charge + C2 sin io sin 1 . The is +qP {t) —^ ).<7o* j sin &t + io cos fct + 146 sin 2k ij and ci + 02k cos kt + ( is — I io/fc. The —107/(100 — £7^ B cos ft = + and cos e. ' EoCf . Substituting gp (t) into the conditions q{0) The current = 100sin7(. . i £bC / 1 in resonance the 1 initial equation differential EqC/(1 — —VIC ==cos —j== VW . Solving O. (cos 2 + coefficients initial find ^-7sin Wt) 1 Q sill in the text . . — . sin q The - charge (t/\fLC) cicos + c2 sin * and 90 5= Substituting qp {t) into the differential equation Equating ^j — 7 2 ).4 16. The current is . 1 When Thus. Substituting gp (t) into the fct imply + cj = —— (&£cos fcf + sinfci) = g .^v = ' the circuit 1 —VLC ^= 1 ci g'(0) = is . LC7 2 Thus.Exercises 5.+ qp (t) cos lOt 1" 10Q 17.lq" qp (t) = + lOq + A&inyt = 100sin7t we obtain (100-7 Z )^sm7( + A= Equating coefficients we obtain initial conditions q(0) = 1 q (0) ~ (100 100/(100 imply = . „ „ 2 . c\ -f 2 — c\ - q(t) and the current = q{t) Bcosft. = Bcos7t.4sin7( we = vlsin7i A= we obtain - see that q{t) . 7 t-cosl0t). cos kt. then -0. From mx" 14. If ^= then 4*+tt/6 = x/2+nn . 8 2. 3. = even and = 12.. < \0\ < 16. If x' n n odd. > or < m < 2. so that the maximum displacement is x = 5e . = subject to x(0) x (b) exist. True 5. = 1. = 1/2. From \x" + \x' + 2x = 0. is 2/ir. 2. is tt/2. 1. From \x" + x' + x = 0.Chapter 5 Review Exercises 19. 9 and — \ cos4f 2 The amplitude 20. n= 1. ft. -tt/4 9.£e _4t From x" + £x' 4. since 6. x(0) = 4. since x 4. x(0) = 1/3. The motion is upward 13. 9/2. = for and sin4i = x'(0) + w/2 n = 0. False 7. 5/4 m. 3. + 4x' + 2x = we see that non-oscillatory motion results if 16 — 8m. 2. period is 1. for + —V sin (^t \ 6/ is 2tt. 2tt/5. + 3ir/4 = 7r/2 + titt or i = 3tt/16 + nn. an external force may = c\ Solving fx" cos </2k t + 6x = + C2 sin \/2fc The amplitude (c) If x (d) If x = = 1 then then downward (e) i x'{3tt/16) for is \/2. n + obtain 3tt/4) . (a) 4. . = 2^3 fl'(0) + we obtain -^sin4t = 2 and frequency U6 = 0then4t+7r/6 = nn or t = j(njr-ir/6) or f = |(?r/3 + mr) for n = 0. = -4 we \Z2sin (At and frequency -tt/8 nn/4 1 for 1. Chapter 5 Review Exercises since k = 1. False. = i — period rnr/2 = t. Prom 0" + = 160 8(0) 0. and x'(0) = 2 we obtain x = 4e _2t + 10fe _2( If x'(t) = 0. and x'(0) . .. 2. ..64x = we see that oscillatory motion results if 2 . since x 10.Owe obtain x = |e _2t .. 2. and tt/16 + t cos4i — is tt/2.2 ^ 4 094 i = 1/10.256 < or (f) for then 4f . = — cos4( + |sin4f. since 3.. overdamped 8. 11. .. \x" + 6.25x = 0. Z/4W. l . §cos50t. = 2. speed of the weight we compute i . and x'(O) cos 2^ = and x = e ~« = restoring forces satisfy k\x\ + t The fci + .. to obtain the desired result. g(0) sin lOOt 20. — j^e '. ~j^ mx" + kx = x(t) The = ^2x2 so xa k ^xi From and a state of pure resonance in 18. (a) Let k be the effective spring constant and x\ and 22 the elongation of springs k\ and (b) = suffices. 2 xp = A/u 2 16. 2.. Writing gi" §£ = + sin 7* cos yt w = 64/3. we in Section 5. . and and g/(t) = ^4§+| = + i- we obtain qc = cjcoslOOt + |a/(t)| = \\l*9 C2 8inl00t. (0-2y|aln2^|( 19. 2^|t. From Example when7 = \/6473 = 8/\/3. The solution 1/Jfc ci cos and 100sin50t. - is lfyfig.3 8 cos 7* + 8 sin yt we see that the system is identify A e _! x(0) . Differentiate 1 From {k\[ki)x\. Clearly 17.2W and £3 = 4W we find differential equation (c) conditions x(0) initial To compute the maximum = we obtain x c ^ sin = 2v^t) e~ 41 (c\ cos + ^e~ 2\/2t + €2 8^2^2*). = x zj = k = 1 _ fcl k{x\ + X2) = k\x\ x'(0) = + J_ J_ C2 sin when + = 2/3 imply c\ 1 and C2 = + |coB2^t = 0. Then k = 4W/Z = + (4g/3)x = 0. . .Chapter 5 Review Exercises + 15. 1. and L dt 2. From lx" + x? + 3x = xp 4 + ^x = form x" in the cos50t) + R ~ + ~g = = i?(t) or t = jmt/50 for and use ^{t) dt 148 — 71 = i(t) 0. and -|cosl00t The *2 then becomes x" ( fcj- we have = I/2W+1/4W . = From qp = 9" + 104 ? y% sin 50f (a) 9= (b) i= (c) g = . sin50t(l 4mg/$. + ^ sin 50t. 2 = m 2 + 2m . we obtain = (m - l) 3 = 0. Assuming that y 1 ^2 [ci differential (\/21nx) . 2. The auxiliary equation is 25m 2 + 10. .6 - (m . Assuming that y = so that y c\ 2 . The auxiliary equation is 8.2m + 2 = so that = x cos(lnx) + casinflnx)]. The auxiliary equation is 12. 3. The auxiliary equation is m s + 4m + 3 = 7.4m .1 is m 2 . + c 2 lnx. The auxiliary equation is 4m 2 — 11. The auxiliary equation is 3m 2 + 3m + 1 = so that y = x~ 1. 1 1 = (m + l)(m so that y = c\x~ l)(2m + so that y x"1 and substituting into the m(m - c\ cos l)(m - 2) - 6 = ^ 1+ly (| In x) = 1) c\x~ l + cax~ 3 4- c 2 x~ + C2 so that y 1 "^. m 2 + 7m + 6 = (m + l)(m + 6) = so that y = cix -1 + C2i~ 6 m 2 . The auxiliary equation is 5.2) = is 4m — Am + m = 2 2 = 1 (2m — = l) = 2 so that y = so that y = cix 1 '' 2 + C2X 2 1 c\x + cax 1 '' 2 . Thus 18. The auxiliary equation is m 2 — 2m — m(m — 2) = so that y = cj + C2X m 2 + 4 = so that y — c\ cos(21nx) + 02 sin(21nx).m — 2 = {m + l)(m .. + 2) = 0./2 16. The auxiliary equation is 2m 2 - 17. The The auxiliary equation auxiliary equation 6. 14. + ca sin (5 lux)]. (-^lnx)]. lnx)].4 = 9.3)(m 2 +c 3 sin(\/21nx) substituting into the differential equation l)(m - 2) +m- 1 = ro 3 . The auxiliary equation is m 2 .3m s + 2m . The auxiliary equation is m 2 + Am + 4 — {m + 2) 2 — so that y = cix~ 2 + c 2 x~ 2 lnx. . In 1. 6 Equations with Differential Variable Coefficients Exercises 1. 6.3m 2 + 3m - 1 . "^ + c2 x 2+ A cjx Ji m+1 = 2 = — — so that y so that y so that y = (2m — + 3) = . The auxiliary equation is 4. The auxiliary equation is m 2 .8m + 41 = so that y = 15. The auxiliary equation is 13. (5 In x) = c^x 1 / 2 + C2X~ = x4 [ci x 1 ^4 cos(51nx) = x m and m(m - |ci [ci lnx) cos cos (-^lnx) -f c^sin + qjsin equation we obtain m 3 . Exercises 6. Thus y 19. Assuming that y m(m - l)(m - — xm c\x and substituting - 2m(m - 2) = + C2x\nx + C3i(lnx) 2 we obtain into the differential equation m - 2m + 8 = 1) . 3 2 - 5m + 2m + 8 = (m + l)(m - 2)(m - 4} = 0. Thus 20. Assuming that y m(m - = xm l)(m - 2) and substituting - 2m(m - -1 = y cix + C2X 2 + C3X 4 . we obtain into the differential equation + 4m - 4 = m 3 - 5m 2 + 8m - 4 = (m - 1) l){m - 2 2) = 0. Thus y 21. Assuming that y m(m- = xm and l)(m- 2){m- 3} = c\x + C2X 2 + C3X 2 In x. we obtain substituting into the differential equation + 6m(m - l)(m- 2) ^ m* - 7m 2 + 6m = m(m - l)(m- 2)(m + 3) = 0. Thus = J/ 22. Assuming that y " = xm + C2l + C3X 2 + C4X Cl and substituting 3 . we obtain into the differential equation m(m-l)(m-2)(m-3)+6m(m-l)(m-2) + 9m(m-l) + 3m+l = m 4 +2m 2 +l = (m 2 + l) 2 = 0. Thus y 23. The = c\ cosfln x) auxiliary equation is initial = ci 4- C2%~ 2 and 24. The = 2, C2 = -2, and y auxiliary equation is m 2 y The initial — — 2cix~ z y' . + c2 = -2c 2 = ci + C4 In x sin(lnx). conditions imply C! Thus, cos(m x) a: m 2 + 2m = m(m + 2) = 0, so that y The + C2 sin(ln x) + C3 In = 2 - 2x — 6m + = cix 2 8 4. -2 . = (m — 2)(m — + C2X i and 4) y' = = 32 conditions imply 4ci + 4ci + 32c2 = 16c 2 150 0. = 0, 2c\X so that + 4c2£ 3 . Exercises Thus, 25. The ci = 16, c 2 -2, and y auxiliary equation y 26. = = is m2 + initial The auxiliary equation is m y = cix initial conditions imply (-oo,0). c\ - = so that 0, = 1 and - 4m + 2 2 conditions imply c\ In the next two problems 1 l&x 1 2x A + C2Sin(lna:) C!COs(lni) The The = C2 and = + C2X 2 lnx = we use the . X + = 2 2) = = —x t + 2sin(tna;). + C2(x + 2x\nx). 2c\x = Thus y 3. X cos(lni) so that 0, = 1 y 10 substitution — Thus y and 5 and C2 = -ci-sm(lni) + C2- cos(lni). y' 2. = (m - 4 5x 2 - 7x 2 ln;z. since the initial conditions are on the Then dy dy dx dy dt dx dt dx and dt 2 27. The differential equation and 4( The dt\ dx) dt\dt) auxiliary equation is 2 initial initial conditions imply y 28. The = 2t» differential equation a c\ = 1 2 and = (2m — 1 y' = initial conditions _. _„__,_„„_ -4. [=1 2 l) \c = y'lt) x = 0, so that t'^ + c 2 (t^ 2 + \t~^intj 1 '2 - 5(-x)V 2 ln(-x), auxiliary equation is m 2 — 5m + 6 = y The initial = c\t 2 = y'(t) S, tfw Uj (m — 2)(m — + C2t 3 and y' 3) = conditions imply Ar, * < become t=2 The dx 2 dt + C2 = — 4. Thus -K^Ini = 2(-x} and 2, [=1 and dx 2 dx dt = y(t) t, 4m 2 — 4m + [y) become conditions y-citW + C2t 1/2 lnt The dt ^f + = 0; 6. -\- Rm — 8 = 0, 2cit so that + 3c2f 2 . 0. 0. . ' interval 1 Exercises 6. from which we find c\ = 6 and — c<i = y 29. The auxiliary equation is m2 — —2, Thus 2 6t - 2t = 3 so that yc = 6x 2 In 1 < a: , + ci In x c\ = W(l,lnjs) 2x 3 + 0. and 1 a: 1/x Identifying /(x) 30. The = 1 we obtain y — c\ auxiliary equation = — xlnx = c\ + C2lnx + ^x 2 — c\ is m a — 5m — m(m — 5) = x 3 we obtain y 31. The auxiliary equation = c\ Uj = — 5X 4 + C2X 5 — and so that yc x 25 is 2m 2 + 3m + 1 j/ = cix 1 — S we ol5ta^ n u 'i = = The x" 1 is 112 — xeX — = m 2 — 3m + 2 = x 2 e T we obtain u\ The u\ — «2 = 5 lux, and x -2 _J r = so that yc = c\x~ l 2 — x3 ^ 2 cix"" 1 ~~ I '^2 Then ui - = jx 2 + cax-W - \x + ^-x 2 x x' 1 2x 2) and = so that yc = x2 — xe x — c\x + C2X 2 auxiliary equation = cix + c 2 x 2 - x3 e x + 2x 2 e x - = c\x + c2 x2 + x 2 ex - is to 2 — 2m + 1 = (m — W(x,xlnx) - l) . Then uj = —x^e? + 2xe x + x3ex - x2 ex x 2 . = 1 152 so that y c xlnx x 1 , . + In x and . < 2xe — jx 3 15 (m — l)(m — — —x 2 e x C2X~*/ 2 and -3/2 6 = 4- eX y 33. and . Then 1) x-'/ 5 lV(x,x 2 } Identifying /(x) + l)(m + c 2 x -V2 + I T - It 2 + \x 2 -\x = auxiliary equation 5x 4 l/5x. £ — x 2 and u 2 I 32. . 5 = (2m + , ; , & 5 ) = = u'2 + C2X 5 = 3X 2 and 1x2 —x s + 7X S inx = ci + C3X 5 + ^x 5 lnx. W(x-\x-V 2 = Identifying f(x) — ^x 2 lnx, + C2lnx + ^x 2 — ^x 2 lnx + ^x 2 lnx = x. 5x 4 = jx 2 ui and u 2 1 Identifying /{x) = Then u'j = cjx + C2ilnx and 2xe x — 2e x , Exercises — Identifying f(x) 2/x we obtain u[ = -2lax/x and x(lnz) = 2/x. + 2x(lna;) u'2 Then m -(lni) 2 = , 112 = 6. 2\nx, and y 34. The auxiliary equation Identifying f(x) U2 — x\nx — y x, = c\x + C2x\nx — = c\x + C2x\vlx + x{lnx) 2 m 2 - 3m + 2 = is = xinx we and cix = + C2^ 2 + we use the i = - l)(m - . = 2) X' 1 2z = and u 2 - -x 3 lna: + x 3 lnx - x 3 = 3 so that y c x 11 -rx 2 = c\x + c^x 2 and .2 X = — zlnr obtain u[ *4 In Problems 35-40 (ro 2 2 . = cja: Then Ins. 3 1 *± £. + cix 2 - -x 3 + -x 3 \nx. £1 following results derived in = |x 2 — ji 2 lni, i/j Example 6 When the text: in x = e* or lnz, dx~ 35. Subsituting into the differential equation The auxiliary equation undetermined is coefficients y m2 + 9 + 8 we = try yp cje" 1 x2 dx 2 x dt dt we obtain = (m + = Ae dt 2 2t . l)(m + = 8) = so that yc This leads to 30Ae = 2t e 2( , cie so that + c2 e- 8( + ^-e 21 = c^" + C2X~ S + -^z 2 -t + A= C2e~ 8t . Using 1/30 and 1 . 36. Subsituting into the differential equation we obtain The auxiliary equation undetermined ,4 = 1/3, B= is coefficients 5/18, m 2 - 5m + 6 = we try yp = At (m - 2)(m - + B. 3) = so that yc This leads to (-5.4 + = QB) and y = cje"" + c 2 e Jt + -(+ J — = ci^ + 1(5 c 2 :r J + - lnx O + — 18 cje + 21 6At + C2e = 3t 2f, . Using so that 1 Exercises 6. 37. Subsituting into the differentia! equation we obtain The auxiliary equation coefficients we try yp m 2 — 4m+13 = is = A + Be l . so that yc = 2 e '(ci cos3f+c2sin3i). + lOBe = 4 + 3e', 1 This leads to lZA so that Using undetermined A= 4/13, B = 3/10, and y = e = x2 zi + C2 sin 3t) (ci cos 3t [ci cos(3 In x) 4- —+— e' + C2 sin(31nx)j + + io 38. Subsituting into the differential equation ^-x. 1U we obtain 2m 2 — 5m — 3 = (2m + l)(m — 3) = so that yc — cie - '/ 2 + C2e 3t Using undetermined coefficients we try yp = A+B^+Ce 21 This leads to -3A-6Be -5Ce 2t = l+2e'+e 2t so that A = -1/3, B - -1/3, C = -1/5, and The auxiliary equation is . t . , y = c\e~ t/2 + c2 e3 ' ^e* - \e 2i = c\x~ 1 ^2 + c 2 x 3 -\x- ^x 2 . 39. Subsituting into the differential equation we obtain The auxiliary equation undetermined is coefficients m 2 + 8m — 20 = we try yp {m + 10)(m — = Ae~ st . 2) This leads to = so that yc -35Ae~ st = = 5e~ and y 40. = Ci e~ m+ c2e 2t - \e~ 5t = ax' 10 + c2 x 2 - ±x~ 3 From dx 2 x2 l dt 2 154 dt c\e~ iat . 31 , so + c^e 24 Using that A = -1/7 . Exercises it 6. follows that dx 3 x 2 dx [dt 2 ~ x2 dx\ x2 dt 3 dt dt 2 x 2 dx J dt 2 d2y dt 2 x3 dt + x3 dt x 3 dt 2 dt)' we obtain Substituting into the differential equation 3 dt [ dt J x 2 dt 2 \x) \x) x 3 \dt 3 d y x* \dt 2 j dy dy\ fd^y dy or d3 y The c$e 3t auxiliary equation . A= -17/12, y The m 3 -6m 2 +llm-6 = (m-l)(m-2)(m-3) = Using undetermined so that 41. is = BI cie* = coefficients we try yp = A + Bt. T* !t m 1 m = m(m + uq and u(b) = m yield the system Cia~ — = /iio-ui\ ab \ , Thus b , u(r) = The auxiliary equation and u(b) — is m2 — ui yield the system c\ Cl = u\ In a ^H ~ J t — «i \ ~Tb-a C2 In c2 a = = ci uo, c\ „M U{r} ~ t Ul lnq -"0 lnfc — uq In 6 ln(a/6) + , = +C2 = u\b -"l MVfe) = cir uq, cib" — — 1 -1 + , In a:. C2- +C2 = The boundary «i- Solving gives uqo. a u ib - «oo — b-a ii +C2lnr. The boundary conditions u(a) + C2 In 6 , and ln( a /&) Th„ hUSB i o so that u(r) + and } Ziifl \ so that u(r) . , —a = 1) 1 9 •> + ci 42. This leads to 17 17 — - -t = cix + c^x* + c%7? - — - - 9* 2 is = c\e +C2e 2t + {HB-6A)-6Bt = 3+3t, so that yc -1/2, and + c 2 e' + c3 e M - auxiliary equation conditions u(a) f.tPy ... dy "o _ 111 = C2= u\. Solving gives tto — u\ M^}' _ «oIo(r/6)-ui ln(r/q) ~ " tafc/fr) = uo 1 Exercises 43. Letting 6. i^i-lwe obtain dy = dx dy dt and 2 _ d y dx 2 d dy dx Substituting into the differential equation The auxiliary equation is m 2 — 3m — 4 = y 44. Letting t = 3x +4 = cii 4 d?y dt _ _ dt 2 dx dt d?y dt 2 we obtain (m — 4)(m + + c 2 f-' = d(x - I) = 1) 4 + so that c2 {x - l)" 1 . we obtain ^-^ - s dt dx dx — dt and dx 2 dx\ Substituting into the differential equation dt 2 dt The auxiliary equation is y 45. Letting i = i + 2we ' we obtain air 9m + 21m + 9 = 2 dt 2 dx dt 3 (3m 2 + 7m + 3) = so that (-T+^)/6 + ^((-7-^1/6 _ cl( = ci(3x + 4)<- 7+ ^>/ 6 + c2 (3x + 4)(- 7+ ^>/ 6 . obtain dy dx _ dy dt and £l2 ~ dx d2y (d}L\ - d2y — ~ ~ dx \dt ) dx dt dt 2 dt 2 ' Substituting into the differential equation we obtain The auxiliary equation y = c\ is m2 + = cos(in t) 1 so that + ci sin(ln () = a cos [ln(i 156 + 2)] + a sin [ln(x ^ Exercises 6.2 Exercises 6.2 1. = lim n—too On x n+1 /{n lim n—too + 1) = lim c"/Vi CO The = — 1, on (-1,1). At z aeries is absolutely convergent ^ -n the series j is the harmonic series n=l 00 which diverges. At x = 1, given series converges on (—1, On+1 2. lim n—too = ^ (—1)" the series converges by the alternating series z n+1 /(n+l) 2 = lim 74—t OO X = J lim on {—1, series is absolutely convergent alternating series test. At x = Thus, the 1]. 00 The test. 1, At x = -5 a convergent 1). the series is —1, the series —— (— 1}" p-series. converges by the Thus, the given series rt=l converges on [—1, 3. 1]. p±l = n—lim too Urn 2 n+l x n+l /(n+ 2"l n /7l 1) 2n — iim 111 = 21*1 00 The < series is absolutely convergent for 2\x\ or 1 |x| < — At x 1/2. —1/2, the series ^ (—11* fc fc=l 00 converges by the alternating series test. At 1 = j ^— 1/2, the series is the harmonic series which fc=i on [-1/2, diverges. Thus, the given series converges On+1 4. lim n—too The On lim rt—too The lim n—too 5n+1 i" +1 /(n + 5 n z n /n! series is absolutely convergent On+1 5. = On = lim l)! series is absolutely (a: Iim "-co on (-00, (x-3) n+1 /(n + n—too — l) - 3)V«3 convergent for 1/2). n+l \x\ = 00). 3 = |x Urn — 3| f-^-l < 1 |i-3| = |x-3| At x or on (2,4). = 2, the series £ —n - n=l 00 converges by the alternating series test. At x = 4, the series ti=i the given series converges on [2, 4]. ~ j ' s a convergent p-series. Thus, the series j —= ^2 is —8. .2 On+1 6. 10. the series a divergent jL- the given series converges on [—8. the series —5. Thus. 00 The series absolutely convergent for is \x + 7| < OO converges by the alternating series At x test. the series converges on (—5. = hm (n+l)(n + . 6). 1 x (n series the series 5. At 1 x — 00 converges by the alternating series comparison On+1 9. lim n—»oo The with ^— (n lim n—'oo + . = 11m n—-oo (x On + 7)" +1 n /Vn+T = lim oc V n + n— + 7)"V" lim rc—*oo (x |3 + = 7| + 7| |x 1 V" (-1)" — p-series. the (— — — = ^— °° ^ = = At x l)*10 fc -j diverges by the n-th term test.Exercises 6.. or 15). — = on (—8. = 15. «n+l lim = On (n lim n—>oo l)(x-4)" +1 /{rc + n(i-4)™/(n + 2) 2 + 3) 2 . 2 3) x 1 £ ^7— 00 The series is absolutely convergent for \x — 4| < or on (3. it=i 8. nx" +1 /{n + l) - 2 <"+ 1 > — l)i n /n 2 " ^ — the series 1)1x1 ' 1 1 = 00.5). x(n + 2) 2 -4 = i-4 111 . Thus. 15). nn In (n-l)(n + lim 1 ' 2n = lira "-«(n-l)(n+l) 2 Vti + x| I/ 1 — 1 lim «—»(u-l)(/i+l) 2 (^)T = lim n^(„-l){ n+ l)2 [(l + l/ n )n]2 Ixl 158 = -Ax\ = -r^ diverges by the limit # l) 2»+ 2 (-l) fc k fc converges on [3. —6). lim n—too The — On (x - 5)" +1 /10n+1 (2 - lim n—-oo — 5)"/ 10" series is absolutely convergent for ^ (_— 00 series l) k - ^(—1)* (— 10)* — — lim n-ooio |x — 5| — |x 5| 1 = < \x 1. = n— lim 2<n + too n\2 n x n lim n—too Thus. test = At x test.. the l)!2 series converges only at lim Ti — On fln+1 10. On+1 7. — |x - 51 10' 1 — ' < 5] on {—5. „. At x or 1 —6. 3.5). 00 1 diverges by the n-th term At x test. —"OO = On n+1 x n+1 = 0. " Exercises 6.. e .. series is convergent 3 2 . 2x 5 4x 7 3 15 315 3x 2 4x 3 1-- 1 4 . .X + X -. x5 ~ — x6 120 ' 720 X6 X4 1 ex — + ^ + 2* ' 24 6 • 720 X4 x 3 1 \ h —+ ^+^+ + X6 x 105 3 —5 T4 + T T s [ 1 — 44x 8 12 ~3 + 4 X3 x 2x 6 " 720 24 ~ 6 + 2 12 1 i2 5x 4 61x 6 2 4 48 1440 24 —_ x3 x5 4 3 H llx 4 + + _.= (l +«+ -b x5 x7 120 5040 1 —X 4 x2 z3 2 o 2 + e. 15 I?* 7 315 + ' _6 . * x1 00). X4 .. „ —— x3 x5 ^ / 504Q 12Q g X' 5 7 4 xe 3 _ 16..j - 4 \ / . — x . e X3 x \ ( + + + '" 2 x4 x6 x 2x3 . 4 x3 x T -+ nx=|l + x +y + .x „ x 4 x 1 19- x2 1 x4 x6 -T + T"T + - ^ \ =x . S1 on (-00. -„=|l-x + T T s 12.. X2 l- W ^ -r h X5 x " z x7 h -r — + + -7T 6 X . = l-x 2 + '-/ X X2 x 3 / + + X2 x 4 .2 The 11. + ___ + — 1 W *3 ^ . . + 120 2 2x 23x — + — .— +— 45 7 x- 15... =!-. 17 a 1- x x — +2 3 ^I= ^ tan ' r 4 2 \ x3 6+ = cosx ~2 e* + e.. 5 + . \ " \ X X T 5? —. sinxcosx . w 13.) — + Xx24 6 — x^xjT+ + 2 18.— + V =z*. ln|d = — x + c => y = c\e~ x . k 1 and = Ar we find r> 5 1 1 1 24* so on. 2.[(fc + fc=0 k=n fc=n-l = 2 n=0 n-l l)cfc+1 -2c t fc ]x' = *=0 160 0.-cox 3 I 1 + ^cqx*4 = cq l ~ 24 6 x+ 1 x 1 Q 2 . 1/ Substituting y = E£L CnX n into the differential equation leads to * t/-2y='£ £ c«x" = £ f. Separating variables 1 1 2 i -cqx . Iterating l)Cfc+l +C fc ^ k—n it=n— 1 Thus {k + l)cjt + i + Cft = and ck +i =— for + 1 Cl - -Q> C2 ." Exercises 6. V Substituting y oo y' + V=J2 n=l = ££Lo W 1 "1 CnX n into the oo + £ CnX" «=0 equation leads to differential oo oo + l)ck+l x k + fc=0 oo £ CfeX* = £[(* + A:=0 ft=0 0.cqx + 22.c3 = .~2 C1 = C3 = "31 C2 = Ci 1 = . 1. Therefore y = co . Separating variables we obtain — = — dx =4. ~r + 1 x 2i we obtain — = 2dx =^> ln|j/| = 2x +c « = 3 cie *. (* + !)<*+!** ~ 2 £ k=Q ck x k — 4 ]x fc = 0.2 21. . x 2 y= = ci = C2 = 3 00 E +c 0. find 2co § = 2°° c2 = -co C3 = 24°° 16 2 «= 4 and so on.— Exercises 6.n ' nto tne differential equation leads to EJJLo cn oo j/ - 3 1/ i/ 2l obtain — = x ax Substituting £ i(2*)» = ^ £ " + 3)c* +3 . (fc ci + 2c 2 x + oo E ^" +2 = E - oo + 3)c t+3 x i+2 - E cfcx^ 3 *=n fc=n-3 Thus ^ 3. Iterating we find .fc + 3 — Cj. k = 0. = — — and C3 = 3~C0 C4 = C5 = 1 c7 = eg -Cf. 1. 1. 2.. Separating variables we ^(2x) 2 -cox" + 6 — cox* + 44 l + -{2xf + -^(2x) 4 + --]= o = lnjj/| = .c^*" 2 = 1 [(* + 3)c.2 Thus {k + l)c fc+1 - = 2ck and Ck+\ = T~T7 ch> ci = C2 = c3 = = for we Iterating 0. Therefore = y co + 2cqx + -cox* + 2 = co [l + 2x + 23. for 1 = 1 0. 2. Therefore V = co = e-o + J jcox° 1 1 x3 + T + 1 * 6 pCoZ" . 3 12 1 V 4 we obtain dy y _ dx 1 -x In |y| = — In |1 — 162 x\ +c y = l-x' Iter) .™=o °n £ n=l + xHy = = ci CO QO OO */ ' ' we obtain 24. Separating variables Substituting y + + nc^"" + 1 £ c^ 3 £ = n=0 00 (* + 4)c* +4 x* +3 - fc=-3 £ ck x k+3 it=0 + 2c2 x + 3C3X 2 +£[(* + 4)cfc+4 + c fc jx fc+2 = 0. Separating variables s 1 i/4\2 4 = 1 + 2V4 2 3 / . 2.Exercises 6. Therefore 1 1 t CO 1 ~T 25. + 4)c k +4 + ct = and c*+4 = ~ fc ^4 & c fe' = 0. it=D Thus ci = C2 = C3 = (k 0. + ^ 1 /x 3 \ 2 T — 1 -^CQX 9 1 + 2 fx 3 \ 1 + 2~3 (t ^- = -x 3 dx => ln\y\ = y = ~x 4 + c => 4 y = ce" 1 ^4 . find 1 C4 = --CO C5 = c = -8 C * = 2-4^° C6 = C7 = 1 8 Cg = Cl2 = 1 1 = Cn = Cio = -_.2 and so on._<:„ and so on. xn ' nto tne differential equation leads to Y. 1. 2y = equation leads to oo CO E tkwe"- 1 n=l ^ + E - ^ k—n oo l)c fc+1 s* + fc=o C! £ fi=0 k=n oo = 2 n=l Jc=rc— 1 .E 2 jfc=0 jfc=l 2co + + z) 3 £ K* + l>«*+i + - 2)c*)s* = 0. £>. = Cl = Co C3 = c2 = Co on.2. Separating variables = co [l + x + z2 + i = E^o + • •] = co => In|tf| = 21n|l+a:| + c => y = E ci(l ~\~ c^x" into the differential OO (1 3 = 7^- we obtain —y . 1.E(* + ~ oo E ***** .2 Substituting y = J^L cn xn into the differential equation leads to oo (1 . iJpfc+i - (* - + l)cjt l)c*+i (* + - o- = a = co and = Cjfc+i Iterating and so EP+ -^+ we c fc = A .T-7—d x x Substituting y • + x)y' .x)y' oo -y="£ nenX*n— 1 oc E n=l ~ 1 _ ~ EW n=0 k—n *=u-l k—n oo 00 = £(* + £ " lJOfc+lZ* n oo £ cti* - fcCfcX* OO Thus = ci ci-co = + (ft 0.Exercises 6.3 find ci = co. . Therefore y = co + cox + cox 2 + cqx 3 + 26. . .. . Substituting y — ^ n=0 differential equation leads to oo oo k=0 k=0 oo oa n=2 n=0 ( k=n-2 k=n oo = O* + 2 )(* + l ) Ck+2 k + Ck\x = 0.2 Then - ci {k and — ci + 2co = + (k-2)c k = l)ck+1 2co — k 2 = .Exercises 6. find ci = 2co 1 and The = 0c 2 = c4 = co (l + so on.. 2. so y — c\ cos X + C2 sin £.l. Therefore = 27. . 3. cA auxiliary equation is co + 2coi + coi 2 = 2z + x2 ) = m 2 + = 0.jT^J c *> Cjfc+i Iterating we = ft 1. fc=0 Thus and (k Cfc+2 = + 2}(fc + l)cfc +2 ___i__ Cic 164 . n c„3: int . 1 cq(1 + 1) 2 .2. +c fc fc = . The 4-3 so on. 1. i + + . 1 -(agr + «iE m2 — 1 = 0. 2. Therefore !/ 28. fc=0 Thus {* + 2)(* + l)£* +2 -cfc = 0. * = 0. 1 1 a= so y = (2n cje 1 Is 1 =coc OS * + Cie n *.2 Iterating we find C2 1 = Co _ 2 C3 1 — C 3-2 i and = CQ = CO + ClX - ~ C2 2^ 4-3- 1 1 fc5 L — C6 = -eV c7 - "7^ C5 = 5-4.3.2 = 4 <=1 1 -ei* 1 1 o auxiliary equation is 1 -cox 2 - 1 . 1 — cix J + —^ cox 4 1 + C1 -**i. 1) . C2e s Substituting y .Exercises 6. Cn 1" ' nto tne . = ^2 differential equation ieads to n=2 CO 00 00 n=0 k=0 k=0 fc=n = £[<* + 2){fc + l)Cfc+2 " ck \x k = 0. Therefore. Substituting = £[(* + -1 00 2)(* 2)(* + l)c fc+2 z* + l)c* +2 - - (fc £(* + l)c fe+1 x + l)cfc+1 ]z it=0 Thus (k + 2)(k + M ~{k + l}c 166 . 1 5 x2n+1 m 2 — m = m(m — 1) = 0. 5 i 1 = 00 C03h = so y 00 - v' = E 00 n(n - £ iKx«.2 - n=2 ci 1 W 00 = £<* + . 1 into the differential equation leads to y" . y 29.Exercises 6. l)c k+i = fc = fc 0. + Cl 9inh x + C2C 1 1 n=l . 3/ = Cn^" . Cy 1 * + 2 + 1 2 C31 .2 Iterating. we find 1 CO 1 ci C2 1 C3 Ci — and 2 = co + c\x + C2X = co + c lX + \cox 1 r co = The = 3 + i - 1 = Cl 7-6-5-4-3-2 7-6 ' so on. ^x + ^cox + ^^J^* + 3 4 S» is 1 r °° + Cl 4 1 2 auxiliary equation . ~--c 3 = 1 1 22 3-2 1 1 2^4! = C1 Cl fc = 0. 2) find C2 C3 C4 = = 1 "2 1 C1 2 1 "2 C2 = 3 1 1 .1. = so y + cue ci x^2 . Exercises 6. + 2) (k 0.lW" + n=l 2 n=2 = 2 £{ft + 2)(fc + l)cfc+2 T + £) fc *=0 = £ (Jfc 2(k and Iterating Ck+2 (2(fc + 2){fe + we + 2)(k + =- \)ck+i x k k=Q + l)cfc+2 (t + l)c i+I ]x Jb=o Thus + l)ck+2 + {k + 1 2{k + Cfc k + i.2 and c*+2 = we Iterating = * cfc+i. .1.0.2..2 find C2 1 = C1 2 1 - C3 C2 = C3 = 1 3-2 3 1 = C4 1 Cl i 5i and so on. l)ct + i = 0. = 1) 0. Therefore 1/ = co + c\x + 1 2 -ctx* cq -C + The V t gj-cia: = l 1 + jjCi^ 1 + x + -x 2i + -x 3i + is 4 + 4 + + ci Co 1 ci auxiliary equation — £n^o 1 + 1 -x-' 2m 2 + m = m(2m + . Substituting CnI ™ int0 tne differential equation leads to 2y" + y' - 2 £ nc^xn-1 £ n(n . " - equation we have CO n(n - n=Z l)^" 2 .-Cii + - 1 3 /x\ J3 1 2U) -3iU) -Cix 3 + 3! + ^^ | ^(|)" = c + | I(^)^c + c Cl l 1( o -2/2 o Exercises 6. it=l Thus c2 (fc and Choosing ct+2 co = 1 and c\ = we = + 2)(* + = l)c fc+a __l_— -c _i=0 ^ find I C3 =6 C4 = C8 = C5 = Tio 168 fc fc = 1 .3 1. Substituting y = E^Lo ^^ Cn^" i nto the differential OO j..c^Jx* = l)c k+2 0.-CiaT 222 2 2 2 ' /x\ /xv2 1 4 = Co — 2ci = -iCi 1 1 1 1 = C + Ci .. 2 .3. Therefore 1 1 1 2 Co ci 2 x = Co + Ci = -2 + l .2 and so on. c *- ..2 Exercises 6..£ cx^ oo 1 n=0 fc=0 f k=n-2 oo $> + 2)(* + l)<*+3** ~ £ = Jt=l fc=n+l oo = 2c 2 +E \{k + 2){k + . For co = and ci = 1 = Cg = C7 = C7 = Cg = we obtain c4 = C6 = k + £ k=2 fc=0 oo = l)ck+2 x * 2 ' M 0. For cq = and c\ — 1 we obtain c3 = 1 C5 = Cg = and so on.Exercises 6. Thus. C*_ 2 z* . Substituting y — Y%Lq Cn^" into the differential equation we have y" oo 00 CO + x 2 y=J: l)^"" 2 4- n(n - n=2 £ n=0 k=n-2 oo ^x" +3 = + 2)(fc + 2c 2 jfc=n+2 + 6c3 x + + 2)(* + + ljcu-a c*_ a ]»* - k=2 Thus c2 (A ^ Choosing Cfe+2 Co = 1 and ci = + 2)(k =~ (fc + we + 2)( fc = C3 = l)ct + +D 2+e*-2-0 Cfc = -2 ' find 1 C5 and so on.3 and so on. two solutions are 2. Substituting y + — Y^=o y" - W Ci»^ B = .Exercises 6. two solutions are m= .1*6 + Jj^ - and n into the differential equation + If = we have CO CO £ n=2 ~ 2 1 ~ 2 £ n=l A=n— 2 = £(fc + W 00 n + £ n=0 k=n fc=n + Oct+ai* ~ 2 )<* £ 2 it=0 + E c* x * fc=0 fc=l oo = 2c 2 + co + £ [(A + 2)(* + l)c fc+2 - (2fc k=l Thus (k + 2)(k + l)ck+2 -(2k-l)c k = Q and 2k Ck+2= (k Choosing q> = 1 and ci = we — 1 = k Ck + 2)(k + ' l) find c2 = -- C3 = = C5 C7 = - • • • = 1 06 and so on. . For cq = and ci = 1 7__ ~ 336 we obtain C2 = C3 = Ci = Cfi = 1 6 C5 = 1 24 4 C7= 1 m 170 - = 1 M-- - l)ck ]x k .3 and so on. Thus. Thus. Substituting y = and „i = x + -J + - a* + -.^120 0.x r + Y.2)ck \x k = 0.™=a Cn^" i nto the differential equation we have y" -xy' n(n - + 2y=Y. £ nc„x" + n=Q £ c^" l)^"" 2 - 2 n=2 n=l r 4=n fc=n-2 oo = £(* + + 2)(* l)c t+2 £ ^- Jt=0 = 2c2 k=n oo fc=l § + 2co + [{* + 2){* + l)ck+2 Thus + 2co = 2c3 (k + 2)(k+l)ck+2 -(k-2)c k = and c2 < Choosing For co = co = and 1 a = and ci = 1 ^ we a = -co " (*+*2)(ft+l) k=W '*' find ca = -l C3 = C4 =0 Ce = C5 c$ = Ct = Cio = - = = = = = we obtain C2 = C3 = "61 C4 = eg C5= . .3 and so on.x 4.Exercises 6. oo +2 £ QfcX* fc=0 -(k. two solutions are __x m = l-. Substituting y = . oo = + 2c2 (6c3 + co)x + + 2 )( fe + *Cfc_i]x = fc + iJcfc+a Jc=2 Thus c2 + co = 6c 3 (* + 2)(* + = l)qt+ 2 + kcfc-i =0 and C3 < = -^ *+ a = < -(Jfe Choosing co — 1 and c\ — we + 2)(Jfc + ' find 1 <* = -g ^4 = = C5 1 and so on.1 l) C6 = 5 252 172 2 3 4 ' ' -" . For co = and Ci = 1 we obtain c3 = 1 C5 = C7= *- *.1 X t + ^ Cfc. Thus. £ nc^ + £ c^ n=l .x2 1 and y3 = - x \x* - ~x 5 Y^^Lq r-n xn i nt ° tne differential equation we have oo co on / + *V + xy = £ n(r.jX ifc=l .3 and so on.l)^"" 2 + 1 n-2 1 n=-0 fc=n-2 it=«+l fc=n+l oo oo = £(* + 2){* + oo + £(*- l)c t+2 S* fc=0 fc=2 l)cfc .Exercises 6. two solutions are = yi 5. Substituting y = 1 .= . 1..and so on.t +2 + k=l Thus 2c 2 (ft + 2co = + 2)(* + + 2(* + l)Cfc = and 2 c*+2 Choosing and so co = 1 and on. 2{k + ft l)ck ]x' . find 1 c3 = -l C3 = C5 = C7 = 1 ' = we obtain C2 = C4 = C6---. K* + 2)(* + l)c.™=o cn I " i nto tne differential equation we have 00 00 00 n-2 n=l n=0 00 CO CC = £<* + 2)(* + l)a +2 ^ + fc 2 A. +232* -" Y. 1 5 « = 1 "6 I + 45 X 6... two solutions are 1 .=0 £ + lfc=l 2 £ k=0 C *x* 00 = 2c2 + 2co + Y. Thus.j-jT^ c = fc. fi 5 4 ""I .3.2. For cq = c\ — and we c\ = = . . Thus. = i .3. C4 _. . and so on.. £ n=2 2n(n - l)c„x"- 2 + 00 £ nc„x" . £ n^x""' n=l it=n-l it=n-2 fc=n-l oo oo = £(* + l)kck+l x k oo ~'£{k + 2)(k + 1)^+2** + E(* + oo = -2c2 + ci + ^ [(A + k t=o Jt=0 fc=l l)c k+l x - l)fcc t+ i (A + 2){fc + l)c k+2 + (k + \)c M \x k = 0.. = and ci = 1 we obtain .n=0 £ n=l t k=n— 1 fc=n— 2 fc=n 00 00 = E (A = 4ca -c + + ljft^+jx* + £ fc=n 00 2(fc + 2)(fc + ljcfc+a** + 00 E ***** . Substituting y l-^ + i^-ix 6 + -- = and - l)y" x -l x> ± xS-JL x + ? + ..E 00 EP+ !)ftcfc+l 174 + 2(fc + 2)(Jfc + l)c k+2 4- (ft - ljcjtjx* = 0. two solutions are = y1 7. Substituting y = = and 1 Cnx" into the y2 = a: differential 00 (1 + 2)y" + xy' - y = £ n=2 1 1 1 1 1 + -x " + -x 3 + -2 * + equation we have 00 00 n_1 n(n .l)cnX + .3 and so on. Y£?=o Cn 1 " i nto tne differential equation we have 00 (x = y2 + y' = 00 £ n (n - n=2 00 £ l^x"" 1 - l^x"" 2 + n(n - n=2 .. two solutions are 4" (1/1 8. = For Co 0. Thus + ci =0 -2c2 (k 2 + -{k + 2)(k + l) ck+l l)c k+2 = and 1 cfc+2 Choosing co = 1 and — c\ we = find C2 I fc + ^ 1 = 03 = C4 1 — = = = C3 = ft Cfc +l' ^TjT2 - - 1)2. Thus..Exercises 6. = j/2 480 24 c fc - (At 6ci = + 2)(k 175 + l)ch+2 = ~ ( k + 2 )( k + l)c*+a] = 0. = ^4 = C5 0.lj^x" - Tt=2 - l)cfci* +4 k=n +4 fc = -2c2 + 2co + {-6C3 + 6ci)a: + 2 £ 2 [(* " * + 4k + 2 ) c* Thus -2c2 + 2co = — 6C3 + + 3A + 2) k=n oo £ ***** + k=0 £ fc=2 2 n ck x k k=\ fe=0 (fc ac £ ncnx" + 2'£c nx n-l n=0 CO . 2. . and Choosing co = 1 and C2 = c *+2 = ci ^ so on. . . two solutions are 1 2 4 9. . ^ we obtain c2 = c3 = C4 = C5 = eg = = 0. 3. « = 1.. Thus..1)^"~ n=2 k=n = cix equation we have oo + 4xy' + 2y=Y. 1 = = -^. 2)(A + fc . 2.3 Thus - 4c2 + {k l)Jfccfc+i + 2{k + 2)(k + = co l)c fc+a + (k- = l)ck = 0. k 1. Substituting i/ = Z^Lq Is If °n xn into differential oo (x 2 - l)y" and n(n . Fbr co = = and l)Acfc+1 2(k we -=1=0. <* \. Exercises 6. . 1) find C2 = C\ + . and + (fc + (fc-l)c „w. 3. .£(* + 2 )( + ifc=2 2 k=n~2 oo £fc(fc ' oo £ nfn . . Substituting y = H^Lo = 1 4- CnX n x2 + x4 H and 2 + l) y" + x3 + x h n=2 oo H n n " l)^" n=2 2 ( OO ~ 6 £ n=0 + =2 . .. find = • = 1 we obtain 1 - - • • = = 1. .4. ^ oo CO = £*(*- H we have oo .3 and Choosing For Co = co = and 1 and c\ — = ci we C2 = eo c3 = Cj Ck+2 = ck ca = l c3 = c5 = c7 = C4 = C6 = Cg = C2 = c4 = ce = C3 = C5 = C7 = = ft .Gy = Yl n ( n ~ Vcn*" + ft x into the differential equation 00 (x = y2 E t=0 + 2)(* + (ft l)c* +2 x fc - 6 £ t=0 cfc:c * oo - 2c2 - 6co + (6c 3 - 6ci)x £ + [(ft 2 - ft - 6) c* ifc=2 Thus (ft - 3)(ft = 2c 2 - 6C3 — 6ci = + 2)c + fc 6co (ft + 2}(ft + l)c k+2 = and c2 — 3co c3 = Cl = ft-3 c *' 176 = 2 3 -4...3. > + (Jfc + 2)(ft + 1 .. Exercises 6. two solutions are j/i 10. 2. Thus. Substituting y = 2 + x*.Exercises 6. Thus. two solutions are = l+ 3x 2 +x 4 -~x b + yi and y2 5 11.lfax"n=2 oo oo £ *(* " 1><*** +3 l)<*+2** +3 k=0 oo £ nci" ..co) + (12C3 + 2ci)x [2(* + 2)(fc + - = + Y. 00 + 2)y" + 3xy' -y=-£n(n- lj^i" + 2 n=2 CO £ n(n . l) Clt+2 + 2 (fc k=2 Thus 4c2 12c 3 2(fc + 2){fc + co + 2ci = l)ct+2 + + 2fc - 2 (fc l) cfc = and Cfr+2 = A: 2 -2(fc + 2lfc — 1 + 2)(i + *- Cfc l) ' 2 >3....3 Choosing Co = 1 and = c\ we find c2 = 3 C3 = c5 Ci = 1 = = c7 - - - = 1 and so on. + 2* - l) cJ x* = 0.k=0 E cjt** k=l oo - (4c 2 .n=0 £ c„x n=l 4=n n k—n co co + 2 E(* + 2)(* + fc=2 2 fc=n-2 fc=n = x n T.4. For = Co and cj = we obtain 1 C2 = C4 C3 = 1 C5 = C7 = C6 = = Cg = • • - * = = 0.^Lq c„x into the differential equation we have 00 (x = E ***** . . IK*"" 2 + n=2 £W n=l n=2 k=n—2 jt— n OQ £ - (-c 2 - *(* - lJCfcl* co) - £ . Substituting y = = I^ two solutions are 52%Lq Cn^" i Qto tne differential equation we have oo (x 2 - l) y" + xy' - J.l)c„i £ + OO to*** (fc a - - £ l) ck ] C x* . 12. Thus. For 1 and co ci = = and we c\ find = 1 c2 = - C3 = C5 C4 = "96 = C7 = " = C6 = -'- = we obtain = C2 = C4 1 6 C5 and so on.3 Choosing cq - and so on. = £ .3.&3Z + (* £ + 2 )( fc [-(fc + l)cfc+2** + 2)(* + -2c 2 - co -6c 3 = l)c fc+2 + = (fc - ^ = + 1)q = and C3 = Cfc+2 = fe — 1 ^j^ c *> 178 2.4 + l)ck+2 Thus + 2)(fc + £ c^" n=0 k=n CO fc=2 -(* - k=n CO = oo oo oo B «(« .Exercises 6.j>(n . n=0 £ cn^ fc=n OO OO fe z* - fc=l £(* + ljqt+i** - £ c*x it=0 fc=0 fc oo = 2c 2 £p + a .Exercises 6. Substituting j/ n J^™= o c„x = = " 1 2 i 1 into the differential equation 00 f" - {x + 00 -y= £n(n.3 Choosing cq = and 1 c\ = we find C2 = --1 C3 = = C5 C7 = • • = 1 and — so on. For cq and — c\ 1 we obtain C2 = c4 = eg = C3 = C5 = C7 = = - = - 0.3.4 find C2 and so on. . Thus.co + - 2)(k + l)ct +2 — (k + Thus 2c2 (k + 2){k + - l)c k+2 ci - co = - (k - l)(ct +1 + Cfc) = and Cft+2 Choosing qj = 1 and ci = we = cq = and ci = fc = 2. two solutions are ! 13.£ l)y' m=^ j 4 8 2 n=2 n=l « we have OS n - 1 OO fc=n— 1 OQ = £{fc + 2)(fc + l)cfc+2 ^ .£ Ac fc=0 n n=l A=n h=n-2 OO £ W" .l)c«* n " 2 . For ^±1^^ 1 = 2' we obtain 1 C3= 6' ^6 l)ck+1 -(k + l) C(t ]x* = 0. 5 + .xy' + y = ih 1 3 = x + -x 4 1 5 equation we have 00 00 £ «(« " IK*"" n=2 1 +12+8 00 -£"(«" l)^^"" 2 . - 2)c* = = c fc _! 0.3 and so two solutions are on. 1 ) c *+2 " ck-l]x k = ~ (* + * - 1. VI = 1 + x2 + \ = ££Lo 14. 2 ) c* 0. and Choosing co = and so on. Thus. and --- differential 00 (x - l)y" .Exercises 6. Substituting y + |«* CnZ" into the + • • y > 2)y = OO £ «(n GO CO £ £ nci" . two solutions axe yi 1 3 1 a i = l+i 29 + -x + -x 4 + 15. Substituting y = ££Lo — 11 Cnx" into the <. For 1 and Co = C2 = co Cft+2 = m ci = Ci =0. C4 5.n=0 n=l . . .. 1 oj 1 Cs = 11 5f we obtain = . .E ncn^ + E 1 n=2 n=l n=0 t k=n— 2 fc=n— 1 180 k=n k=r. = ^. we + (^ + 2)(Jc + A = 1' 2 = -. 2. - 3 --- l)' find = C2 and ci = 1 C2 = CO = = c3 1.l)^"" 2 - n=2 t=n k=n-1 OO - 2c"*" k=n oo £ - 1)<*+2** £ n=0 fe=n+l CO = £(* + 2 >(* + • we have differential equation CO _ xy _ ( T + " = x + \* 2 + \* 3 + \* 1 + V2 • E *^^ fc oo c*-i^ fc " E 2c*^ oo = - 2c2 2co + + 2)(* + £[(*. 3. C5 = i and so on. Thus 2c2 (ft + 2){k + - l)ct+2 (A + - 2q. Thus.. l^x"' n=2 2 - 2 £ n=l W 1 CO OO "1 + £ ncx" + n=0 £ n=l ^ r k=n-2 k=n— 1 OO fc=n-l OO = £{* + l)kc k+l x k A. Substituting j/ = (l and C% = 6. Thus -2c2 + -(* + 2){k + l) CJc+2 + (k = Q) - — l)*Cfc+i (A — l)qt = and Ck+2 Choosing co = and 1 ci = we Co = and = c\ 1' 2 3 ' -" find C2 and so on. For *- -JT2-(k + 2)(k + l)' 1 ff = C4-0 C=l \.t + 2 + (A: CO £ - fcct Z* - l)fcct +1 (k + £ c*x* - l)c k ]x k = 0. P + 2 )( + fc l)t*+2 - (* + l}cfe+ i + (* + ljefclar* OO £ *c x* + £ - 0.3 00 £ = OO (k + l)kck+1 X . + i* 3 + ---)+Q i: c and The initial — —2 conditions imply C\ = -2 V 16.l)^"" + 1 ti=2 OO £ n(n .-2c 2 + + k - 00 £ E H* + (Jfc + 2)(fc + + 2)(* l)Cfc +a i* + l)c. we obtain c2 = Ci(l + ^ = 3 = C3 = C4 = - Thus. fc=0 ck x k .Exercises 6. so + ^x 2 + ~x 3 + )+ 6x = 8x - 2e z . 0. 52%Lo Cn 1 " in*o the differential equation we have (z+l)y" ~(2-x)y' +y OO OO = j>(n .=n OO + £(* + 2)(* + ljct+ax* ~2^(H fc=0 fe=l fc=n OO l)cfc +1 i* + t t=l fc=0 CO - 2c2 - 2ci + co + Y. = Q= 0.)-(x + ^-^ + -) _ 2 x ..x z + (l ) . xn i nto tne differential equation we have Y^?=o °n 2xy' + 8y = £ n(n .l)^"" 2 - 2 n=2 £ "ex" + 8^^" n=0 n=l ( ^ ^ oo = £ k=n k=n k=n— 1 oo (fc + 2)(fc + l)c*+3S* - 2 £ oo fccfc** +8 fc=0 £ c t x* fc=0 GO = 2oa + 8co + £ [(* + 2)(* + l)c k+2 fc=l Thus 2c2 (fe + 2)(fc + + 8co = l)c t+2 + 182 (8 - 2A)cfc = + (8 - 2fc)c fc ]x i: . Substituting y" - + ^x 3 + —2 and t?2 ) + C2 = — 1. + 2x . 2r * ^+^+ . y = C (l-lx 2 -lx 3 + ±-x 4 1 12 + --)+C2 (x + x 2 -]x 4 + --- and j/ The initial = 2 Ci (-x .Exercises 6.3 Thus (k + + 2)(fc 2c 2 - 2c t l)c t+a - (fc + co = + + ljcfr+l + (ft l)cfc = and Choosing cq = and 1 C) 1 C2 = ci Ct+2 = ^ = we ..-co "^ Cfc+1 Cfc * ' = 1' 2 3 - find 1 2 and so = on.±x conditions imply C\ !/ = 17. For cq and ci = we obtain 1 C2 and so = C3 1. on. Thus.-. so = 2(l-^-^ + l^ + . equation we have differential n(n . conditions imply Ci y 18. 00 00 n=2 . - (l n(n .3 and C2 C* = -4co 0fc Choosing co = 1 and c\ — we — 2k +2= + 8 Ct 2)(fc+l) * ' = 2 3 1' ' find C2 = —4 C3 = C4 = c5 = = c7 = cio = eg = • = • 4 3 = eg For cq — and c\ — 1 eg = 0. we obtain C2 = C3 = -1 C5 = C4 = • = I5 and so on. Substituting y — T. Thus.l)c„x"" + £ 2nc x n £ n=2 2 1 2=1 k=n k—n-2 k=n 00 00 = £*(*fc=2 l)c fc x fc + £ 00 (fc + 2)(* + l)c* +2 x fc=0 fc + £ *=1 2fccfc x fc 00 = 2c 2 + (6c3 + 2<n)x + J^[k(k fc=2 + l)c k + {k + 2)(k + ljc^+a]** = <> .Exercises 6. s/ = C (l-4x 2 + 1 (x-x 3 + ^x 5 +--) ^) + C 2 and = The initial (-8x C.l)cnX n + -) so + ^x 4 ) = 3 - 00 (x 0.%Lo Cn^ = + ^x 3 ) + C2 3 and C2 = 3^1 - n into Ax 2 = 2 + l)y" + 2xy' = £ 3x a + ~x 4 + 12x 2 4x 4 + . ^coj x 3 ) H j = 2c 2 + (6C3 + ca)x + (12C4 + ci)x 2 + 184 (20c 5 + c2 - 3 ^co) x + = 0.^x 3 + ^-x 5 * = x 2 6 ) 120 i + 20c5 x' + + cox (co + ax + ca x 2 + + CiX 2 + . = 0. .. 3.3 Thus = 2c 2 + 2 Cl = 6c 3 k{k + l)cfc + 0./ The initial conditions imply qj = Cl = (l-x 2 + and = Ci 1 19. Substituting y y" = £J^o + (smx)y = dii" into the 1.. = For Co and c\ = 1 we obtain 1 C3 -3 c4 = C5 = "5 C7 =7 = ce c$ = = 1 1 and so on.l)^' 2c2 + 6c3 i + 4 -x 6 + --). . + 2)(k + l)c k+2 = and e2 = C3 = "3 C1 a+2 = "-j^2 C^ Choosing co = 1 and = cj we find c3 = = c4 A = c5 • = 2. 3 £ n(n . 4. . {k 0. Thus and . so 1 1 5 7 differential equation 2 12ciX we have + (x.Exercises 6. Cn^" into the differential equation we have („ _ 1)^^-2 + + 6c 3 x + CO + I2c4 x + CiX + cq) + (6c3 \ C2 2 + _ Ix2 + 20c 5 x . ci = 1 c2 = 0.Exercises 6. 1 = =0. For cq = and = "12 C1 find C3 0.3 Thus 2c 2 6c 3 + cq = 12c4 2flc5 = + ci = + c3 - = \cq 6 and c2 = C3 . = = c5 "IS' and so on.^c + ci)x + 3 + 2 + ^ x (l2ci (c 3 ) (cq - + c2 - x3 + ^cq) x 2 + ^cj*j - + c lX + c2 x 2 + = 0. two solutions are Vi 20. C4 = C5 120 we obtain c3 = C4 0. Thus.--CO 1 D C4 Choosing cq = 1 and = ci we = C2 and so on. Substituting y» + j/ = £J^L _ gn = [2c2 + = Thus (2c2 = 1 - ^+ 1 + and J/2 — —x £ — 4 + • - - . c3 x 3 + -) . + + + ' Exercises 6. For co and q= 1 we c2 and = l/' + -3 = C4 0. Thus.cq)x + {12C4 + c 2 . C4 = two solutions are so on. <S and = so on. 21.c\ + Then 2c2 6C3 12C4 + co = + ci - + c2 - ci 186 ' ' ) co + = ^Cfl = - cq)x ^co)x 2 + + (c 2 - - - ci = 0.3 12c 4 + c2 - \cq o = and C2 _ __1 C3 = -TCi 6 = -12^ + C4 Choosing Co = 1 and 01 = we find = ~. Substituting y C°- 72 tne differential equation £)n=o -n^" e-^g^-lKx"- we have 2 n=2 + 1 i ~x x2 \ = [2c2 + 6c3 x + = (2C2 + Co) + 12 Ci x 2 (6C3 hx ~ l x3 * + 20c5 x 3 + — + Cix + C2x2 + 03x3 ]+ [cq + (ci + ci . + . = ~. i = obtain = c3 0. 3 and 1 Choosing cq = 1 and ci = we find C2 and so on.Exercises 6. Substituting y = l- — 2n=o -x* 1 + e x y-y='£n(n-l)cn x n n=2 |2c2 + = (2c2 y2 = x. ] . two solutions are yi 2.co = 6c3 Uci + 2e 2 = + 3c3 + c 2 + -c l =0 2 + + c2 + - [co + cjx + c-zx 2 H 2 ^ci) x + • • = 0.-X s + ci + 1 + ci - co) 2 + [ci + 2c2 i + 3c 3 ^ci x 2 a: + 4ciX 3 + •)- £ c„x n=0 n + 20c 5 x 3 + + ci)i + + 4 we have J 12c4 x (2c2 1 00 \ + x + -x 2 + -x 3 + + 6c 3 x + —x -2 (1 1 = and cn :c " into tne differential equation 00 y" 1 + (6c 3 ( 3c3 + 2c2 + + 2c 2 )x + (l2c4 + ) 3c3 Thus 2c 2 + cj . For cq = and = cj 1 = 03 ~r 1 we obtain + -x* = = c4 6* Thus. two solutions are m= + 23.Thus.3 and = 02 2°° ~ C1 2 1 1 1 C4—4C3 + -C2--C. + "' fc=1 > 2 3 <' > . Choosing and so co = 1 and on. Substituting y V" = I^Lo .xy = £ n=2 1 2 o 1 X ~ X 6 o 2c 2 n(rc + V2 1 X ~ X 2 2 + 1 X ^ 1 6 ~ X 4 24 + Cnx" into the differential equation leads to . For co = c\ = and we c\ find — 1 1 1 2.. o we obtain and so on.l)^"" 2 - £ k=l [(* £ n=0 ^ 00 = £ + (* 2)(* fc=0 + 2)(fc + l)c k+2 - ck ^)x k = 1. Thus 2c2 (* + 2)(* + = 1 l)c fc+2 -<:*_!=() and C2 = 2 Cfc+2= (yfc + 2)(yt+l)' 188 + l)c ft+2 x fc - £ fc=t A=n+1 fc=n-S = = .Exercises 6. 1 -coi- 1 . l)c k ]x k .4(* + k=l Thus 2c2 - 4c = 1 and c2 = ^ + 2co ^^(T^A^*' fc = l.2.Axy .!4.. Exercises 6.3.. + —x + 1 .3 Let Co and C\ be arbitrary and iterate to find 1 C5 = = C2 55 20 and so on. + —cix* + 12 1 c5 + " Sn^=o Cn 1 ' nto the differential equation leads to OO y" — 40 .l)^' .£ 4c*x B £ n=2 n=0 2 Ti=l ^ k=n~2 ^ k=n = k—n oo oo £ £=0 (A 2c2 - + 2)(i + l)c* +2 x* oo £ - - t=l £4 Cfc s* jfc=0 oo = 4cq + £ [(A + 2)(fc + l)ct + 2 .Ay = 1 CO oo n{n .£ 4nc*s* . The solution = Substituting 3/ = co + is 2b 1 cix . dt 2 190 in terms . 2 — d e — = and the boundary. Two power Substituting 26. If t of 13 i 17 4! 5! * 261 6! fi 64 105 409 7! series solutions are „ n —2 3! ^y **y 7 16 2" 64 n = 1 -.21c into the second series gives the polynomial solution — into the first series gives yi = L—x fe <w then -p- dx — —— and de ' dt . + 2 Vf n(u-2)-.(H-2fc + 2) .Exercises 6. The solution 14 +s = C7 1 1 c5 5 3! 4 13 6 4! 4 17 7 5! 13 +2CO 4! 16 17 16 8 261 4 409 64 is (5 /261 + a^y 4 \ + Qi /409 fi t^y - 64 \ + .. 1 — <Pe -p-^ dx* 2i j/2 = z> whereas substituting 2 .3 Let Co and ci be arbitrary and iterate to find + C2 = 2 C3 = 14 + 2c° = 3 5! 14 + C! 3 3! 14 11 + C2= + +2C0= 14 4 = C4 C1 4 4! 2 4! 1 14 and so on.value problem becomes at* t. 15 1 25. For co = and ci = 1 C7 = Cg = C2 = ca = we obtain A2 C5 = C6 = A3 C7= 504 CS = C9 = and so on.3 Substituting B = ^ Cnf into the differential equation we obtain A2 Choosing cq — 1 and ci = we find c2 = C3 = "T = C5 A2 <=4 = A4 and so on. Thus C2 = and .Exercises 6. Thus Now or the boundary condition S'{0) — implies C2 = 0. . 5. Regular singular points: x = —3. . collecting terms. Substituting y 2xy" -y' = X^o + 2y= (2r 2 cn^" - +r 3r) 2. 7.. For r ci For r = - 3/2 the recurrence relation Cft and The ci general solution on s (0. is w -2co. = 0. Irregular singular point: x = 5. Regular singular points: x = 0. and into the differential equation which implies The = C2 = ^. Irregular singular point: x = 0.) +p l ^(l-|.).4 Exercises 6. c2 2cq. Irregular singular point: 11.. 2. Regular singular points: x 3. x = 3. Irregular singular point: x = —1. 9. ±2i.Exercises 6. ±i. = 1- 2 3 - Cs -- =- 9l5 D0 - is -Oi( 1 + 2 . 1 coz^ + 2r and (A indicia! roots are r = and r 2 - 3r + r){2fc + = +r- £[2(ft = - 2r 3)cfc = 3/2. = c3 4 -co- is = t -(2TT^' = oo) - + 0. —3. 3.-tf + | a! - + . Regular singular points: x = 0. Irregular singular point: x = 1. Regular singular point: x = —3. Regular singular points: x 10.1 ( 0. 6. 4. the recurrence relation «-m?*jand 3) = + r)c k - l)(fc - r(2r ±5. 2.4 1. + 192 ^-^ + . (fc = + r)c + 2ct _ fc 'X - - 2cfc _i k = we obtain 1 ]x* +r . Regular singular point: x = 0. 8. Regular singular points: £ = 0. Irregular singular point: x = 0. Regular singular point: x = 0.. —^co...Exercises 6. C4 = ^co. is .4. 5{k we obtain r + r)ck + c fc _ 2 ]x k=2 = 0.jr) co^. + 5) c\x 7r l) Ck + + 3) = 0. which imphes and r I (*»- r(2r + 7r + 5) ^( _^ + ^ + ZSJLo Cn£ +V= c2 2 = + r)(2k + 2r + 3)c* + c —3/2 and C* 3r 4r i(/c 2 7 — -r — f r ( 4r — 7\ = -J + r}(8fc + 8r . = gjgCo.7)q + Cfc_i = 0. we obtain ..4 12. which implies 2r 2 + (2r and The (k indicia! roots are r = and C2 For t = = and The general solution on . n+r into the differential equation and collecting terms.^.1 + £ + r)(* + r - [4(A l)c fc 0.)...3. —3/2 the recurrence relation is ^2. cs = 0. +j ^. -(2T3tjP -^co. + . = 0.1 . = = 00) is = 0... c3 = 0.. Substituting W = the recurrence relation + j/ \v' = Ci (0..) + ft l ( . Substituting y = H£Lo 2xy" CnX n+r + 5y' + xy= and into the differential equation (2r 2 + 3r) cox £ P(* + + r_1 + (2r 2 + r- r)(k + collecting terms. 13. + + r)c k + c fc _i]x* +r . so c\ a = ci . fc _2 = = For r 0. 2 (2r - + r) ci = ~ 2r so ci 3) = + 0.1I + I | 2xV .(i-i. l}c k t 1) ' is . . we obtain differential - 2r2 ( ^. -^h)> c3 -~co.3. For r _ " Cfc and = cox r + + E[2( + r )( k + fc (2r 2 + r) - - r r+l t CiX*" + 0^ + (* cfc + ck . rt- + 7)*' (8* C2= 4 -^ggCO- s| „ -. 7/8 the recurrence relation = 7/8. 2ct-i = c3 _ .) Cn£ B+r into tne EfJL = C2 -^> oo) i.. 2. is Jfc(8fc-7)' — " Cfc the recurrence relation 2 £fc-l -2co.. 0.. 0i 1 ( 14. = - l)(r = 1/2 the recurrence relation = = c*-2 "*(2fc + fe 1)' 194 = 2 3 4 * 1 i^g^ is Cfc 0.. which implies 2r 2 - 3r + = 1 2r and The + r) (2fe + [(A indicial roots are r = 1/2 and r Cfc and For r c2 = 1 = the recurrence relation = = 1. + l) V _ is = + 3r l) l . 2 ]x k+T k=2 = o..4.. C4 0..Exercises 6.-^.5 s? + ^/. + ct _ 2 = For r = = 0. + . q '"^'---' 1 C0 = Ca ' ~m5 C0 - is .4 The = indicial roots are r and r ci For r = = and The ci general solution on „.zrf + (i 2 + o i i 2 -co. Substituting y = = (0. + .. equation and collecting terms.). .cix«»(i-^ + 1 L^ + .). is .. oo) = c3 -tttco. c The - 1 = 1/3 the recurrence relation and r + r)(3fc + 3r - and and For r -i I 0.... Substituting y = 2\ -r+-)cffL T + ^ r '52 [(* + r)(* + r 0..Substituting y + (0...). which implies indicial roots are r 360 co^"* £ P(* + + The I into the differential equation (Zr = r^Q>. = 3xy" + (2 n+T Z^Lo CnX -y=* - 2 - r) 3r and (Ar — ci = = r general solution on .) & C3 1 - + ftl . / 2 \ = we obtain is ^^ (2 x-Q)y=(r which implies 2 -co.. we obtain + -c* . ^+ = X)^o ci - 00) is » + L«./. 10 = c4 0.) + a. + ^ + 5 ^» + crt x" +r into tne differential equation and collecting terms.( 15.ck ^ 1 .. + + — — co. s and collecting terms. r 2 -r+\ = (r — ^) (r - ~"\ = - 2 l)c* . r = + r)c k + 2(k + r)c k - — l)c k - = = c2 — r(Zr 1/3. For r + r)^.3.. 1.( 1 + "j^co- J.4 and The C2 general solution on (0...Exercises 6.2. 1 - . ot 1 ( 16.. 1 C2 -co. the recurrence relation 1 = c3 JqCO' — is 1 co. — ifc . + .]^" (k 1) (fc = + r)c k -i = 0. 1. . [2(ft + r)(ft + r - we obtain 00 2xy" - (3 + 2x)y' + y= (2r 2 - = and The (ft indicial roots are r = 3{fc 2 r Cfc = - For r ci = Ck - 5/2.. . + 5) C2 4 = ^CO. 1/3 the recurrence relation = £*_i 2/3 the recurrence relation ft . Exercises 6. 2r which implies - is 1 . +k 3k 2 ~ Cfc l60 general solution on {0. For r = c* and = — is CS = C3 = 560 OD - Co - is and The 0.2. 6 is 2{ft+l)ct-i -co.4 and The + r)(fc + r-l) + -j <fc indicial roots are r = 2/3 and r Ci For r — = = 1/3..-CO.2. oo) is 1/3 17. C3 = . the recurrence relation 3) Cfc _! - 5) o fc(2fc and r(2r - = ^o. 5) C2 - + = 2r - 3)ct_i = 1.3. . 196 = 1.. For r = (2k - ci = £ + 2(fc +r- — (2* l)ct _! = .3.. 2. ^CQ.jjx^*- 0.. C3 = 32 693°°- l)c ft + ct..... 5/2 the recurrence relation ft 5)cfc fc(2ft and r-1 + r)c - — 5r = + r)(2ft + 2r and 5r) eo* 0. 3. Substituting y = E^Lo +r Cn%" into the differential equation and collecting terms. 1 -rco.. .4 The general solution on . * . $( k + ifc=i r )( k +r- + 2cfc + 9(* + r - l)c*_i]* is ._ Ck 1 + ( 18. we obtain V + T. is = ~C0. Substituting y 2xy" oo) is |.9r + 2) cqx t equation and collecting terms.4) -. C4 = j^<*. we obtain differential - 1 (r 2 + 2r + + £[(* + r )( + r ~ fc !> c * jj) + Cl z (* r+1 +r jt-2 = ) c* ~ 5* + Cfc-a]^ y 0.. so ...Substituting y 9a: V (0.) +ai w( f) q.4. = For r = —2/3 the recurrence relation 2... (x 2 CnX n+T into the - = y (r 2 .Exercises 6. which implies T z + 2r+-)ci=Q. is = J^Lo CnX n+r into the differential + 9x + 2y= for 2 .3.0. and The indicial roots are r = —2/3 and r ~~ 3k(3k and C2 For r = -jCo.^_^ + — ££Lo + xy' + (0./ + + | I+ + equation and collecting terms. 9 128*°- C3 = °' C4 = C3 = 0. 2/3 the recurrence relation and The = C2 general solution on 19. oo) a 2/3.. ~ C3 7 ~Y20 C°' is 1 --co. is into the differential equation = - 2 4- (3*-2)^ " 2/3 the recurrence relation and The = 1/3 and r and 9r + r)(k + r - Ct For t - 9r 2 which implies (2r +r- 2 and collecting terms. we obtain r l) cox CO + = and The [(jfc indicial roots are r = -1 +r- 2 and For r ci = 1 = (2r - + r)(2fc + 2r + 1) = r _ and + r)(fc + r - [2(fc l)cfc + 3(fc + r)c k -ck + 2ck _ 0. oo) ZJJLo cn i Tl+r + 3ay + (2x = (3r - l)(3r 1) + 2}c k + 9(k = 2/3. = 2) +r - *- fc(3fc-l) - gco. c 2 == —5 = c3 cq. For r c2 = l)cfc _! = 0. = 1) + 3)' 198 fc- 1 " "*'---' 2 is l ]x k+r . 1/3 the recurrence relation 1 ' 2 3 ' is '"- ' 1 1 --co.4 and The [9(k indicia] roots are r = ci = = ci general solution on 20. -— 1 cq.Exercises 6. For r l)(r + Ije* + 2c = -1 <= 2 fc _i = the recurrence relation = -2c C3 . _ 2cfc_i 2co. is C *~" *(2* 0. = 4 qCq. 2r which implies £ k=l = 1/2 the recurrence relation 1/2. Substituting y 2 2x y" = = (0. x{x - 00) (0. + l]c fc For r 1... t + T){k + r - + r){2k + 2r - and = l) cqx (2(* Ck For t -— 945 0. Substituting y collecting terms... which implies The r E + = - C3 Co.2.3. = 02 5 The general solution on = 21.) +CM ( 1 into the differential equation + y'-2y= (-2r 2 + 3r) cox + i. is ^ = 7^7. 2k Ck-l c2 = ' c^o. is 00) = C. 2k + 3 The general and l)ck k= ' -co. c3 = ^co.X^(l + i. 1 = ci 1 cq. relation is 1. c3 = ^cq. and V solution on x(x - (0.4 and c\ - --co. - +r+ .3. + 22. * = 1. Q-^co.Exercises 6.. is n+r [2t — 35 ^+i I3 + .. 2)y" = we obtain + (k + r)ck -c k -(k + r- l)(r - 1) = l) C*_!]x fc+r Y1%Lq CnX n+r - = (A. Substituting y 2x 2 y" . +r— l)2c fc _! = —1/2 the recurrence 0. + ^ + 3L I 3 + . 1 1 -co. Yl^=o Cn% l)y' -y= into the differential equation - 2 2r and \{k indicial roots are r - = —1/2 2 - - r and r the recurrence relation = = = 1 1) (2r - ~..2. ) we obtain r~l CO + + ifc=0 +r- l)c* - 2c* (* l)(2fc + 2r- l)ck+l ]x k+r ' .. and collecting terms. . oo) Y. -2co. the recurrence relation and 13. x - c2 2co. = = --co.1..4 -2r 2 + 3r = -r(2r - which implies and The [(fc + r)(fe + r- indicia! roots are r = - 1) — - 2]c t j ci +r+ = For r 0. ci general solution on Substituting y = xy" (0.. 2.2. 0.Exercises 6.. r 2 = —1.. For n = the (k indicial roots are ri = and c2 = C3 = C2" = gjCo C5 = C7 = = CO (2i^TIj! 200 - recurrence relatic . + r)(k + r + l)c k .%Lo cn x" + 2y' +r -xy= =0. = _ C2 c°' 03 ~ "i^g 00 ' is Ck+i The + 2r - 1 32 For r l)(2fc ill 3/2 and r 2k- and (fc = 3) = k-2~ = ^k _ ct = A: .. is into the differential equation (r 2 + i r) c r" 1 + (r 2 and collecting terms. l)cfc+1 3/2 the recurrence relation is *»0. 0.c fc _ 2 ]r Jfc=2 which implies r 2 r and The and +r = 2 + r(r 3r + + = 1) = 2) ci 0.c k -2 = 0. + 3r + 2) ax + £[{* + 0(* + r - + l) CJt 2(fc we r + r)c* . 1.. c3 = 0. so ci = 0. For r*i 1/2 the recurrence relation is . oo) is OO i 1 „to<2n + i)r = — [Ci sinha: + C^cosha. + £[(* + r)(fc + r - and collecting terms. = 2.±) + (r a + 2.1) A-2.. and C3 = C4 = = C5 C7 = = OD 4i = C2n (2n)! The general solution on (0. which implies r 2 + 2r and The ^ + Ck indicial roots are ri = 1/2 and T2 = —1/2.1) y into the differential equation = (r 2 .. 1) ci = 0.. 24..Exercises 6.4..* \w *(* .. Substituting y x = S^=o %£™ +7- V + V + (V . + Cfc_ 2 = c\ £ 0.3..].3..4 For r2 = —1 the recurrence relation is <*= J. we obtain + |) C^+l l)cfc + (A fe -c* 4 fc=2 = + r)c - + c*_ 2 ]z* +r 0.4. so *(* + . = = 0. %Lo Cn% 1^2 [Ci sin x + C2COSX]. Substituting y x(x-l)y" = x T. oo) is x 2n — 5.4 and c2 = -jjjco C3 = c5 C4 = = = c7 • • • = C0 ^! (-1)" C2n= For T2 = — l/2 the recurrence relation C0 - (2^TI)! is °k ~ 2 c= and - l C2 — — ^rco C3 = C5 = C7 = C4 = 2 3 4 2! = CD 5f The general solution on (0. Ar which implies and equation and collecting terms.2c k ~i}x k+r = 12)c t _! - (k + r-2) -2]e _! = +r- + r)(k + r - 1 0.Exercises 6. n+T into the differential + 3y'-2y = (4r - r 2 ) cox ". l)q . we obtain -(fc — + rJffc + r.^cjj + r 2 — ttAr r(A — + r202 r) l)(fc = fc 0.1 1 + £ [(fc +r - + 3(k + r)ck . is + ^x + ix 2 ) + Ci + (x 4 + 2x 6 + 3x 6 + Ax 1 + ^ + ^ )+C n=l f nx" 2 6 6 2 ' +3 ._i =0. =s- c3 = . . .6. - fc = 1. . 2.4)ct + = r<i the recurrence relation . 00) Ci (l =C 1 ('l K ci = c2 cs = 2c4 ce = 3C4 C7 = 4c4 = c3 = . 3. = -co C3 = C4 = C5 = - - = 0. .= Taking Taking The cq cq ^ = and and q= y = ci = + C3 = ( oj is arbitrary *-^ = . we obtain C4 ?^ general solution on - = + 0c2 = Ocy and 2cq (0. . .3)cj. . 2. 7 we obtain c. .The = indicia! roots are ri = 4 and ti -fc{fc For 0. or — - 4)c* + - (A: 3)e*_i - 0. Then 3ci - 2c2 ca c. k(k k = 1. 3. 5. Substituting y = y" + X)^Lo £n% -J n+r into the differential equation and collecting terms.2 + (r 2 + 4r + 3) c.)' 1 /. the recurrence relation is .. we obtain -2y= (r 2 + 2r) cox'. 1 9 7 4 19 6 \ .4 26.Exercises 6. =0 2) cj = - 2cfc _ 2 For r\ - = 0. so c\ 2c fc-2 _ + r(r 0.fe f _ dx (1+1^ + f ^ ^ + r 204 ..^" 1 fe+r-2 fc=2 = 0.-234 and = -co C3 = C5 06 The A 1 c2 = C7 ^li 00 = = - ' result is second solution is „-/(3/*). which implies r 2 + 2r = (r {k The indicial roots are n= and 2 + 4r + 3) + r)(k + r + — r% 2)c k — —2. we obtain + r X* + r " [(* ^ + r)ck -(k + r)ck - + 1 ]x k+r . = 0. 00} 19 -x* 2. = xy" + (1 £?|Lo .Exercises 6.i E + indicial roots are ri solution = f"2 = +C2 y2{x). 00) 3. = + r)c k +i +c k = Q. 1 3 second solution cq (l = + x + ^x 2 + ^x 3 + coe*. = we obtain 0.2.1 and the recurrence = £ Ti * - = relation is 1.304 . . C\y\{x) differential equation and collecting terms. ck One 4 and (*: The 19 2 is y 27.7 1. Substituting y .x)y' -y = = which implies r 2 = r I "+r into the W c.4 f ( 1 The 7 1 general solution on 19 1 7 2x 2 96 (0.3 is = yi A + + r) a Cfc-(* + r)i*_] =0. is .-±-x 3 + L The i*'-s*' + Cn^" = -\ e x lnx - e x J + T ^ ?-s** + -)' K 1 n-nl . 28. Substituting 1/ = E^Lo xy" -)*-'/(j- + -±-x 2 . is = +r 3-3! 1 Cie 1 + C2 e x [lux - £ J 1 n into the differential equation + y={?-r) ^ +r+ + \){k J" n\ and + r)c k+l + ifc=0 which implies r (fc 2 -r= + + t- l)(fc r(r - 1) collecting terms.-/(i/*-i>fa »2 = yi IS y -^/i('-* + = e 1 Ins - x general solution on 2-2 (0. x n . ^co C2 = 03 = C4 = 3T2 result second solution _ 7 /... A 2 =C m +C c„x" +r into the V"" + lnx + 2 yi differential 1 r ~x + ^x + £[(* + r )(* + r ~ ^ + (* and indicial roots are n = r2 = + r) 2 c k + c k -i = Q. we obtain oo (k The - is yi (x) xy" . W is -*J J-J + foa*- The 00 is 1 A is C0 'm 5!4! The the recurrence relation 1 * 206 = 3 is +r ) c* + c t -i]x fc+r -1 - . 2. and the recurrence relation ^-^fi1 1.. equation and collecting terms.)..) 38 \ .. general solution Substituting y 2 ^ dx f f 1 9. r J ^-^ / 1 7 1 + .Exercises 6. 7 . 1 / = S^Lq + y' + y = which implies r 2 1 1 3 = _ + .. ' 2 + . and ci = .- 38 1 4 5 19 dx ( yi ^(1-* + J \ .4 The indicial roots are n= and 1 — — For r\ 0. lr^' which implies and 5 . {k l)c k = 0. * + VI oo) /„ I Substituting y — xy"-xy' + $2%Lo Cn x y= (r 2 'i+T - r) c =C r"1 a: The indicial roots are n= 1 23 — x .6. + 2x £ 2 — [{k r +r+ — r(r — l)(fc 1) + r)ck +j - + r)c k +i -(k + r- = For ri = 1 + r)ck + ck ]x k+r = = l)(k 0.. 1 4 .(l-^ + H-l^+H**-. r% we obtain 00 + r {k + tne differential equation and collecting terms.4 One A solution is second solution is = VI — = — ! V\ — dx dx .j [ = The VI In general solution on (0. + l^r>x* + +r+ and 1 y l (x) + C2 y2(x).) / 5 Z 1 [. . is y 3. the recurrence relation is . _ l2 + 2l + 677 o 23 .. !riI yi ————— yi / f VlJ -/J = = 23 i 23 2 677 5 „ 5 . i 677 . \ 3 _^ + 677 + _. __ x3 + ..Exerc/ses.. C 1 1 2 3 is " +r +V= 1 .. second solution is dx 1 f ( 1 1 i The 1 1 1 2 general solution on 1 oo) (0. is 1..(r(k second solution + r- e- -^.2. = l) is yi A 2 l) c k l and the recurrence relation 1 ck One -2r + + r- [k indicia! roots are r\ 2 yi \lnx +x+ general solution on (0. j/i J > is V = Cm(x) + C2 y2(x).^x 3 + /XG X ^-^dx = y f(l-l/x)dx 1 + ^x 3 + +x+ etc 3! The 2 - = c xe" is .3. 1 . 208 1 + 1 + 2* + 1 3!* 2 + dx +r . 4 y 1..Exercises 6.4 and one solution is = yj A cqx. into the differential equation P+ + 2r and collecting terms. -x* oo) + —x 1 J + = f 1 1 x j -e dx . 1 3 \ 2 = T^Lo Substituting y xV + ~ cn .. = C\x + C2V2{x). = = l)ct _ 1 *= = 0. we obtain l ) £P+ +v- r)(k l)c k -{k + r-l)ck + (k + r- l)c*_i]z* k=l = which implies r and The = r% = solution + . = cox(\-x + ^x 2 . ) = M /(i-2.3. l) ci + r) 2 c k - — Vl = 0. + §*»-fx' + .Exercises 6..)« fa = n The 0.ic^x^- 1 *=2 = which implies r (r and The (ft indicial roots are n = ri = 0. ft = relation is 2.™=o cnX n+r mio tae + y' ~ 4xy = 2 r cax r ~1 differential + (r EP+ + 2 equation and collecting terms.. 4 = ax 5 l(1+2ia+ yi £ ~ I WK'-^r -?* +-> J +tl .. we obtain + 2r + r)(k l) +r- cn r l)ck + (k + r)ck . + - 6 gx 6 + ] . + .x2 ^+ u is j/ = Ciyi(x) + C2y 2 (x).4 32. One A solution 2r = + 0. and the recurrence ^. is I W = = is second solution 02 = + ~ so c\ Cjc 2 2 yi In x + yi [-x + general solution on (0. oo) 5 A - 8 23 [in x . Substituting y = xy" Y.. 4 — 33. yi = = and c%x 2 . Substituting y n+T into the differential equation and collecting terms. 2. 34. 1 T y x + 4i 1 x a ) 1 2 ^ln*-. 3 Then = => cj = = = ci = + 0c 2 = = c3 = - -ci — 0c2 3c 3 2co ci -2co and C2 is arbitrary and Since ci Thus. + 4r + 3) l)cfc - (* we obtain r+2 c 3 x'"' + r)cfc + c k ^)x jfc+r-1 ..+ *--^ + -*<-.Exercises 6. = A // 1 have co second solution ^— n = ^7 . Substituting j/ = EJJLq Cn£ xy" -y' +x 3 n+r y = into the differential equation and -1 r (r 2 - " 2r) cox 2 + (t + £ + [(A 1 + 2r) c 2 3: (r r+l + r)(fc + r k=4 -0 210 2 - + - l) c : x (r 2 collecting terms..1 = which implies r 2 = and + r)(k + t - (k The indicia! roots are ri = = T2 fc(* + 2)c k (k + r- and the recurrence " 2)c fc + (* - relation =0. = — 2co. we obtain Yl^Lo <^x + (x. 3)cfc_i 3)ct _i k = 0. is = 1.. 11. C2 = we obtain eg = C4 = C5 = - - is ^ J 1 dx ~ \ 1 x j ^V~ X+ 2 X 2 .IV - xy" 2y = r 2 ^ - 1 £ + [(fc +r- + + r)ct + (* + r . we ct dx =x 1 1 = Taking 0.3)^_ (* l)c fc I ]^ ft+r . is and ci arbitrary we find C3 = c$ = C5 = _ * = 24 C2 A second solution = C10 = C8 = C9 1^ C2 = - is tin y2 The C7 general solution on (0..Exercises 6... {r r 2 + 2r = (r and (A The indicia! roots are r C2 arbitrary. fc _4 Also. 0. 0. r = 2. 0.. but when r = 0. oo) = fr. ci = 0. when 4. c 2 \x / 2 — -±x 6 + -±-x 10 . = 2) + 2)c2 = r(r + r)(k + r- = - Ciyi{x) + C2 V2(x).920 is y = 0. = +c 2)ck 0.5.6. . For r is = = 2 and r 3 = cq arbitrary and = C3 the recurrence relation C2 = we c3 0. = * 1 One solution Taking co = = Cfi eg = cio = C7 = — en — 0. r 24 1. find C5 = is «--*?Sri)' Taking = + 4r + 3) so cj 0.4 which implies t - 2 2r = r(r 2 - l) c. 8 - = Cjx -4 + C2^ (r + 4)(r - 2 - a regular singular point so that oo CO n=0 Multiplying both sides of y" + P(x)y' x + V+ 7i=0 Q(x)y = by x 2 we have 1 x (xP(x)) y 212 + 2 (x Q(x)) y = 2) = 0. c2 so on. Substituting y xV - we obtain +V= if = + [c k (k +r- + r . fc=o Thus r = and the recurrence relation is =k + — Ck+1 *- °k Then and Q =0.8 = r 2 + 2r . Therefore.8cn ]x n+r 71=0 00 = [r(r - 1) + 3r - £ + 8]c [(" + r){n + r + 2) .4 = 35.%Lq ^ into the differential equation and collecting terms. we have + 3r . .8Kx" +r 71=1 = Taking cq / and c„ = for r{r The genera] solution 38.2)c l)(k lt _ 1 ]x fc+r = 0. -(k = y(x) is £ A=l 0. is Cf> into the differential equation = ^[{n + r)(n + r - l)c„ + and 3(n collecting terms. Assume x == is is y - 0. one solution = y(x) 37. we obtain 00 It follows x 3 y" +y= = and that co cox r + Ck The only solution = En^o 36. rcox 1 - we obtain CO 1 £ + + r)(k + r - {k ([ + 1) l]c k - {k +r+ l)ck+1 }x k+r - 0. Substituting y Y. Substituting y = x*y" JZ^Lo Cn2 + 3xy' — n+r 8y 1' 2 - 1 = ' ' i 1 -co. . 3. 2. + r-l)(k + r-2)c k ^. Cn x7l+r into the differential equation and collecting terras. = -co. . c3 c4 = 7 -co. . we obtain + r)c . n= 1) 1.Exercises 6. . \/x.(n + r)(n + r.Exercises 6.l^z"*" 2 + x n=0 o. Exercises 6. we obtain / 00 £ £ p„x" \«=0 \n=0 \ (n + r)cn z»+ '- 1 / / n+r \n=0 / \n=0 CO \n=0 = The coefficient of the lowest indicia! equation is 39.x n+r into the differential equation CO \ / 00 x 2 Y. c\Jb / 2 ( x ) + c ^-&/2l x ) . is 1) +Por + becomes aw 4y = a singular point at co.5 1. the dw (b) Identifying \Tt=0 °- w 2pLi + 2w %Land we / 0. Since v 2 = 1 the general solution y — c\J\{x) 1 = 25/4 the general solution 3. 2.5 *y PnA y' + (f: <hA * = / \n=0 / \n=0 Substituting y ~ J^Lq On. Identifying po The = 40. Since v lojco- 1/3. 2 we see that oo is a regular singular point. P(w) = 2w and Q(w) = — 4/u. obtained when n - 1)cq then r(r 5/3 and go indicial roots are \n=0 = —1 and + porco + qoco = - 1) + po^ + 50 = - is 0. r(r The \n=0 / is is y = + C2Vi(x). Since v'1 = 1/9 the genera] solution is y = ciJy 3 (x) + c^J^^ix). (a) Substituting w— / Vi=0 power of a:. the differential equation see that there — indicial equation is —1/3. 2nx~ n 214 -1 J'n + n{n + l)*""" 2 Jn . 2 = 2 the general solution is y = ci J2(3x) 2 = 1/4 the general solution y = ci ^1^3(63.^ + 92^-1/2(^3. If j/ = f 1/16 the general solution y is is — ciJi/ 4 (x) + + c^J-i^ix).2n)x n J'n + 1 (n J£ 2 2 + xj'n . c\x~ x ^ 2 Ji/ 2 {Xx) + C2X~~ 1 /2 J_ 1 / 2 (\x). - (1 n 2n){x J'n . Then c\J-l 2 {^ x ) j + ni"" 1 ^ z n j. find x-"J'n -nx.n . From y = x~ n J n y' = is = y = x"" = 0. 0. and x2y" + Multiplying by x 1/2 + X2x2y = 2xy' 10. l)x we have Jn + + = xn J% + y" and a solution of the we _1/2 find + 2nxn J'n + n(n - Therefore. C2V2(3x). Since e2 = the general solution is y = ci Jq(x) + C2Yo(x).Exercises 6.Jn l and y" = x~ n J'^ . x n+l Jl V + x^v' + (aV . 2 — 4 the general solution is y = ci + C2l^{i).n Jn + x Jn JZ + xJ^ + (x 2 -n 2 )Jn 1 (since * 2nx n~ l J'n + n(n - n .).x. x n Jn 3 + xv'+(\ 2 x 2 -±)v = Substituting into the differential equation. Since i/ 7. y' = x~ y" l ' 2 v"(x) . = 3/ we obtain 2 x v" whose solution a. .5 4. 2 l)x + n~ l (2n ) v.b l 2 v{x). 2 Since v = 5. + C2^_jy a(6x).2 J„.) 6. FVojn y = is v = x n Jn (x) we y' - x = x n+1 n+l J% 2 [x -x n+1 lx 11. Since 9. ar-V2 w {a.3 ' 2 v'(x) + ~x.) then = x-Wv'(x)-\ X -V%(*).2n 2 )x n ~ l Jn + x n+l Jn ] ] a solution of Bessel's equation) original equation. . Since i/ 8. Jn is 2 + nxn ^Jn + x n+l Jn ) -n + n. with n = -1/2 we find 11 with n — —1 we find 11 — x~ 1 J-i(x).-x-V 2 MXx). From Problem n = we find n — — 3 we find 11 with 1 J_i(x).5 Substituting into the differential equation. we yfx J„(\x) find l = X^Jl(Xx) + -x- y' l'2 Ju {Xx) and = y" 2 + Xx~ X Vx~J'J(Xx) '2 l Jl{Xx) l . = xj-i{x) = From Problem y = A) „ = ^ = -Jx = 0- a solution = n 1/2 = n 10 with 1 we find we 10 with l n = -1 we y = find y + XxJl(Xx) + Jn is 2 (X x 2 . = 3 we we find y ±3/2 we find y = = find *Jx Jo(2x). Substituting into the differential equation. xy" + Therefore. y = x s Js{x).n 2 Jn ) 2 \ a solution of Bessel's equation) is a solution of the original equation. 2 {x we have J% + x J'n + (since (x Jn .Exercises 6. From Problem 14. y = [X -xJi(x). From Problem 10 with n x J-$(x) x 2 J'J(Xx) (since find y From Problem 3 2 of the original equation. 12 with A = 2 and v = 17. From y = n (1 Jn + 2n)y' is + xy = aT"" 1 — x~ n~ l = 0. we have x 2 y" Therefore.u2 ) J„(At)] a solution of Bessel's equation) x x l 2 J1/2 (i). From Problem 12 with A = 1 and e = 18. From Problem = From Problem xJ\{x). 16. + (aV - -JxJ^Xx) 13. x~ 12. x~ J\{x) = -x~ l is 10 with + „2 = -x 3 J3 (x). yfx Jyz{x) and y From Problem = yfx J-z/2( x )' 11 with . From Problem y 15. 1 E ^ 2n!r(l + c + n)V2.Exercises 6.1 + n)rff + n)V2 2n!(e E ^ + f(-l) n 2n!( I/ 1 216 + rt)r( + n)V2 l/ .5 19. ( 1 (_l)n W*) = E / x n 2n+f-l we obtain ^[x"7. (-l)"(y . ^nirfl + (-l)"{2n y + y (2) + n) + n) V2 /x\ 2n+( ' f) + n) V2 20. Using •M*) = = (-D E {2n n 'x\ 2" + + v)(-l) n ". The recurrence relation follows from -vju {x) + xJv -i{x) = -J2 00 2n+i> (-1)"" 2n+f-l (-1)" n) (x\ 2n+ » (-l)V = -E = E + ^Jn!r(l g.(x)j i^-VvM = z"Jj(s) = + i>)(-l) n / x\ 2 " +t -1 *"E n)V2 ntj2n!r(l + + + ' (2n i/ 2n+u (-1)" + (2n „tl2nl(y f)(-l) n /asN 2 ^"" 1 + n)r(i/ + n)V2 - 2n+i/ „(-l)»2-i (I)" (I) E ~ + (2n i/)(-l) n /x\ 2 " + ". Using Problem 25 with n jx n n j x -\xJa {x))dx = j x ^~{xJ^(x))dx n- . 25. in JJx). relation follows from recurrence relation follows from Example 3 in the text and Problem 21: 2Jl(x) 24.5 _ Alternatively.(n- = x = x n h{x) . 2 Jo(i) - 4 - J I) 2 / i""V xJ (x)dx 4xJi(x) + c. 23. . Using Problem 20 and 24 and integration by parts Jx n J {x)dx = 26.y r -- \2vJ*(x) By Problem 20 By Problem 19 .Mx) —Ji(x). .2J„ +1 (x) = \xJu+l {x) rJ (r) dr = J„_i(x) = xMx). (x) <*c - J„+i(x). The pf"^- x". An integrating factor for this equation The recurrence 22.Exercises 6. rJi(r) r=0 J-\{x) and by Problem 22 (x) = J-t(x) . so is is 1 a linear first-order differential equation — ax v \x J„(x)} — x"J ir -i(x).2J„ +1 {x)\ = - -^-fxJi(x)] dX we obtain = xJ J'n {x) 2J so that J'q(x) = (x) so that [* JO = + xJ„-i(x)] .(n - - xVi(x) + — 3 we have l xJi{x) (n - 1) I) ji jx n-1 Jb(a=) and Problem 23 we have J (x) xJ]{x)dx (-Jo(*)) -1 lja:" 3 dx n-2 - (« = x 3 yi(x) + = x 3 Ji(x) 2 + 2x Ja (x) - 2a. (-1)" that the formula in Problem 19 we can note 21.Mx) = JSC*) . z:) pZ /cosx . j 30.5 27. + xJ_ 3/ 2 ^H™(^ . .7_ 5 / 2 (:e) = xJ_y 2 (x) !(l) \ . By Problem 21 we obtain . we obtain -SJ_y 2 (x) = xJ„i/ 2 {x) . . ^ . (^) so that -*t*w--i™{— +sm v- ZJj. By Problem 21 we obtain 29. / . (n-l)!22"-l J + +COSX the function 35. so that 34. — .— J Exercises 6. hiiW 33. 2^1/2(2. = i~"J„{ix) then S/i=i-" +1 218 ^( l x). 5— - 2 — s 6sinz + xJ_ij 2 {x) — /— 15cosx \ so that 15cosa. By Problem = . - so that 3sinx + xj3 / 2 (x) 15sinx feh / + xJ_ 5 / 2 (x) /3cos:r 2 we obtain — 5. — 2 . By Problem 21 we obtain By Problem 21 By Problem 21 21 xJ^j 2 {x) + .J_ 1 / 2 {x) Ji/ 2 (a:) = xJ 1(/ 2 .) so that /3sini 2 / we obtain 5J5 / 2 {x) = ^ so that (. 3cosx \ a: 7)/ 2 = (x) /15sina: /15sina.ji{x) JT 31. If yi = Iv{x) \ - -^-^ {-l)"i 2n fx^+" 6cosa: ^. 32. Since 2 we obtain = xJ3 / 2 (x) + xj_y 2 (%) 28.+ is real. Since -viC \ 1 \ SiQX )- . V» mK x4 x (I = M*)lnx + (. (l- (a) ^(w) + Jo(x) then using equation (35) on Page 299 in the text gives V2 39. . x2 = T . . " (-1)" (-l) /S^'" (-1)" n^n!r(l-m + fl) W m J (a.Jj»!r(l-m + n ) V2> = 38. m m we have ' (1 + i + . and the general solution is Jrn^ dx dx dx j f . the differential equation.(x) = 2 = i~" [(w) i"J.y (ix) = Mx) = Mx) satisfies + {«) a ((«) 2 .). Using (8) with v 7 =m ) T + 3x* — — 128 23a: 5 23a: . .5 and *V + aVi " (* Similarly. ." ) M*c)] = i " = 0. . Jo(a:) In a: hx 3 x2 5x 4 2 +— 4 37. If = t/i + VI = I-v{%) — + C2l~v(x). 2 — 16 (231a: 8 . 6 --~ + x6 x* 11a: = -£ d r \ 2 fx + ^_ _+_+ 5a: 23x 6 . . . a: . y 36. 6 13824 flN*"".315x 4 + 105x 2 - 5)' =(-')•«. Using (7) with f x2 xe we have ^.Exercises 6. t/2 ciIl.) (-5) i- i r£ ^ + i t .| (-i ) Using the formulas on Page 315 in the text we obtain P6 (x) = and 4 4- . J. z 2 )^| + (~ 2x )^. (\t\<i) 1./2 = we have . By are shown the binomial theorem [l + + 44. 42. and sin#S( + cos(?^( do That do + n(n l)(sin0)y = sin# -cos 2 S) 1 ^ . and Pi{x) = x = (1 . 3t 3 + . - 7 693a. 16 P6 (x) (b) 40.t Exercises 6. 3 - 35a.5 Pj( x ) = A {429a. Letting (t x 2 -2xt)y = 1 in (1 (1 - 1/2 2t on Page 316 in (18) in the text. .2cos$^ +n(n+ l)y = 0. The polynomials 43.x 2 ) y" - 2xy' + 56y = 0. 5 + 315a. cos# then dy Sm0a . + n ( n + = (1 - l)y x 2 )y" - 2xy' + n{n + l)y = 0. .) . (\t\ < 1) t 00 = £'" 71=0 From Problem 43 we have n f. we have 2 =l-±(t 2 -2xt)+^(t2 -2xt) + --. . We - satisfies (l 2 a.. y" - 2xy' ) + 42y = d_ If satisfies (l .= l + xt + ±(3x 2 -l)t 2 + ---. 2xt + t 2 )" 1'2 + t 2 ). is. we - see that Pn (l) = have +2t + 2 y 1 '2 = (i +1)-! = i .t+t 2 - = L = "T* i £(-i)V=£pn (-i) n=0 n=0 220 t ". Similarly. use the product rule for differentiation: + n{n + dx 41.. d9^- (fiy 9 sin dy Tx' Q-r-z — cos dx* de* dy — dx . - (1 t)" 1 = jl— = l+t + t 2 + t 3 + .pn (i)t = Equating the letting i (i-2* +t a coefficients of corresponding terms in the — — 1 we (i r l/2 = £t n two series.. 3^ 231 8. - _315 6 2 . recurrence relation can be wrtten ft+iW = 3 k = 1: k = 2: k = 3: w = k = = 4: n— 0: = 1: n— 2: 71 "- ^ C 1 = -?*•-!)-! 35 ort 3 1 Pl(x) = 1^/2 (X — l) = z ft(x) = 0.6x 2 + l) dx = J | and j P${x) dx ^ =j i (25x 6 - 30x 4 + 9x 2 ) = In general. m. 1.5 so that 45.. "2 MkFT k 2 2* = 2. Exercises 6. 4.35 = n #» . All integrals of the form P*(x) <ix = 2^ Pa (x)Pm (x) dx are for for n n= ^ 0. For n 3 "2* P6 (x) = 5: . 15 x x+ T .4] " 3 *' + 312 " 3 = » 48< 16 = -x 2 3 1 2 2 ' 5 5» 12to3 and 3 we obtain ^P a 2 (x) dx = (9x 4 . 2. . 1 ?)-!(?"- W W .. 1.' 16 . 5 /35 \ 15 3 = 3: 47. 3.- . 8 35 1 <P Sdx 2 ^ ^ 2 " J (l2 8 &x* 30 4 " »' " -2x 2 + ( - (1** . £ 48. 2. ^(12x 2 a\ -.1-) l) = T f 2 T "fjUT . The P„(-l) = (-1)". X4 + " 72l > = 105 IF ? 3 1 " 5* . = and a particular solution is x so that auxiliary equation is ro 2 — c\x 2 + C2X 3 + x A — — 2m + 1 — (m — l) 2 = y 5. Since P(x) 2 2 4) (x + 4) . cix~ 3 — 5m + y 4.x)(l + x)y'i - 2xy'2 + 2y2 = 0. All others are ordinary points. and a particular solution + c 2 x In x + -x is yp — jx3 so that . The auxiliary equation is m— = 6m — = y 2. so that 1 = c\x x 2 In x. 6. — Q(x) ^ v ' and and x — —1 — 1 (at - 2)(x = and Q(x) = 222 (x 2 - + 2x + 4) ^/Si.X Exercises 6. Since P(x) = the singular points are -2x (x-2)(x 2 + 2x x= 0.5 49. x = —1 + + 4) \^3i. 1/2 cix 2m 3 + 13m 2 + 24m + 9 = (m + y 3. The auxiliary equation is The yp auxiliary equation = x4 — x2 In is m2 The = + c 2 x~ lli 4- C2X~ 3 In x = (m — 6 . 3) 2 (m + + ^x 3 2)(m — 3) = 1/2) so that . Chapter 6 Review Exercises 1 . Let = y2 + x)-ln(l-x)}-l -x$n(l that = J/2 1 t: x 1 + l+x + I- + x)-ln(l-x)} -\ln(l and 1 2 (l + (l + x) 2 x) 1 1 r + + (1-x) 2 1 + + (l-x) 2 1 1 2 l+x 1 + 1 + l-x 1 1 2 1 +x 1 + 1 1 -x\ 1 l+x + l-x Then (1 . x the regular singular point is x= and the Alt others are ordinary points. Since the regular singular points are x = 3 and a: is i = 0. Since P(x) = —x 12. the regular singular points are x 9. Since = 8. c = —i. Since ifa^+l) the regular singular point 11. = and x — -co < x < . ^ 5)' Q ^ =0 irregular singular point is = £ 5. interval of convergence is oo y" + -2 < x < 2. [*(* - 1)<* Choosing co = 1 and ci = we find . Substituting ?/ = and (x 2 = S^=o x is - 4) cn x 'i = 0. — x — 2i. oo. The irregular singular point There are no irregular singular points. . Since P{x) = —2x/ 13..k-2 k~2 + c k ^\x and c k-3 x equation we obtain it=3 which implies lj i . = — 2i. = —3. afo - —2.Chapter 6 Review Exercises = the singular points are x 7. and x = = and and i <?(*) = = 5. Since P(. 2. The [X^ irregular singular points are 6 the interval of convergence and Q(x) mto * ne = 9/ (a: 2 - differential 4) an is + xy = 2c2 C2 = +Y. 10. 3.. 4. . . = 2 3 C6 4 = 45 and so on. For co = and = ci 1 we obtain c3 =0 Ca = 1 12 = C5 C7= = fk and so on. two solutions are and 14. Substituting y = ££Lo 1 dj^' into the differential equation y" - 4y = £ [*(* . find C2 = 2 C3 = C5 C4 = = = . = .Ik* - we obtain 4ct _ 2 ]x fc -2 k=2 which implies Ck Choosing Co = 1 and c\ = wc = 777 rr fc i -D' = 2. Thus. .Chapter 6 Review Exercises and so on. For cq = and ci = 1 we obtain c2 = C3= C4 = ce = 2 3j C5= 2 15 5 4 C7 =3i5 224 = .. 4. For £ + 3co/2 and ck Choosing + = 1 = c3 0. -. Thus. and — C * +3 Choosing co = 1 and c\ = we =(A + 1 3)0t + * C *' =1 2) find 1 Ci = C7 = C5 = c8 = en = C10 =" = 225 • - -- = . ..---- (* - l)c*]i fc+1 = = .Chapter 6 Review Exercises and so on.. two solutions are and = G2 V2 = 16. two solutions are and 2 I » = C ( I+ 15. 2 . 2 + X » 4 * + X YH 3 into the differential equation Y5£=$ 7 a: 3i5 + \ ---)' we obtain oo - (x l)y" which implies + C2 3y = = (-2c2 cq = 1 3q. 1} find = C3= =2' 2' ^= 8 = - we obtain 1 c2 and ~ (* k{k _ C2 and so on. C3 = — cq/6. Thus.5.) and c\ = _ - (fc - 1)(* en we " 2 H-l ~ k{k - + 3ck _ 2 }x k l)ck ~2 = and - ci + 2)0^1 3ck „ 2 _ fc-3. c4 1 so on. .3. Substituting y Cn*™ ' nt0 { I + ^x 3 + ^x 4 + equation we obtain diiferential CO y" -x V + xy = 2c 2 + (6c3 £ + co)* + [(* + 3)(* + 2)c*+ 3 - it=i which implies C2 = 0. Substituting = ?/ .. Chapter 6 Review Exercises obtain C3 = ce = Cg = C4 = c7 = Cio C5 = eg = cn = - • = = • = . C2 c = -~2 C linearly independent solutions are and 226 <>' c3 = "18°°* is c ft _i]x* .r- 1 = (2r + l)(r- 1) - ck = and [(A The indicial roots are r = 1 and + r)(2* + 2r r = 1) —1/2. two solutions are = yi 17.%Lo Cn% . equation we obtain differential + xy'-(x+l)y = for 2 -T-ijcox'+Y. 2 3 ' c3 = ^co.^x 3 - Co (l n+r into the ~x 6 and ) y2 = c Y x. For r - = ^mrry SO For r CI = —1/2 = c2 \co. Thus. and so on. the recurrence relation = 1 ' ^co. the recurrence relation Ck Two c\ = 1 fc ^ = 0. is = fc(2* so = \\ck - 3) * ' = 1 1 2 3 ---> > 1 -eg. |2(* + r)(A +r- l)c k + (k + r)c k -ck - it=i which implies 2r 2 .. Substituting y 2x 2 y" = T. -1 = .Chapter 6 Review Exercises 18.(i-. + 1 -^-~.. linearly independent solutions are yi = c . ' C3 = -^CO. + r)ck + ck ~i}z k+r (k fc=l which implies 2r 2 - = r - 1) \)c k + r(2r = and + r){2k + 2t- {k The indicial roots are r = = and r For r Ct = = 1/2 the recurrence relation Two = =1 > 2 3 > is -" 1 = -—c c3 ^co..2{fc + r)c k it=i - (* +r- l)(fc +r- 2) Cfc _! + c^]^" 1 = which implies r 2 . b . Cfe the recurrence relation = ~k(2k~-1)' Ck so = 1/2. 3 + . 1 1 . 90 is = ~*(2fc Cl * 1 = c2 -co. Substituting y — n+T into the differential equation we obtain EJJLo cn x CO 2xy" + y' + y = (2r 2 - r) cqx'- 1 + £[2(* + r){k + r - \)c k + = 0.) and ?/2 V 19.3r = r(r - 3) = and (* + r)(k + r- 3)c fc - [{k +r- l)(k +t- 2) - l]cfc _! = 0. Substituting y i(l - = 630 30 3 / n+r into the differential equation we obtain EJ^Lo Cna: x)y" -2y' + y= (r 2 - 3r) coaf' 1 + £ [(* + r)(fc + r - l)c* . For r -loo. + C2 fc = 1)' = 2 3 1' Aco. + c + ^_ 2 1* fc .Chapter 6 Review Exercises The indicia! roots are n— 3 and = t-i 2 + 3* + (* _ For 0.2) + so 1. / 11 209 o \ 3 SI A second solution = V1 -/ is ^(l-2x + ^)(l + |x + f§^ + / / 1 =m J = 1 1 3/ = yi Cn^" xV -xy'+(x 2 +r . into the differential equation + l)y= 1 91 ^+^+ (-^3 + + _ 1.lni V^" fl r3? + 4? + lfa 36 + ""J 1 l?-2?-16?-3fe— -^ lnz + 20.4 fc _2 = 0...) \ 1 (r 2 - 2r £ + + [{ft l) cox we obtain + r 2 cjx r+1 r + r)(* + r - l)cfc - (A + k=2 = which implies r 2 -2r + l = (r-l) 2 = r 2 cj [(ft The indicial roots are n= = = + r){k + r . c2 -co.-). c3 = —co- is 5 . C* c\ = = 1 . n= 3 the recurrence relation is l)c -i fc so Cl One solution = 209 11 5 = -co. Substituting 1 |i| 3 x 1]^ +c and -^T' * 228 = 2..3. 2 ]i fc+r -1 . and (fc + r) 2 c fc (2k + 2r- !)<.Chapter 6 Review Exercises Thus - C2 -4* 1-5 and one solution A Ci - C6 = — — l*T (-7 1 61°° 2. x - + ct . 23 4 + ) = 2 6 equation we obtain differential T.1 .(2x = ££Lo - l)y' + + E c„:r (x [(* - n+r into the l)y - 2 r cox + r " 1)^ + + {(r (* -\ .*_! + ck_ 2 = 0.304°° is second solution is e^/ 1 xdx f vi — - y ^ f V /• x(l _l x 2 + ^__ _ l6 + i f /1 5 1 23 5 2 21. I)co] xT ^ ck . + 2r + + r)cfc - l) ci 2(k - (2r +r- + l)ck ifc=2 = which implies r (r + 2 l) 2 = 0. Substituting y xy" ... ci-(2r + l)co = 0. 3. 4. k = 2.5. and [(* The indicial roots are ri -2ci + Co — + r)(k + r - = = 2 and r2 and 1) fc(fc - - -(k + r.. so c\ = -Co.. Substituting 1 1 = C3 is 1ft A 1 = £™=0 2 (* - y—^ e cn x" 2) y +r = = dx . For V2 - 3)c ft (k - 0.xV + —Co.^ e = e 2 + - - r £ r 2) CoX + +r- 2 +r- + [(* [(r 2) ci - l)c fc lnx e - we obtain into the differential equation (r Jx 2cfc - - rco] (A: + x r r+l r4 - Ijcj^ +c fc fc=2 = which implies r 2 - - t- r 2 = 2 +r- - (r 2)(r - 2) ci + rco - 1) = 0.ljc^ + c k _ 2 = 2}c k = -1. _ 2 ]z fc + r . is y> 22.— Chapter 6 Review Exercises The indicia! roots are n = r2 = 0... 2)c fc _i + ct _ 2 = 0.. -1. . . = —Co C4 = co (l + x + -x 2 + -x 3 + ii 4 + = Co e dx = x . — and cq Thus C'2 and one solution second solution i i/ = V . Thus -2c 2 cs and - — c% + c\ = cfc = c\ 1 = ci ^co + co = =? (ft-2)cfc_i C2 = -c fc => = Ci _2 .6. 0. 230 C2 = ^co and A = |co C3 is arbitrary 4.. CB 2 = 03 we have 1 1 ci=c2 = c 0. 2 l) ci = 0. and the recurrence -^' ^ = relation 2. Choosing co 1 2' = and 1 = C2= C2 -. 4 64 l—x — 2. c* = + r) 2 ck + cic-2 = = 0.304 6 + is . Substituting y xy" + y' + xy — = = Qx- J^nLo cnX n+r ' nto the 2 2 r r CQX + (r + 2r + 1 (x 3 1 + -x 4 + 3 ~x r+i + £ + equation we obtain differential l) Cl x 5 [{k + r)(k + r - l)c k + {k + r)ck + ck . 4=2> C5 3 = 20' and so on.3. Two solutions are and Vi !3. = - = -^. and (k The indicia! roots are r\ = ri = so ci 0.4 Thus c2 = --co C3 = C4 = = C7 = = 1 C6 = C0 64 -2^4 C0 and ?/i 3/1 = rs\ co 1 ( 1 1 — \x 2 + ^-x_ i .Chapter 6 Review Exercises Taking cq = 1 and = C3 we have = Cl and so on.2 ]x k+r = which implies r (t + 2 = 0. 7. #{/(*)} - 3.e {sint)e.~e~ at 0.l)e~ - \ 3 \ e -st S 232 — = 2 -je"'. f. e ^2 .e 7 Laplace Transform Exercises 1. ~ St s s s 6.st dt S s o J™ e- ie 3 -St s o e" st \ — = #{/(*)} — . e_s e (e-" + 2 1 8. s e~ st smt 1 */2 >0 1 1 = -e = J™te.flt ¥{f{t)} r = (cosi)e- st = = (--^ye-^sinj- dt = ( \ < t (> 1 0.^° - . - -ji-ye-^cost 00 #{/(*)} = > s J 0-(o + ^e-"' )=-Vrre" 2 s + l' «a + l /(() "s e e e Jn/2 7. l . S = #{/(()} = et te- J* -7 e 4. s > . + /2 .(--t e ~ s - . s>0 .e~ at ) 5 JO 5.St dt = (-\te~ at . - 2 Hit J\ t '. < — f(t) = —-i—e^'cost + + S £ 1 J?{/(t)} <t< > 1 2t - 2. \ dt "s /'(2t + - S — + l)e- 3t "s ) dt = J* e st - = dt (° (~l -i) te ' s ~ St - -| e £ .dt = JO %{f(t)} e' st dt ~e~ st dt + S 2. s 1 l)e~ st dt = 2 st (--{t .r ' S + s>0 l).1 e~ a s>0 . = s St 2{f(t)}= f Ae. s ! s)* 1 — oo e-* Jo 3-s 15. 2 l l e P-)* S s . #{/(«)} e 1 (4 14. 2 —^ w) "<it (3-5) 2 = 1 4 = jT t 2 e 3t e~ st dt = (3-s) 3 ^{/(*)} -^—te^ \4.(smt)e-" dt + l) 1) = £° e t 2 2 s . s > 1 -5 + a . t e.Exercises 9 -' ( f 1 - Ho. 13. *</<<» - t< o < 1. (s-3)3 t J™ + + 1 - s2 t s) 2 + 3 (s+l) 2 + (cmt)e-* dt + -s 2 3 -')' > l 1 (l-s) 2 < = J^ismt^+^dt + 2s + 2' = J^°(cost)e^-'^dt 1-5 (1 - - si so -e- + cosi l* (l-s) 2 + s> 1 a-1 1 s2 - 2s + .£ll e -(»+2)t /o s + 2 = rte^'dt = dt JO s . = jo ' -(e" sa b °° / - e / 00 > -2 1 7. .st dt = 1. 1 >0 t /V < f 0. = 0- e^^'df = 7 -h. {/(*)} = #{/(*)}= 1 WJ #{/(()} 1 - t> e' + V'dt [°° e- 2t [°° te 4t e- st fe-t'+^dt = _. 11. - > " 8 '* -tV 3 ( - 2 e( 3-s _JL~f e ^(3 (s+ 16. - < f - t)a-' + ^ = « "2 e a t -st 10 0. 12. e 3 o (4 - X °° s) 2 — S > . /ao 18._-i -2 + -i ^ 5 - i = 3 . *{t B } = 22..^-^ + ^ 3 3 -5-2s + 9 5 28. + st 2 l) 19. ^{( 25. dt * 2s (c08t)e + 1 1 (s 2s 7 (P + if{4(-10} 23. %{coshkt} . ^{l+2e 2t + e 4( a if{4t -5sin3[} y{sinhA:(} i ^ + 3-| + 4 + - = l} —-!- 2 -l + 1) {s 2 1 (sint)e . ^{7( + 24.1 Exercises 17. 2 2 -*_ «». 5 1)' s + ' 4! 4 if{2t 29. ^{l + e -"} = i + } = 4 10 — 4- + 4-^ s s s + 6(-3} = J 2 + 3f + 3( + s 33. ) = 34. i? ( c „ s_5. 31. Jf{/(t)} = 7. + sin5. t(coat)e- J st st s + s 1 (s -l 2 -5 2{f{t)} = t(8ini)e" 5C jf -1 2 2 + 1) > S . - 5 -^_ + ?A-5 a .i} - 234 { 9 ^+^ so.+ 5}. 4 s = s 35..i„ h <) . y{-4f 2 + 26 ^{8i 3 -12( 2 +6i-l} = - s + ^ s — 2 2 4 + ^-r s . = 2^ 21.-.a + . = = 20. -fc 2 ^.. } * + 2 s>0 .4 - 2 ^} -* 16f 2 + 9} = -44 + 16 8^-124 + 4-" y 32. ^{t 3 27.if {e- 3}-^ + -s ^-. the Laplace transform of 1/f 2 does not exist. 41. i?{co S 39. Let for 3 u= } = Ys s + l 2^ +4 =if j^cos3f + ^cosfj = ^{ S m(cos2t} =i? y {sin cos2t l =J? {^cost - if{ Sin( S in2[} = j^sin4fj + {l = Is ^ du s dt 2 s2 9 + 1 \ - ± ^ ( = Is + : | (^l C0S 2 )}=^{i f}= ^{ sin( st so that cos3f} \ . r(i/2) 7. If r(5/2) 3^ s 5/2 4s 5/2 we attempt to compute the Laplace transform *^ If s = = for of 1/t 2 atdt+ we obtain rk stdt then which diverges. -^r{a + 1) .Exercises 37. Thus. i?{sin2tcos2f-} 38. 5^_l^ = . ^{costcos2r-} 40. 2 t}=i? = if 43. If s < then which diverges. If s > then which diverges. 42.) .du a > — 1.^in(} = {|sin3t and %{t"} = Si n / e~ t-i(ism3 i -i S mt)} = a st t = dt ( In V s r{3/2) 47. i 1 Exercises . se- Us + xto 1 ii. 7. and N such that Then = for c. Exercises 7. Since / and g are of exponential order there exist numbers and < Ne \g{t)\ dt for > t T. 5s - i*— + 4 s s& s \4 S+1/4J 2 \5 s J '{?T49} = ^ _1 + At + 2t 2 8_ lJ - 2/5 4 5 J = {? ' I si sin7t 7 2 236 3. and is M. 12. 8. d.2 24 s5 ' 3! . if-lL.1. |/(t)| < Mea . 48. < Me<*Ne dt = A/TVe'^ 1 \f(t)\W)\ of exponential order. t >T.i + 9. Exercises 7.2 13. iy ' " ^ 26 1 25 = l/4j 2 10cosh5f h I s2 U 2 + 2j \s 2 +2 U 2 + 3§J U 2 + *- + a2) \( s _2)( 5 -3)( 5 - 2 9 3 S s2 + 1 + l)(s-2)J \(s-2)(s 2 + 4s + 3)J S 3 + 5j"9 9 + s-2 ft ( s-3 + 2 1 1 1 12 s s-1 3 1 15 2 sin 3* s/2 1-jf-if 1 * - -^ ^-^}" ^ 12 fl2 2cos3i + 2j 9 }-^"4 (a3) \s(s-l)(s = + 3/~3 5-4 -6)J + 9)} - s2 \/2 * \9 20 J ^1 (. g-i/-il-} = .y-iJ 2 I4s 16.1 / X a* + S 5 s-2 3 s + 1 5 s + 2 1 6 1 6 1 1 + + a8 '^' -6/~2 e 1 s M s-2j + 3J G 15 3 C 5 . i?" 1 !— I 17.-a^ + X + J 2 1*3 23 — \s 2i lj U|^|| + 9J i?. r Utf-J * 4 1 + 4 )( s + 2 )}=^ (s 2 20 ( 7( } = > 10! (s + 2) 2 + 16 exist and .6) 2 J I 3! 21 _„ hV3t..3 1.. jf/ te -«} s ( y { etsin3( } = + 2)4 4. J£ s s-l 1 {s + 1} J i. 5 - ^{(* 3 e- - 10) 2 = - -sinf . 2 (s -4s)(.J = = In Problems 35 + 4j s 1 2 1 f s2 8 5-4 36 ' s _^L_ J_. ^ = #{/(*)}= s 13 + 4)J e~ st dt = 7 + + s 1 2 + ) > 1 J 12 2 cos - 1 r e^. (73^ ^{( 1 238 = 10 e- _J + fs s 4. U S- 33. 36.— Exercises 7.^L} = J_ * + 4)j +3 6* \(s 2 f{x) „_i \6\/3 * + \(s 2 3? -if 34. 1 ! -9J .-5t 36 = 1 — f — cos t + sin t I 1 + 20 5J r^-4'^T2) = 4 COs2i+ 4 6i/3 -t^-t 2 s + l + 4 ~5 s 1 45 2-^ + ^-2 + + and 36 we use the J + 5 -i 2 1=^-1/ g(z) for at most finitely 2{f(t)} s 5 2 l4"^?4 ^U .-Lsin^f Exercises 7.^dt=-^~ — 3 s jo - 2 s 6 — s2 1 + 4j + sin f — fact that 45 1 „ 1 1 1 s 3 1 2 cos 2t sin 1 S1Il2i = ^ (s 3.i 2 2 s s W 32.2 s+1 28. l)(s 2 l)(s 2 + 35. > fors>3 3 2.sin2t 3 6 3 s 2 2 1 +4 2 s 2 +4} — — sin 2t /™/(x)di = Jo°g{x)dx if both integrals values of x.s + 5)J U .. -4 e 6v3 3/ t many for a s 1 . (11 MIs s2 4) 30. 11. 3 (2e 2 J (C + 6* 6 _« .*{*»«» - I)*} J 12. = S ioh3t} 5 { [ \(s 1 - I_ s + 2)3 - 2 —^2—^1 t (s l + 2) 2 + l 2 = e_2t cos _ 5 + 5. l s + I) 1 1 1 . it.s v-J* 5 _ s+l (s + 2)3 _ (s + + 1)* 3 2 (s 6(s + 2)V 239 - 1 52 j 1 4 * - 2e_2t 9in( / 3) ot2 2 3) ! l22 2e..^e. t coa W ._ 4te _.3t coS 5 I -»-t_ I 2 + 5t. 18. ls 2 W 2 ' 4s + 6s + 34 J 1 _ = s 2 (s l(s 1 „ 1 . } * " (s-l) 2 * l = ±tV \6(s-l) 4 J 6 — V + ^ = Win 5 U [ 22 1 * ( + J 3t}=y|^e + i y-i 9 .3t sin5t 1) fe 5 J t „-t _ 1 _ 5e _.3 7. +4 2 s-l + 36 1 2 7^ " 1 2s-l + ^e cos6t} 1) . _ (fi 10. ^ - 14 .Exercises 7. t = _} + 5J ^. \(s W v-J^ J J + 2)^-^ 2) ( 't.le-'costt) = 1 J.1 12 2 2 + 2 (s+ y|L e -' sin 2 t} = y{e 5)2 . s-l) 4 J -J 21 2 _ + V' . [ J-lJLi} = + ^ . 2 "I? + 3) 2 + {. + 2) 2 + ( (S 2 (s + " 1 . ** _ + sin(i-7r)%((-7r) = ^- 1 f e~ 25 e {-— -- ^ = --f ' d$W + 4-J -28 55 p~ 2s + ) r^} S (s2+4) 2 240 = -'«(*-2)-{*-2)K(*-2)+«*->*(*-2) . 36 yj ^" - 37. — = 2)) 3)} = ~ + sl 3 s .3) * (t - 3) + y* (t - 3)} 3s tt) It (t .3 23. # {tK ^{(3i - (f + = 2)} 'X {(t . 30. 26. - 2) m It - . - 1 f e" 2a 5)} = 1 {ivrTyj y{icos2(} = if f (t -^ - ? l)< (S - SJe 1 "5 « (i - 5) + 5e*" 5 * (i - 5)} = 6 .tt)} = 4f=P » *H*(«-!)}-*M'-i)«K)}-j£ 29.Exercises 7.^(t-l)} = ^/(e 1 "5 (* 33.3)} = 3 ^{ lj'tt (t if{cos 2t K (f - tt) } - 3? {cos 2(i - - <U (t 10e-. - + s + 1 + 2°U (t .2)} = 2) + 3e~ 3a 27. 1 ^{( ( -l)V. a?{(t-i)qi(t_i)} = i^ S 24. if{e 2 -' «K (t - = 2)1 2 J? /e-f" ' (t <3i - 1 25. (e) 2{2-4<V(t-3)\ = ---e.if ^ -. (a) 50.-^-y - a J-^-j- * (t - 1)} . (f) s 52./ s 53.2(3 + 1 )„ 1 + _2) 2 + 36f [( 5 l) <. (b) 57.Exercises 7. t-smt<ft(t- if{/(i)} 2tt)} + i 49. (d) '' — — l- + s l) 2 2 + 1\ 2t - s l] K (* - 1)} - ^{[(f - l) 2 - + 2(t - 1) + l] _s J?{t-*qf(t-2)} = #{t-(t-2)«(t-2)-2'W(t-2)}= s -^2 i?{sin 1 1 l. > V 2 \2 J [(5 + 3 l) + 3) a + 9] 2 . (s 5 + 1 j-cf-if »-'f 1(^ + 2^ + 2)2/ ^ 1 45.3 2 2 1)} = y {[(( / 2 55. 2 3 - 2 a i?{sin t - = -^{^(i-a)-^((-6)} = sin(f p-as - 2tt) K (t - — .(-J—'U + (s 2 + (s-2) 2 + ds \{s ' \(s 2 44 + 46. 1 ds 2 Vs 2 ' ^ -j.2 12 [( s + (s 2 y 2 1) + 2 1] 'dsVs 2 + l// 48.3 38.-6s 2tt)} 2e e s2 .f\2 -'f / s 56. 2 I J + + 3) 2 + 9 J 2 l) 1) ds\(s 2 l/ 6 41. 2 s 2{l-<V(t-4)+«U(t-5)} = 2{t 2ail(t - 2 ^ 36. (c) 51. l) 47. gft B mh3*} = 2 / 6s ^ 1 -^(p^)- 39. Y( S ).I)* Y(s) (s 6.s) - y(0)\ + Y(s) = 2 (s - 4s + 5)Y(s) -s + 5 0. y'(0) (s 2 -2s + l)K(s)-2s + = l 2s. [Jo o ds \s s 2 / 1) m^} -^ £ sinTcos(f-7-)dr[ = ifU + +3 +l l J 14. . 2[j\e^dr} = 12. s 2 + s+l 1 J (s+l) 2 d s+ + r* 'o -t.S y(0)-y'(0) + Y(s) (s 2 + l)Y(s)-2s-3 VI r(S} ~ 1 7.<M J s ( s2 l 1 + 1 2s + 2) 2s 1 ^ cfes 2 +l./ s (s 2 + 2 2 l) (s ^ ++ l) 2 + = if {sin (} if {cosi} = ^ ifn/'sinr^U^if/rsinrdxU-A^.if{tsint} = .4 4. ( } s[s X{ tco&rdr) = -¥{cost} = 10.1 Exercises 7.1 . if|^ e-r cosrrfrj = X\ fTsiRTdA = s(s 2 + l) ifje-'cosf} = J . / d« Jo <* \ Te- T <M = f —d (1 - 1 + X 1/ s 1 ^ ds\s(s + l)2) =- 2 (s 2 3s 2 a (s 2 + l + l) 3 2 l) . T 1 Te. if{ 13. We solve = s 2 Y(s) %{y" -2y' + S - if {y"} 2 - y} 4if{y'} - sy(0) = Y( S ) - */(0) = if {0} - sy(0) + 5if{y} - 4\aY{a) - y(0}} + 5Y(s) - - 2{sY(. if{/V«fT} = -if{ e [Jo 8. X{y" - V+ by} = 5. S (s* sl 1) t 9.(a \ 11.— -f. —^~ - = a ) 2s + > = . We solve %{y" + y} = if {1} s 2 = 1/s. y- 22 if.sin 4^ + - — . + l)[(s-l) 2 l)J } \s(s 2 + (s > 1 e I cos2f] ' . (s „ 2 = + 4) 2 J = -/ t /V * e -'= ' r e-<*- T >d7- Jo cos2f * 2 1 = I /'cos2rsin2(t 2/0 — cos 2t sin 2r) <ir = 1 rl 1 l + -sm4r = -sin2i f-f 2 \2 -( sin 2i + 4 = ^(sin2i — 16 + = ^sin2i + 4 + 8 + — cos 16 [sin2i(2sin2*cos2f) -^cos2i 16 2 [2 + l) cost 2t ~ 3T dr = -± 1 1 o 3 . = 5 )} + e = .T = 1 *sini = /Or) cos 2{( J - / sinft -rWr = cos(t. y -1 j .2t-3r ( 3 ^ tWt . 5-1 I 21. = <s r) dr \=e-Ue = f e^e^dr = f e \(s + l)(s-2)J Jo Jo g-*J jif o JO l)J = -ri s(s + 2) ) W Jo I .3t *{ e «* 20.cos At cos 2t 4 o = (s-2)(s 2 lo 2 -sin2f y-T —J f dr^te"' e"' 2 Jo = -I Jo isin2t cos2r(sin2( cos2r = 1 I v ' 2i * . 25. — cos 2((cos4t - .5(1 . 26. • ^l^}-!* ^{i* 16.}=e\(s + l) 2 f i? -1 S j . 27.sin2( cos 2r dr / — cos 2t Jo 4 sin 2t 2t cos 4t 1) —— cos 2t 16 + cos2t + cos 2 2t - sin L (cos 2 2t 2 - - 2f ll J 244 sin = 2 2t) - -tsin2f 4 / -sin4r(ir Jo 2 16 / sin 2t sin 4t . {^F( 23 + (s(s 24.5l l] t */(t) = /(r) e |o cos2t*/(()= .Exercises 7.4 15.l {-j^—F(s)} .t) = if-if—l-. s *{/(()} 33. #{/(()} /" / e~ <dt- #{/(()} JO - -r- dt at T te / ~ s(l dt = _ <it + 1 fff e + _ ee( 3t Bint ^ A- (2 - t)e _s( 1 #{/(*)} s— / -e.T )dr=- f{r)g(t = tf(rMt-r)+h(t-r)]dr= JQ = f f(t. 2{f{t)} 34. 1 ~~ e~ aa ) 2 jf te~ -^y 35.cost — -sin ((cos 2r) — e 2t e .r)dr JQ = [f(TMt-T) + h(t-r)}dT = f*g + f*h 31- 32. f f(r)g(t .2s ) W+ ^——^ e J o 36.u)g{u)du g*f.} Exercises 7. Let u —t~ i — t so that du h} e 2t sin t .t) dr = sin(t ' = 6 -2i — . (1-e-"8 ) 2 s(l-e. x f = e" j£ 2r 8inTe" s Tl 1 sint / Jo 2 = e * '{[(TT^TTf i* + 5) 2 } + 4s «2 -2( ''" r .2as ) dt r 1 _-2a 8t Jq 6 e" 2 "8 Jo e- 2 e ~"s/2 1 - + l l~e-' s j coth — 2 .r) dr + JQ/' f(r)h(t .2"Jo e "sint rft = 1 „ s2 1 dt " s(l -e" + e-<") -e" s 2 (l-e.4 28.cos t — sini(cos2i — ^ sint 2 (cos 2 ^sinf (-cos 29. - -(1 V / sin sinr(smtcosr Jo 2 o ^ -2 ' /' 1 sin2TdT .cost sinr)dT cos2t) dr sin 2t 2 — - t + sin 2 t + l + 2cos 2 () -tcosf = ^e" 2 '(sint - 2 sin - ^ sin2t t t - + ^ cos t(2 sin t cost) tcost (cost) dr and rt f*{g + ~ \ 2 t * ' ~ [t — ^cos t 1) f*9=f 30. of the differential equation is 3 Solving for X {y} we obtain „. is s Solving for f£{y} s 2 + tete~ 1 _3 e -a. 4. 2.Exercises 7. + 4) 2 y The Laplace transform 11 3 As 2s As + 2' = 2 ( is 1 + 4i?{s} = s +4 2 + + 4' u+ of the differential equation 2e _4t : . 3. ^_J + Thus.4 38- 2{f{t)) = T^f\-«costdt~ 1zLri Exercises 7. The Laplace transform Solving for %{y] we of the differential equation is obtain s Thus. l of the differential equation *tf{»}-|f(0) Solving for ?£{y} 11 ' l we obtain vr 1 1 ( 52 + l)( s 1 _l) 246 s+1 1 1 + 13j-1 2(' . (s Thus. The Laplace transform we obtain t£{y} = 1 -75 -. y The Laplace transform s — 1 = -1 + e'.5 1. 1-s 2 2 s {s + 2) . y The Laplace transform a Solving for 2 %{y} we = X{y) - s+b + 5s + 4 2 4 2{y) - - sy(0) 13 X{y) = 0. s ^(cosf of the differential equation a2 6..(0)] 1 + 9 X{y} = we obtain 1 .2. The Laplace transform s Solving for 2 f -27 e + of the differential equation X{y} - sy(0) - y'(0) 3 1 - 4 [s 10 T te is %{y) - y(0)} + 4 %{y} 6_ »4 we obtain s 5 -4s 4 + 6 3 4 (s - 2) 2 4 s 9 1 8 s2 3 2 1 4 s3 4 Thus.y'(0) + 5 [sX{y} - sy(0) Thus. 247 3! 1 1 4 s _ 13 2 8 (s - 2) 2 ' .-e of the differential equation s2 7. 2 2 1 ^27 + 9 8. . obtain 3 -6s + 13 of the differential equation *{y} - *s(0) - 2 (s-3) 2 2 + 22 = --e 3t sin2(. Exercises 7. T/ The Laplace transform Solving for ^£{y) 39 + 4' . 5.6 \s2{y) . 1 -4t .5 Thus. — ~e* - j/(0) 6 [s is %{y] - j. + s2 2 s (s ~ 1112 2 3) 2 27 5 10 1 27 s 9 - 3 9 1 (s - 3) 2 ' Thus. + 42{y] = y(0)} 3s+l 4 -t = j/ %{y) we is obtain The Laplace transform Solving for + sin(). 2 11 1 is 3 s 0.(0)] + y'(0) Thus. 2 2{y} - %{y) we Solving for !£{y} 2 ~ ay(0) i/(0) is .5 9. ^ {3-2) 1 5! ^ 6 = 4 The Laplace transform s 2 = J_ 1 16 a 115 00344 + 16 16 of the differential equation 2{y} - sy{0) + . 3 Solving for %{y} we obtain 2 W _~ s 2 s( + 2a + 1 _ ~ s 2 + 16) Thus. we obtain a 2{v} = 3 ( - s 2 +a s2 + 1 )2 y = cos t s a2 1 + s2 1 + 1 ( a2 1 + 1 )2' Thus.y'(O) - [s 15 a 16 a 2 +4 + 2 4 1 2 a 2 + 42 ' 1 + sin4( " 2 is X{y) - 3/(0)] 3 = (s 248 - " 1 l) 2 + 1 ' . .y(0)} + 5 X{y} = ~ + ^.4 [s 2{y} . of the differential equation X{y} - - _ = j^e 21 .y'(Q) + X {y} = . 12. 2 \sJ£{y} we obtain iVi _ ~ 2 4s s 2 (s 2 7 1 25 s +s+1 _ ~ 2s + 5) - 7_ 1 25 5 7 1 1 t 5 s2 25 (s + 1 _1_ -7s/25 + 109/25 - 2s +5 + 2 5 s s s. is y'(0) + 16 = 2{y} - .I) 2 + 2 2 2 51 2 - | 25 (s l) 2 + 22 Thus. The Laplace transform — ^ sin t — of the differential equation s 2 2{y) - sy(0) - cos t. is . The Laplace transform ^^ e ' COs2t + l ' e Sin2 *- of the differential equation is a Solving for ££{y} 2 X {y} . V 13.y(0)] + 1 = Z{y} obtain The Laplace transform a of the differential equation sy(0) - y'(0) Thus.Exercises 7. The Laplace transform s Solving for 10. y .sy(0) . y= + 2l 11. V(0) . l) l)(s-l)(2s 8 1 + + l)(s 2s + 2) 1 9s + 18 l 1 . 13 -20 e .74 e 7 4 we obtain Solving for w 4 of the differential equation . is = [s^{y} . 16.y'(0)) - [s 2{y} .!/'(0)l-3[ S if{j/ }.y"(0) + 2\s 2 X{y} . !/ 15.S (0)] - (s Solving for Z£{y} we obtain v/ s 2 (s - 4 s 2)2 Thus. = y 14. The Laplace transform 2 \s 3 X{y] - 11 = (s + s 1 it 4 a - 1 2t + te 7 4 2 - 2 1 1 4 (s - ' 2)2 is 2 y"(0)]+3[ S i?{t/}-sj/(0).y(0)] - 2 <£{y) we obtain a 2 + 12 >-l)(s+l)(s + 13 60 13 1 s - 1 2)(s2 16 1 + 20 s + 9) 1 39 s 1 + + 2 3 13 s 130 s 3 + 9 65 s 2 Thus. _ 111 1 „ s - 2 v + 2s 2 s 2) s-1 2 (* - 1j 1 1 2 + - 2 (a 1 l) a + ' 1 Thus.y (0)]-2^{y} 115 s-1 +3 2s .V(0) " 2 s (0) 111 1111 1 . 1 9s + 2' l/2 Thus. 2 + 74 ( . 13 ^60 e . The Laplace transform s z ^ - r 4- ^e* cos of the differential equation X{y} - - 53/(0) - j/(0) 2 ^e' sin t. fl .Exercises 7. l = 7 T . The Laplace transform 5 3 £{y) - Solving for 2 s (0) of the differential equation is . + „ 16 C + 39 249 3 „ CO83( l30 1 - Sin3t 65 - +9 ' = 3 ^+ .sy(0) .5 Solving for % {y} we obtain . cost. The Laplace transform %{y} we -t It 1 1 + -e --e -t + -smi. y 19. is . %{y} we obtain The Laplace transform s Solving for 4 %{y) we .sV(0) . The Laplace transform = 5%t-l)-5e-^°U{t-l). X{y}-y(0)+2{y} = obtain bests = 5e" +1) S 3 + 1- 1] Thus.V(0) - f"'(0) . y 20. £ £ of the differential equation S Solving for = . The Laplace transform = l. of the differential equation is s2{y}-y(0)+Z{y} = \3 Solving for %{y} we V«.e -'-2[l-e-('- of the differential equation 1 £ '] W((-l). y 21. is -€-*.Exercises 7.5 17. 2 - sy"(0) .2{y) = \.*V(0) - 3 s y(0) is g %{y) = of the differential equation 2{y} - A s y(0) = Thus. 3 obtain *{y} = 2e" s($ + l) s(s + 1 l) -2e -a 1 8 s + 1 1 . y .ii 4 s 1 _J 2s !+ 4s + 2 250 6 -t 1 1 4i + 1 2 .3 1 3 + l] Thus. 3 obtain ~ sV - 1) 111 + ~ 4 s 11 - + 4 s 1 + 1 11 2 a2 + T Thus. is a o we obtain Solving for !£{y) 1 l " _ ~ 1 s 2( s + S 2) _s s+ 1 _ sHs + l)~ ii.2{y} = y"'{0) 0. The Laplace transform of the differential equation s*X{y} Solving for 18. 3 2 1 i s2 + 6s 2 + 4 1 Thus.6 25.5 Thus.Exercises 7. y 24. » = + . of the differential equation s Solving for ?£{y} - i 2{y) - 2 sy(0) - j/(0) is + 42 {y} = e" 2* 8 s2 + I we obtain i . y 23.>-k 2 22. The Laplace transform i of the differential equation . is - [s y(0)] + = 6 <£{y} obtain X{y}=e = e a(s-2)(s-3) + 1111 6s 2 s -2 + 1 (s-2)(s-3) 11 ^ 3 1 .2 2 s X{y} - - sy(0) y'(0) 2<1 -" is + 4 X{y} = l-tl s Solving for a %{y} we obtain 2{y} = 1 s(a 2 s Ills 1 — e + 4) s(s + 2 4s 4) 4 s 2 12 +4 +4 2 s2 _. 3-2 s~3 + „1 s Thus. 3 i s 4 s2 +4 . »=-i4 + k '-H<'. The Laplace transform 2 - 6 1) + - 2( e is g-jre 2 31 3 of the differential equation s e 2{y}-sy(0)-y'(Q)+X{y} = i Us g— 2irs .(0) i/(0) - 5 sin 2(f - 27r) (t *tl - 2tt).r i-icos2(i-l) «(t-l). The Laplace transform s Solving for 2 %{y} we = coe 2f + \ Bin(f - 2jt) - of the differential equation ^{y} - - aj. = The Laplace transform ~ cos 2t ^ sin It . ri — e c Thus. s l.(0)] + 3 X{y] = e~ 2a - - e~ is + e _6s obtain 1 %{y} = = tj 3 s 2 s — e -4a + 1 65 + l — e -2s 3 1111 2s+l 3s + 11 11 3 s 2 s 11 - 6 s + + + . 27. y 26.-6s e~ 3 111 + + 3.5 we obtain Solving for ?£{y} '1 3 ITS S . 2 1 + + s l. 2tt) is 1 + 4[s %{y) . = y(0) we . y{t) = c^" 1 { -1^} + (c + 2) if" 1 1 252 (J^p} = OB ' + (c + 2)te- ( .Exercises 7.cos(t . = . 6 s l 1111 + 2 s 3 s + 1 11 T +3 6 s Thus.n)\%t - (1 The Laplace transform s 2 Solving for 2{y) - - - [1 of the differential equation - S y(Q) ^{y} we it) y'(0) cos(t - - 2ic)]°tt{t + sint.e~ 2™ 1 - s rl s .3 + 2 . Taking the Laplace transform of both sides of the differential equation and letting c obtain 2{y"} 8*2{y} - sy{0) s 2 - y'{0) + X{2y'}+X{y} = + 2s 2{y) - %{y} -cs-2 + 2s 2y{0) X{y} (s 2 2c + 2s + + X{y} = + %{y) = l)%{y} = + 2c + 2 cs +2 (a + l)3 2c ( + 5 l)2 + 1-1 (s + l) 2 c s + (a c + l ' +2 + 1)3 2c s (s +2 + l) 2 ' Therefore. y 1 1 3 2 1 -i i-A e -('-» + Ae-"('-« -st 6 A + 3 2 "!/((- D Z + <tt(t-4) 6 2) A _ A--(t-B) + A P -3f'-8) <B(t-6). + 2 l Thus.2. W) 29. To find c we compute B '(i) and let y'{l) = 0. 28. 5(4 - ' 5e) . 6 20 4 The Laplace transform + 5 4-5e^ I 20 + 4(4-5e) of the given equation is y{/}+if{t}i?{/} = y{(}.5e ' Thus.} + 20if{y} = J s (s 2 -9s + 20).5 To find c we let y(l) = 2. Thus.c (4 e *-5e Bt ) Then 0=-e 4 + e 5 -c(4e 4 -5e 5 ) and C ~ g 4e 5 4 ~ - e 4 5e 5 _ e. = _ e * + e «.y'(0) s 2 %{y} + 9s X{y] + r- c 9y(0) 2{2Qy) = ^{1} + 20if {y} = . Then 2 = ce y(f) _1 + (c + 2)e _l = (e-l)e^ + = (e 2(c + lje^ 1 and c + =e- 1. l)(e-'.S?{i/} = . Taking the Laplace transform of both sides of the differential equation and letting c = y'(0) we obtain %{y") s 2 X{y} - ay (0) .+ c s - s(s 2 + 9s 1 s(s-4)(s-5) 1/20 a s2 20) + .9s + (s-4){s-5) c_ 1/4 a —4 20 c s — 5 s — 4 c + — 5 a Therefore.9s ^{*.Exercises 7.1 ~ 4. /(*) . 2 (s %{f) we M obtain (s - l)3( s + 8 s i) /(t) The Laplace transform 2 + |* l . S +2 {sinf}.Exercises 7. /(() if{l}.5 Solving for Jf{/} 30. l) 2+ % ( is = Thus. = e"'. +2 _ + 5) of the given equation . + 5' is - + V5 8 1 5 a2 113 + s Thus. 31.lte + 4iV if{/}+if{l}y{/} 34. - we obtain Solving for Solving for 12 4 (a + J(V - _t 2{f} + 2X{cost}%{f} -4J?{e } 33. e* of the given equation 5%/5s 2 sin \/5 t. we obtain if{/} = The Laplace transform of the given equation = Thus. (s + 1) 3 ' + 1 .2 ^ 2 = /(() The Laplace transform Solving for s is if{cosi} +if{e- we obtain 254 t }i!?{/}. 32. is #{/} = 2{2t} . 1 of the given equation 2{f} = Solving for if {/} T + . smt.4%{smt}Z{f}. Solving for %{f] we obtain 2s 2 yrf1 _ XUJ Thus. 11 l) 3 8 a is + 1) = /(*) The Laplace transform 3 5 4e~' of the given equation + (s 1 %{f} we obtain The Laplace transform #{/} = . 4 1 te * | 1 (s - l) 2 + (S Thus. 5 Thus. is {/} we obtain v/„l j3 ~ g2 +s 2 + l) s(s Thus. 2 4s 2 + 4' 4 we obtain U' Solving for 2s 2 + 2 2 of the given equation jfr 37.Exercises 7. sX{y} . = is . 8s + + \e~ 2t + a 36.1 3 |e 2t | i(r) dr = 100[1 is -°U(t- i(0) 1)].i(0)] +if{i} + 50 5 = 1 1 s 3 100 e 0. + ( l 1 2s 2 (s 2 + Solving for if{/} we obtain %{y) = —+ 7 (s Prom equation (3) in 0. 1111. l) - 16 /(f) = s 4 8s-2 The Laplace transform Solving for if {/} a 2 ^ f1 ~ f{t) The Laplace transform of the given equation + \ sin2t. 4 11 1 ~2s = ±t - 1 2 3! 12 5 4 ^t 3 ' . cost of the given equation is = #{1} +%{t} + Solving for X {/} we obtain s^s + Thus. = f{t) The Laplace transform + sint. y = te~ 3t Jtf {1} . Ji? \ cos y The Laplace transform = ^ 2 fl* sin* of the given equation — ^ .005[sif{i} ' .} = 39.3/(0) + 6%{y} + 9y{l}^{j. 35. . 2{t 3 }2{f}. 38.005 The Laplace transform —+ i + 50 of this equation 0. Thus. 3) ft / 3 is the text the differential equation di l) sin*. 2t is 2s" Thus. -U. The differential equation 10Ot - 2K(t - + 2e _100 < t . = or — + 200i + 10. = 2 - 2e- 100t - 200(e- 41. 100) 2 Thus.1 '^ - is fl^ + The Laplace transform 1) of this equation lg= £be- ta fl(O) 1 -0.000[£ i(0) 1)].Exercises 7. From equation (3) in = 20. is + 200 y {i} + 10.l > <W(t - + 200(t - 1) lle" 100 ''. i(t) 40. is = m[t - dT Jo -100 " -1 " 1) (t - " * (* " *(0) 1)1. = 0.000(t - the text the differential equation rf» 005 dt r* 1 + + * i{T) O02 lje ' K(t - 1). + 100) 2 (s Thus. is we obtain Solving for E /R EC ' > + k)(RCt + When 1/RC 1/fiC (* + Jfc)(* + 1/ftC) ^we have by partial fractions W=^ When 1) = ( V l/(l/RC-k) _ l/(l/RC-k) \ _ Eo s + a + l/RC R } fe A we have £b R 1 (s 256 + k) 2 1 (J- l/RC-k\a + k a + l/RCJ' .000 The Laplace transform / i(T)rfr of this equation = - 20.5 Solving for !£ {i} we obtain *»-^iU-.000 we obtain Solving for ^W 200 "s(s+100) 2(1 6 J + s s 100 (s + (l-e" s ).000 - s - = 1 1 20.000fe- 100i - 20. g(t) 44. g(0) = 0. The differential equation is 2. The differential equation = — f 5 3) — e | 5 -B(t-3)^^ _ 3 )_ is or 50^ + The Laplace transform of this equation 50s 2{q} Solving for %{q) we = 100$ + Eo[^(( - 1) -H(t - 3)].5 Thus.-e" 3a \s 3 obtain „-3 S 50 [s(s + 2) s(s + 2) 50 12 s + 2j 2 Is s + 2J ' J . 5 1 1 _ H s 5 s + 5) e" 3 *. 43.5 9 = 5 s!/(t-3). The R R differential equation is 10^ + IO5 = - 30e' 30e'^/(f - 1.5^ + The Laplace transform Solving for %*{q} of this equation 12. _ ( . 42. is we obtain X{q} 2 = + 5) s(a e -3. is 100 ^{5} =E (-e~ s .5). Thus.1 Exercises 7. OX The Laplace transform of this equation is 3e 3 Solving for if {q} ls _ liBa we obtain ^-(-'i)-^T + i^"(^^)--Thus. The W= 1 e ll K - " 2(t " 1. )*(* - !) - (! ^ - 2(£ .3 ™/ 2 1 we obtain Solving for 1 y) (a 2 ' + l)(a + + 10 + 10) (s 2 ' 101 Vs s2 + + l)(a + a 1 + 10) W N + 1/ 2 1 101 /_^ + Vs + 10 W.uw-^ffl + 46.3) )^C " 3)] differential equation is I + lOi The Laplace transform = sin + cos t of this equation (t . I_e (•+!)*«>-£t a(l + e" 258 s ) - *) .4. The differential equation The Laplace transform (<-|) + *. = 0. is se. i(t) = rjrr 101 (e~ 10t ^ . (« - is of this equation 8<£{i} Prom Problem 10 c„s 31.~j^(t - t(0) .5 Thus. Exercise 7.cos t + 10 sin t + _1 (. we have #{£(()} 1-e . ? 45.Exercises 7. .' = s(l + e Thus. s 2 + 1 + 1 s 2 + 1 Thus. is + 22{i} = j-X{E(t)}. 4. Therefore = i(t) -[1 -1%t- I e R « 1) + 2%t- -Rl/L + 2e .e -«/£) + £(-1)" (l . = 1 -= + 1 1 Thus.. | Tl=l ' 47.5 and 1-8- 1 = - L s{s + i L/R L 1 1 R s + e" L/R + R/L s 1 s ) (1 - 1 e-")(l - e" 8 1 1 + e~* + e~ 2a . R\ „ / f . we have 33. is s^{i} + |y{i} From Problem t ).2%t - 2) -R(t-l)/L _ 3) + 1} _ 2e -R('-2)/t ^( ( I (l .e~ 3s + -3s is {\-2e~ 3 + 2e~ 2s -2e-' + + R/L s -e h s(s + R/L) 1 R/L){1 ) ). = -iy{£(()}. The differential equation is f + fi=^( The Laplace transform of this equation i(0)-0. Exercise 7. 11 Jl/L) 1 111 and + L s*{s 1 fL/rt 1 +- L rt/L) s(s L 2 /fi2 | L A R s \ 2 1 _ L/R + s s + L 2 /R2 \ s + R/L) L/R s + R/L 1 e s (L/R | L s L/R + R/L / 1 - e _2 ) .Exercises 7.e -«t-«)/£)^( t _ „). ltt .^e" 10 silll0( j_ ' 2) - 3 _3 e -io( t -2) <W(t-2)../L "e 2 Z{q] g(0) = = ?'(0) 0. then 150 s(s 2 9 (i) = | _ C o S lot _| e 49. i(t) < For f < 2 - 5 (' " 5 + e_nt/L I + ) = s + - (' e_i!t/L i ^ + ) U('-*+* a_ * /L )+A( The e ~ fl!t " n)/L )^ " ») we have t(t) 48.-*) of this equation **X{q} + 1009 = 120 sin 10*.+ ^| at* at The Laplace transform sin lOt (1-.5 Thus. is 20 S #{9} + 100i?{9} 260 = 120 - c03lo(( _ 2} . 4 (8 + 10)2 + 102' lOt and i(f) Hcj If E{t) = 150 - 150 %t - 2) . is + 20$. 9 '(t) 5 iur I5e- sinl0f.luc —lot . is + 20s^{9} + 200^f{g} = — 150 . we obtain 3 150 s ( a2 + 20s + 200) Thus.nx .3__ 10t 1K cos r _ r3 e -i°' cail0tlot e sin ^ 4. = -10t 10t = (t. ^i = q(t) 3 1 3 4(s+ 4 a LV + 10 10) 2 + 10 10 2 3 . The Laplace transform ~ fl( '~ 1. !) ^*< 2 - is ^+20^ + 200? = 150. 4 4 .Exercises 7. The differential equation -lO(t-2) + 20a + 200) . " 1 t differential equation 1 of this equation s Solving for !£{q} - )^ - .1)/£ e 1 ). = E C (l .* so (recalling that w2 = 1/LC. is ^b - + uj 2 %{q} = 2{q} + 2XsX{q} s or ( s 2 + 2As + W 2 )^{g} = we obtain Solving for if {q} and using partial fractions 2^ = Eo(l/u L \ For A > w we write s 2 2 (l/w 2 )s s s + 2As + J1 = X{q} = for < <j we write = Ek.s Thus (s + 2A/w 2 ^ _ E + 2As + w 2 2 f i. The steady-state current is The differential equation is = q '(t) -60ie.c(ls 2 for A < e- Ai - J1 ) . 5 5 and = 50.' X > u. . + (u> + A) 2 + (w 2 cos i/w 2 - 2 - A .5 Solving for f£{q} we vj obtain 1200 \ (a + 3 + 2 2 10) {s 1 + 5 a 100) 3s 1 10 + (a 2 10) 5 s2 + 10 2 ' Thus. of this equation S 2 5 (0) = = .) X -(A 2 -^ 2 ) ( s + (s {a + A) 2 — .1O( 6sinl0(.e _A * A2 t - ^ sin 1 . f sinh ^A 2 .'(0) 0.A2) 2 ] . . |f + 2A§W 9 =f.w 2 ^ so (s + A) 2 + (u 2 — X2 ) J \/w 2 - A2 j ' w.a Thus + A) 2 (A 2 s2 +X S (s - A) 2 fl ImP \s J A) 2 -(A2-w2). The Laplace transform + 6 sin lOt. q(t) For A + 1 E c{ -.Exercises 7. s + 2X + 2As + w 2 .w2 1 - cosh V'A 2 + 2Xs + J1 = . 1/A 2 fl/X 2 is ^{ q } + —X{ q = E } 1 L + s k obtain E E 1 fl/(k 2 + l/LC) s/(k 2 + 1/LC) + */(* 2 + 1/LC)\ . 1 slug.At ). 9(i) 51. The sin The Laplace transform = sinf - sm(i of the differential equation s 2 < t. x'(0) = [e~ + l/LC) kt - mx" = -kx + Thus. is of this equation S Solving for + s «. initial 2k conditions are . Also. sin £ = sin(i — 2tt) we can = W/g = 32/32 = + 16x — /([).5 For A = w.Exercises 7.2 Solving for !£{x} x" is sin (t/y/l~C)] 0. is 1 %{x} + 16X{x} = . Recall from Chapter 5 that so that k i(0) = 0. Thus for = A + A) 2 and L\ 2 s The differential equation The Laplace transform N £b /l A) 2 / LA 2 ^s 1/A X + (s 1 s + A (s + A) 2 /' %{q} we = -EoC(l-e. the cos [t/^/LC) f(t).A( -A(e. Eq = ?(*) L(k 2 52. s 2 + 2A + w2 = + (s A) E s(s Z. t < 2?r write /(f) - a 2ir) il(t - 2tt). L + (s k){s 2 + L 1/LC) s \ + k s 2 + 1/LC 1/LC )• and 32 = Thus. = 16 lb/ft. since ( and m Now equation differential + k^LC „-2tT3 + J s 2 +1 we obtain „-2tts y+ 2 16) (s -1/15 s 2 + 16 + + 1) 1/15 S2 + l 262 (s 2 + 16) (s -1/15 s 2 + 16 2 + 1) 1/15 s2 + 1 _-2jts . . and k a:(0) + 4 sin 3t = x'(Q) 2 cos = & or 8 sin 3£ ^ The Laplace transform 0. x(t) = -^r sin At + t>U \ x(0) + ^- and x'(Q) = - sin 4(t 2jt)^ - - 2tt) <(< > m Now 0x'. of the differential equation \ - + 24 _ 2 + 9)2-3 4s ~ {s 2 2(3)a 2(3) 12 + 3 (77^5 27 Thus. 55. is dx 4 Taking the Laplace transform of both sides dx 4 and using y(0) El = ^M-V'(0)-/'{0) = j/(0) |^ = we obtain is . 53.21/2 _ _3 ~ s 2 + 7s + !6 2 + 7/2 a VlE/2 7^*15 _ ' (s + 7/2)2 + 10 (715/2)2 ( s + 7 /2)2 + (v^5/2)» Thus. mx" = -kx differential sid(* lo The Laplace transform 0. = —3/2 t 0.„ 2r „. equation is + x" .bx = conditions are 10 2 mx" = -kx + + 4 -(sin3t y ~ 3tcos3t) = 2 -t sin 3t o + 4 4 y o . Recall from Chapter 5 that that k sin -^sin4f + ^sint.5. The 0. Exercises 7. the 2 Ib. 2 54.W/g = 4/32 — + 7x' = I61 ' 5 slug. initial of the differential equation 21 .5 Thus. f = ~ lo = and 4 = 2k so conditions are is 0./ft. The = 2 -ts'mZt differential equation 4. Now 2 m = W/g = x" + 9x = 16/32 = 1/2 slug. Recall from Chapter 5 that so the differential equation ^x" The initial Solving for Z£{x] f(t). + 4 cos 3f we obtain VI x(t) = is + A. . we obtain _ -3s/2 .sin 3t . (t bU t Thus. f {X 27r) 1/ ( t - 2tt) 2tt _ .-tcos3(.3 ^{z} +-s + 7si?{z} + — + \&%{x) Solving for !£{x) - 2-jt. we have so that 1 3 — = -ax 2 + -c 6~*2 x + 1 1 y(x) 1 24 ' To find ci and C2 /i. y . system yields the c\ = 2 /EI and Umax = y{L) e2 = —woL/EI.5 = Letting y"(0) c\ and = y"'(0) we have C2 v/.2L/3)]. i .L/3) -%x . The this ci we = = and we obtain find C2 y(^) + c2 + ci y"'(L) differential 1 1 ^^i 2 and 4 we compute C2 y"(a:) Then ? 1 i a: = equation ^v>oL ^ - c2 + ^ x.sy"(0) Letting y"(0) = c\ and y"'(Q) — c% and using y(0) y"'(0) = |J ± = y'{0) (e~^ — we obtain . c\1 i c2^ .«-»•/») . .f 2L 3 y we compute = ci + c2 x + 1 - — wo *-!) *H)-(*-?) *(-t) and El 264 .Exercises 7. = Thus. Taking the Laplace transform of both sides s*X{y] . ] wou 1 1 w so that i To and find ci = Solving for From 56. (a:) WQ — EI L\ 4 m . is EI^ = vjo[H{x ./ 3/ L\ ( 2L\\. so that .y"'(Q) = ^{y} - sy"(0) = we have y"'{0) C2 = and using y(0) w ^ 1 \ (l y'(0) - e" = we obtain Ls / 2 ) . The and c\ C2 2i-j <> 2 . = . 21.- ~ £7 ^12" 1 3 is E/^ = ti*o[l-<Sf(x-Z/2)]. Thus.x 2 1 + -c2 x 1 1 — 1«0 we compute (x) = ci y"'(x) = C2 y + c2 x + - — and Then = y"'(L) = yields the r 1 . 4 . w system 21 L2 -f4 w (L\ Solving for ci and c 2 we obtain c\ = %woL 2 /EI = Cl+C2Lr + 1 «j and c2 2 £ ___ =0 3 iu £ = — ^wqL/EI. Taking the Laplace transform of both S Letting y"{0) = c\ and 4 sides . .Exercises 7. c2 Solving for system yields the we obtain cj 2L DO + _ ^L = \wqL 2 /EI w L2 6 £/ w L 1 and = — ^woL/EI.5 Then y"(L) = y"'{L) = w 1 i ^ 57. «B[a: I 24 18 equation 1 r Ll / differential 2 £7 x. y{x) To find ci and C2 1 -c. c2 Thus. ^ ds = a 3 . is -|[^M. we obtain The Laplace transform c\ = wqL s /24EI and of the differential equation C2 = —wqL/2EI. dx* we obtain s^{y}~ S y^)-y"'{ Q = ) f 1 -. The differential equation and using j/(0) = y"(0) = is d^y El-r—r dx* = — =— El d^y mjo or r Taking the Laplace transform of both sides . I Letting j/(0) = C] and y"'(0) — C2 wc have " £/ + 55 + S2 S4 Then To find c\ and C2 we compute Then y{L) = = yields the system ^ = -w Solving for cj and C2 59.5 58.y'(0)]-^M = ~ Then and = -4- + This is 3 a first-order linear differential equation with integrating factor e/' s 3 ^{y} = . . Thus. Thus.Exercises 7.(\ds = J so 5?{y} = —2 + —c s and 266 s + c. The Laplace transform of the differential equation ~ \s 2 Z{y} - - 2/(0)] 2 is ^[sX{y}\ + 2%{y} = 0. = = ln^} -~l -21n( S + 2) + c => {»} = ce" 2 ^ = ci^ + 2)" y(£) (a) The Laplace transform . Solving for !£{y} we obtain ^ {V} = as 2 + bs + c- Thus.5 60.J Exercises 7. } is = sin t. V2 1 -21 and cife = 1 f' — (sin I t cos t — cos t sin t) dT — Jo cos t — t sin t -+- (sin i) / Jo cos (In cost!.2{y}\ - ( 2 S X{y} - 2s . From part (b) of Problem 61 the solution of the given initial-value problem - sec ry\ (t / t) dr where y\=!£ Jo 1 1 1 L + . 1 Thus. the Laplace transform of equation (13) gives \as 2 + &s + cj a I w as 2 + 6s + cj a by the convolution theorem. Vl (b) Now if = %?{g(t)} =i?_1 {as 2 + fe + C }- G(s). Tlien -S 2 .Z{y} } - 22>{y} + 2%{y} = and This is a separable differential equation so ^§jjjy 61. of the differential equation is (as 2 + bs + c)l£{y} = a. I dr— sin t (cost) / jo cos T dt . 62. The Laplace transform = ^sm4((-27r) fl l/((-27r). The Laplace transform of the differential equation is of the differential equation is so that 2. 268 2?r) +*U(t - 4w)].6 1. of the differential equation is S -\" 1 bo that J/ 6. of the differential equation is *<»>so that 3/ 5.6 Exercises 7. ( .Exercises 7. The Laplace transform p~ s 2 so that S 3. The Laplace transform = 2e- 1 + e-< i_1 of the differential equation ^(t-l). = sin( t -^(t-f) + sin( -f)^( -f)=-cos^(t-0 + co ^( -0 The Laplace transform ( S i of the differential equation is so that y = cost + sint^fi. The Laplace transform sinf + sini^((-27r). is " ?TT( + e *<»> = 1 a ") so that = i/ 4. of the differential equation is „-2tts so that = !/ 10.3 "! S in3{t - 12.2 *'.{t of the differential equation + 4s + 52 3 (s s2 13 + 2)1 + 3 2 + 4s + s + (s ~ 1)a U(t-l). - 11 1 4s 2 2 s2 + r 1 _1 |.tt) 111 2s\ . The Laplace transform of the differential equation 2 i is + 2s 11 11 2 s 2 s + (l 2} + O so that _ 1 8. The Laplace transform = (t-l)e. -I]*(t-2). The Laplace transform e -2(t-2.) sini 9/(f-27r).6 7. is 13 +2 + 2) + 3 2 2 + 1 3 (s + 2) 2 + 32 so that 2 y 1 = -e~ a sin 3t + e _2£ cos 3t + -e _2((_.r) sin 3(f + ie. of the differential equation %{y} = is + {s 1)= so that y 11. The Laplace transform s 2 i e -2£*-i) <?/{(- of the differential equation -2 8= 3 e (s-2) s(s-2) is 3 1 4 s 1). The Laplace transform X^ of the differential equation 1 (s-l) 2 (s-6) + -2s e 5-1 5 (a + - 3jt). The Laplace transform O Of J. is -is e (a-l)(s-6) 11 25 c 37r) «(£ - 11 l) 2 25 s-6 1 1 5 s- tt)^ (t .Exercises 7.2s-2 so that J 9. 2 + i/'(o)^ + Using y{L) 6EI\ 3 I2 0< x3 )' a. < 4 ^ so that { ^ JfOO ' FOB rO dt = / 9(t)d(t-to)dt = g {t ) fV 3l e- st S{t - ) g(t)6(t = e- st - t ) dt - 6(t / °. X we know that 13 ^iv"(o).. { ( j * e _3. 5 of the differential equation is 13 12.Exercises 7. ) < (>0 then from /°° /(*)*(* J-oo - (7). . If /(0 = f 0. 4) dt = f°° J0 270 <5(f -4)dt = We' 12 .6 so that _A e t-4 + 5 13.-f). Using y"(L) = = and V 14. The Laplace transform l e 6(t-4) m{t - 4). FVom Problem we obtain EI A X 6EI + X = = and Assume io > X 2) 2] \ and g(t) = i§(x-f)\(. -co 16... we obtain !^(lV 15. £/ 6 s" so that 3 . y'{0) = — — eiiiwt. w 1. of the differential equation s + is s 5 so that V = S(t) - 5e" 5t ..3ir)%t - 3jt) of the differential equation = ( e" cost is * <»> = i so that y Note that 19.6 17.Exercises 7. . The Laplace transform s 2 of the differential equation X{y} - sy(0) - y'(0) is + 2{sX{y} - y(0)} + 2%{y} = -e" 3 " so that *M. +5 + e . The Laplace transform of the differential equation is so that Li Note that 20.(s+ + 8 1 1)2 + 1 ( S+ l)2 + 1 and y 18.3.(t . i{0) = 1/L ^ The Laplace transform 0.) sin t°lt(t - 3tt). = e"'coB( - The Laplace transform e~ (t ~ 3w) sin(( . ) 1. 2{f{t)} 2. %{f(t)}= 3. = 1^0.jr)H(t .2a -e.Chapter 7 Review Exercises Chapter 7 Review Exercises = te' . False. False.£{flm2t V(t Theorem +4 ^{tsin^^-l^ 12. since /(() 5. 4s S2 ( . f e- if(e. False. True. if J Is .3 'sin2(} 1 1 11.tt)} = s ~ 5 5 l(s-5) + 22 2 -™ ? | 2 -e (s-5) 2 + 272 1 22 / . et = = 7 dt = -U. co F{s) 6. 6t + 1.tt)} = if {sin2(( Utf-i ^ -103 + 29/ £ 2 7. (See f + 4)2 — .4 in the text. consider f(t) 4. since lim e _. consider /(£) jo dt r°° ^ . e 10i .i3 ) = (e (s t -1 ' 2 10 ( ) — + 1 = . and 3) 2 g{t) — 17.12 (2-t)e~ et dt=-^-^e.s [ 10. Chapter 7 Review Exercises 18 = 20 . cos7r(t - - +sin7r{( 1) - - 1) = H-^(t-^1 = ^^^t ^{ht^I L + nV + tmr i 2s 2 I 21. y{e~ 5( } 22. ^{(e 84 /^)} 23. ^{e°<^*>/(( 24. 1*1= 2 Z/ mr J exists for s > 2 I s (n 2 7T 2 ) /L 2 J -5. "^(s " ») - - k)%t - = k)} e~ ka %{e at f(t)} = e- k *F(s - a) l f dr = t Jo 25. (a) /{() =t- 26. = + = - jt) - sin( = 1) -<U(t _ =t- 4) _ -(a-l) (t - l)*{t - 1} - *(t - 4) _J_ e ~4t<-l) s %t - 3tt) = — -sin(( 1 - jr)<3f(t [(t - + 2]*(t - 2) = 2 + 2) - (i - jt) 2)<K(t 5 *{* »,}.^ + * l ( ( /(£} - -2s 2' S (a) !)+<«(( + sin(t --^- + -^-3- = 2 - 2"tf(t - 2) + (b) 28. - i]H(t _ I e -«- sint"U(* *{/(,)} 27. (a) /{() (C) l) ^/w}-(rh?-(rriji e (a) /(*) (b) - = (b) <«> [(( =t- t°U{t =i- 2(t - - 1) S -l)2 + {2 - t)K(( - l)<B(t -!) + (*- 1) 2)«« (t (2 - - 2) - 2) - 2) - 37r)K(t - 3jt) Chapter 7 Review Exercises - (b) + we obtain 29, Taking the Laplace transform of the differentia! equation (s-l) 2 so that = y l) 3 (a- 2 + ^tV. 5fe* 30. Taking the Laplace transform of the differential equation we obtain ^ {y] = + 20) (s-l) 2 (s 2 -8s 111 6 169 a - s-4 6 - 13 (a 1 169 1)2 (s - so that y y = —e< + - -I( e ( 169 13 —e 169 4t co S 2t 5 + 22 4)2 — + e 338 4t {a - 2 4) + 22 sin2(. 338 31. Taking the Laplace transform of the given differential equation we obtain s(s 2 -4s + 6) (s-2) 2 (s-2)2 + 2 « V +2 so that y = 5W(t-*)- Be 2 *'"*) cos ^2 (( - jt) « (f - tt) + 5 V2 e 2 32. Taking the Laplace transform of the given differential equation My) ws = s3 s 2 (s 6 + fo2 + 1 + l)(s + 5) 1 1 1 5 s2 ' + 3 2 + s 1 1 s 5 s2 1 1 2 s+1 + + + 5) e 50 1 1 4" 1 H> + l e -(t~2)^( t _ 2) - A ? + i)( s +5 s 1 W 1 +5 1 ie- 100 e -5('-2)^(( 274 V2 (t - 2s _ tt) %{i we obtain so that + -*> sin S (s 1 s ( -2* 13 1 1 -+ > I s 2 (s+l)(s < 2). 2B e ) 2s Chapter 7 Review Exercises we 33. Taking the Laplace transform of the differential equation s 3 s 3 (s-5) (s-5) 2 obtain 2 1 1 25 s 2 125 a H 127 1 5 s3 125 s - _37_1_12_1__1_2_ ' 25 s 2 125 s 5 5 s3 + _37 125 s so that y 2 2 125 25 = ( _ -f , 2 + 27 e ,, 5t - 125 5 34. Taking the Laplace transform of the integral equation we obtain + s(s 5) 3 . 5s 2 35. Taking the Laplace transform of the integral equation „(0 = i +t+ 36. Taking the Laplace transform of the integral equation (#{/}) so that /(*) 37. The = 2 = 6-^ or +5 „-« that so that 5 s we obtain |t a . we obtain #{/} = :M * ±6(. integral equation is 10i +2 2F + 2t. Taking the Laplace transform we obtain ^ A (s 3 2 s ^Wl0s + 2 The differential +2 = + 2) —+ 9 s ' 2 s 2 + = -9 + 2t + 9e~' /5 Thus, 38. s 2 s (5s equation ' 45 5.s + = 1 . is + 10^ + 1009 = 10- 10*(i-5). — +" s -* s 2 + " ' a + 1/5 Chapter 7 Review Exercises Taking the Laplace transform we obtain 20 + 20s + 200) 2(s 2 1 3+ 1 10 s 10 (s + 10j 10 10 2 + ~ M sin 10( 10 2 + 10 (s lO) 2 + (l-.-) 102 so that = q{t) To~ e ~ l0t cos 10( " To e ^o 39. Taking the Laplace transform of the given differential equation we obtain 2{y} = (L 4! 15! EIL \48 5 6 2w s 120 s 15! 120 s3 2 s s4 6 so that V where y"{0) = = cx 2wq and L 1 EIL 48" y'"(Q) 120" = c2 Using . = ci u) j/"(Z.) = c2 2 , Q2 3 2 = we find and y"'(L) L 2 /24£7, ci --) ~\~ 2/ 120 V ' = -woL/4EI. Hence wo 1 5 YIEIL ~H + L i2 4 ~ 2 T 3 L3 + 2 T L\* / 1 / L\ 5^2) n*-*) 40. Taking the Laplace transform of the given differential equation we obtain 4 s4 2 +4 + +4 s4 4 wq 4EJ s4 +4 so that y = sin + where j/"(0} — cj a: sinhx 4El and + sin j/"'(0) ^(sinxcoshx — - = ^ C2- Cl cosh Using _ wq ~ El j/(tt) cos a: sinh x) ^= cos and - y'(7r) sinh = we find WO cosh j siring 2 sinhTr 276 - sinhn ~ ^)]^ x — ( Chapter 7 Review Exercise Hence, wo smh f* y - —=77 2£7 . , . sib x smh j - sinliTr + 8in [ (* - —wo— cosh 5 (smxcoshx - coszsinhx) cosh 1) , 4£7 smh ir x cos ( " I) siIl!l (* " 0] * x" I) ( 8 Systems of Linear Equations Differential Exercises 1. Prom Dx = 2x-y 8.1 Dy = x we obtain y = 2x-Dx, Dy = and 2Dx-D 2 x, and (D 2 -2D + l)x = 0. Then = x 2. c\e + C2te Dx = 4x + 7y and Dy = 2 (D 2D - 15)x = 0. Then From Dx = -y + 1 and cie Dy = t and y = - (ci + c2 e~ 3 5i x - and ' we obtain y t y c2 )e* + C2ie'. = ^Dx - x - 2y we obtain y From x= 3. l = - c2 e~ 3t ^c\e ht — t - Dx, Dy = Dy = ^D 2 x - ^Dx, and fx, - 1 . D 3 x, and (D 2 + l)x =1+ 1. Then x = ci y = ci sinf cos t + C2 sin f + 1 + i — 1. and _^ 4. From Dx - 4y = 1 and x 4- Dy = 2 x — c\ V = -7 c 5. From (D 2 + 5)x-2y = and (D 2 + 1 2Cos( — 1.1 = 0. and (D 2 + \)x = 2. Then 2 -ci suit 4 and -2x + (D 2 + 2)y + 6)x = l)(D 2 , + C2 sin t + and 4 +t— - \Dx - | Dy = \D2 x, we obtain y cos t C2cosf — 1 ci sin t 4 - . 4 we obtain y = £(D 2 +5)x, D2 y = £(D 4 +5D 2 )x, Then i = ci y = 2c\ cost cost + C2 sinf + C300S VEt + C4 sin <Sfit and 6. From (D + l)x Dx = -\{D + 2 + (D = ci -C3C0S y/6t — -C4 sin \/6t. — l)y = 2D)y, and (D y + 2c2sint — 2 2 and 3x + 5)y cos v^5 f = + (D + -7. Then + C2 sin V5 — i \ and 278 2)y = -1 we obtain x = - £{D + 2)y, Exercises = 1 7. Prom (P 2 D2 x = 4y + ^ 2 ( l~3 e l and ~3~ C2 D2y = + 4){D - 2)(D + 2)x = x ~ Cl -3e = \ 4x - ( v5t+ cos I I — we obtain y = e' 2 \ -C2 J - Ci 3 i- . smV5i + - |D 2 x - ^e 1 ZJ 2 y , 8. . = jP 4 a: \e l 4)x = , and Then . ci cos 2t 4- c 2 ain 2t + c^e 21 + Cie~ 2t + -e l 5 and y 8. = -Ci cos2t- C2sin2f + 5)x + Dy = and (D-5)(D 2 + 4)y = 0. Then From (D 2 (£> + + {D - l)x a: = c\e y = C4e + c3 e 2c + C4e~ 21 4)y = -e'. we obtain (D - 5)(D 2 bt + C2 cos 2f + C3 sin 2t 5t + C5 cos 2t + eg sin 2t. + and and (6cj so that C4 + (D Substituting into l)x + (D — 4)y = + c4 )e + {c 2 + 2c 3 - 4c5 + 2ce) cos 2t + = — 6ci, From £)(D 2 (-2c2 +c3 - 2c 5 - 4ce) sin 2t = — 5C2, and C5 = 1/ 9. gives st Dx + D 2 y = e 3t and (D + l)y = -8e 3 Then -6cie l)i 4- 5t + cos 2t + (D - = l)y - ^C2 sin 2f. 4e 3t we obtain D(D 2 + l)x = 346 31 and '. y = ci + C2 sin f + C3 cos t — _4_ 3i 1 = C4 + C5 sin ( + ce cos f — 15 -re 15' and Substituting into = ci, C5 l)x + {D - - Cl) + {C4 so that Ci + {D = C3, ce (C 5 - = — C3, from D 2 x - Dy = D{D + 1)(D t and (D + 3)y = -1 - 3t. + 4e 3t gives - C3 - C6 C2)sin( + (C6 e . + C5 + C2 ~ C3)cost = and ci — C2Cos( + C3S111 + 3)x + (£> ci + C2e 1 + 3)y = 2 17 — 15 e we obtain . D(D + Then x and = = x 10. l)y -+- = ' + c3 e 3( - + t ^t 2 + 3)x = 1 + 3t and 1 Exercises 8. = y (D + 3)x + (D + Substituting into + c5 e" + cee~ 3i + 1 ! ci = Z)y D 2 and 3(ci+c4 ) + 2 - Dy = x -t . gives t + c 5 )e- = l 2(c2 2 2 and + c s )«- + 3(3 C3 + ce)e- 3t = 1 (ca — — c\, so that C4 = — C2, C5 ce = — 3^3, = -ci - -l)x-y = and (D - 1)(D 2 + L> + l)x = 0. Then l)x + Dy = y 11. Prom (D 2 {Z? and c 2 e' i - 3c3 e~ 3! +t- we obtain ^t = t/ 2 . (£> 2 - ljz, Zfy = (D 3 - D)x, and ' x = 4/2 +e l cie >/3 C2 cos -—t + \/3 C3Sin -—-t . and 3 / 12. \ \/3 _ tj2 3 From (2D 2 -D-\)x-{2D+l)y = land (D-l)x+Z)i/ and (2£> + l)(D + l)y = \ _ (/2 . v/5 = -1 we obtain (2D+l)(D~l)(D+l)x = -1 -2. Then x = c\e~ y = c4 e" t/2 + c 2 e~ l + c3 e' + 1 and Substituting into (U — \)x + Dy = —1 (-|c 1 so that C4 — — 3ci, C5 = — 2c2, Prom (2D~5}x+Dy = e 1 and 2 + C5e" - 2. 1 gives -^ )e' 4 1/2 + (-2c 2 -c 5 - -3cie _t/2 {D-l)x+Dy = 5e t = Cl e - 2c 2 e _! - we obtain ^ 4t + U l •J Dy = -3cie 4t + 5e ( e -1 = so that j, = --cie 4t + c2 + 280 5e 2. Dj/ Then and ) and y 13. l '' l - = (5-2D)z+e and l (4-D)a: = 4e*. 1)(D 2 + so that D + l)x = 0. Dy 8.ci . = = - c2 + D2 x - 2(D 2 + D)y .1)(D + l)z « 1 2 and Substituting into y = (D . and the system (-D 2 + D From Dx + z l . = -Dy.l)i + Dy + Dz . C7 = V 16.l)x and — c<it l)y = _1 - ^( + cge"' - -f + {D + l)x x = ci+C2t — C3e' . we obtain z = ci+C2e -t cos( + C3e -i sml + 2 and 2 1 7 cos t 4- 5 . + C3 cos — -t 2 e~'/ sin ^t + (^~c 2 - ^ 2 e"'/ cos ^t.2y = 0.C3)e~ sint and Dz = x we x^de' + e"'/ 2 y = Cl e z = cie ( + ^ic C2 sin 2 — t . ™t + and x l)x (-^-c 2 ~ + 2y \c-& + Dz = + Dy = -e 1 e l \ e~ t/2 cos ^t.sin* and x D(D 2 + 2D + 2)y = — sint. ! + f-^c2 + ^cs^ e~ t/2 sin = e (D . Then - C2 (D 2 - From (D . From Dx = y.c2 t 4.1)(D + l)y = 1. 1 2 5 5 -sint .C3e* y = -c\ + te' and 15. .0.\}x + (D 2 + = + C2 1. (ca + (2c g _( - 2c4)e + e (2c7)e = 1 and l)t + Dy = + Qe" 1 ^t .~c^j + (c2 - obtain x c C3)e~ cost + = D2 = D 3x t.c 7 e' - 2) l)i C2 + + + 2. FVom Dx+Dy = e* and (-D 2 + D+l)x+y = Owe obtain y= andD 2 (D-l)x = e ( (D 2 -D-l)x. x = ci 4.1 + c 5 + (c2 so that eg c 5 4. and 18. we obtain 2 D 2 (D . + (D 2 + l)y = 1 and D 2 (D . D 2 x = -D 3 y. t (c2 4. gives 1 - (ce 0.cet 4.te* + + c3 e* + c 4 e e l . C5 (ci = \)y .suit 5 and t 17. = -Dx 4.cost. Dz = -D 2 x + e'. Then Prom « - C2 c\ 2 .e 4. = (D3 — D 2 — D)x. Dy = = z.1 Exercises 14. Then we obtain 2 and (-D 2 4- l)x 1 . Then .. (D . ) = eg C4. Dy = \D{D 2 - = \{D 2 - x— - and [D l)x.3t . 282 1 we obtain + f2 so that the system has . From Dx — so that 6y = x 0. — Dy + 2 = {D + 1)(D .y D+l D{D + l) 2 z = 0.1 Exercises y 8. 3 — 2){D 2 c l e2t — c 2C0St — cgsin t. x = c\e y = --cie z = --ae -t + -c 2 e. and = — 1c\e 2t — a cost + C2sin(. 2 19. and z . + (D - = ( and Dx + l)y Dj/ y and Dx = cie"' + t2 - 4f +5 no - 2Dj/ = t 2 and (D i1 we obtain (D cie- t + l)y = 2f = It 2 - t. Then + 2( 2 -5( + 5 so that x 22. (D + l)y - = z 0.-C3e and 20.2(D + l)y = solution. Prom Da. From (D + l)x - = z 0. Prom 2Da.2r + 5t. we obtain 0-6 D -6 1 -D 1 —D 1 1 1 -D 1 -D = Then 0. = = + - l)x -cie" c 1 + c2 + -r . C\e 2t ^ + = l)x -2e* so that + C2 cos t + C3 sin( + e'.3)(D + 2)x + y — Dz = and x 0. and 21. D+ -1 X — D -1 1 we obtain 0-1 -1 D+l ao that + Dz = l -1 D -1 Then X= C\ + C2& y = c\ + z = ci + C3e * {c2- + C3te c3 )e _t + c3 te~ ( . -c 2 e dl . .-c3 e -2t t + c2 e 3f + c3 e + 2( . Thus 3. value problem —sinv^A + 6 / ^-cosv^f + ^sin v^ij +1.2D)y = y y = x = -e e +3c 2 = -e 3 . The solution of the x + 2 1 - Thus sin\/2(] 0. Thus and Using z(l) = = and y(l) -(2ci + c2 )e~ 3! . Dy = -(D 2 + 5D)x.-3t+3 t( = -\{D-2)y From Di-j/ = -1 and 3x + (D-2)y = Owe obtain x 2 . solution of the initial value problem _ x = y = _ e -3 t+ 3 + 2ie -3 i+ 3 1 and {D [c\ co3\^2* l l + c2 = e -3t+3 2 is .3 = cie ci 2ci Thus 24. 8. Prom (D + 5)x + y = Then Ax and Ax. — 2D + 3)y = Dx = -\{D 2 -2D)y. The 3 + c2 e Then -\{D . + c2 sin\/2t) + so that 1 and 1 Using x(0) = j/(0) = l [(ci - \/2c 2 ) cosV2i + (v^ci + c 2 ) we obtain C! ( ci = —1 and c2 = Cl = e t = e' initial f-^cosv^f 3 ?/ 1 = .2c2 te~ M we obtain 1 3 = -(2c.{D + l)y + (D + 5D)x + (D + 5)x = 2 = we obtain y = -(D + 5)x so that and (D + Z) 2 x= x = ae~ 3t + c 2 Ie~ 3t y = 0.V2c 2 ) + . ci = e 3 and c2 3 = — .3 -2c2 e.= V5/2. 3 is + 2 - .+C2)c.e 1 Exercises 23. 3 Taking the Laplace transform of the system gives 1 s<e{x}-L = 2X{y} s-1 s%{y}-l = 8Z{x}-^ so that eBI . = -e s 2.Dy = {D-\)x-y = Q. 284 173 « 53 _ it 160 s 1 +4 .1 _ *\ 4( 160 1 1 . 2\Vt ~ s 3 + -s + — — - 777 s{s 7s 2 777752 \){s 1 l T7T7 -16) = 1 " 16 7 s 77i 8 —1 7 77 15 s 7 1 + 173 96 53 1 s - 4 and y Then — 1 8 16 15 t 173 e* H e 53 4i « e 96 1. ^ e 3 and y y = 3 + -e -e' 3 . Exercises 8. Adding the Dx . Taking the Laplace transform of the system gives = -2{x) + X{y} sie{x} s%{y}-\ = 2%{x} so that 1111 + 111 + 1 >-l)(s + 3s-l 2) and s(s- s 2 2 2 1 . A system is 25.Exercises 8. 3s + 2) l)(s 3 s - 3 s 1 ' 2 Then a. Differentiating the equation we obtain equations we have x -t- y Thus 2c2« Dx = Dy. Dx = 2c2e 2t and Dy = = 2ae 2t = Dx.2 1. - 5s .sin 3t. 2 cos 3( Taking the Laplace transform of the system gives (s + 3)X{x} + sX{y} = ± s 1 (s-l)X{x} + (s-l)X{y} = - s 1 so that „. t t j and V -+r + 1 = 1 .3.te 5. 5 = — cos 3t — x -x' = — sin 3i. 3s(s- 1111 1 l) 2 s- 3 3 s 1 4 1 3 (s- l) 2 and l-2s - 3s{3 1 1 ~ l) 2 _ _1 1 1 3s-l 3 s 1 ' 3(s-l) 2 Then 1 1 1 I= 3"3 e -3 . 3 4 t t& ' 3 Taking the Laplace transform of the system gives {2s-2)%{x} + sX{y} = 3 (s-3)2{x} + (s-3)X{y} = Z so that X{x} -s-3 \L 1 s(s-2)(s-3) 2 5 1 2 2 s-2 s-3 h s' and 3s s(s - ~ 1 2)(s -3) 1 1 5 6a 2 s 8 1 - 2 1 3s-3' . 7 — . Taking the Laplace transform of the system gives a X{x} + l-X{x}-2X{y} sX{y}-2 = 5X{x}-X{y} so that -s-5 s2 + 9 s 5 3 +9 3 s2 +9 s2 and Then 11 = -x — y 4. Taking the Laplace transform of the system gives + l)2ix}+<e{y} = l 4%ix}-(s + l)X{y} = l {s so that |X| ~~ s2 + 2s + 5 " (s + l) 2 and 286 + 22 + 2 (s + 1) : .2 Then 2 6.Exercises 8. y 2 6 2 Taking the Laplace transform of the system gives + \)2{x}-{8-\)2{y) = -\ {8 sX{x} + + 2)2{y} = (s l so that + s w„i = and 1 /2 s = + s+l S2 ( S + 11 s 2 +s+ /2 1 + (V5/2) : ~ 3/ 2 ~W <*r x] + l/2) 2 l (s + 1/2)2 + (^3/2): Then y 7. x" 1 — 2 2y2 + x = --( + 3 y= sin V2t. 2 2 Taking the Laplace transform of the system gives 2 (s + l)2{x}-X[y} = -2 -X{x} + + l}X{y} = 2 (s l so that * ' y = 113 -2^-1 X lx] S4 + 2s 2 2 s2 2 s2 1 +2 and Then 8. = e~'cos2t + -e _t sin2( and Adding the equations and then subtracting them = y -e~* cos2( + 2e" f sin21.cos i 5 2 X{x}+3(s + l)X{y} = 8 .sint.Exercises 8. 5 11.2t . smt 5 and 1 = 1 .2 +3 + 2s + 5 -s . 1 13! 4! 2i? + 3^ + and v. 9. n l 8 . gives Taking the Laplace transform of the system gives ^> = r„r . 14! .2ro + -e >.81 5 s - so that = 3 X{ X ] + l) 2 + 5 5i 2 + l 1 1 2 211 s s2 s3 5 s- 2 6 *81 5 s2 + Then 4 = -e 2t —-cost 4 x 5 8 5 . y 5 6 . 13! so that 1 3 10. s2 + s + (s l + 22 l) 2 (s + l) 2 + 22 ' Then a. Taking the Laplace transform of the system gives s + 2 T 1 5 s2 + ' 1 . . Taking the Laplace transform of the system gives (s (s - 4) and w „\ - 2)(s 2 23 + 1) +4 s 3 (s-2)(s 2 + ^ + 2)if{x}-2s ^{?/}=0 4 (a = 3 4 14. 2 so that 1112 sj + i) 1_ Then = i +*+ 1 ^f 2 - e~* and 1 »=3 te t 1 1 » -r =r -/ 1 + r -t 1 -3- 12. Taking the Laplace transform of the system gives 2e" -3tf{x} + (* + l)tf{»} = i + ^ so that -1/2 (s-l)( S -2) 1 (s-l)(s-2) 11 11 2 + e" 1 ~7^T 2a-2 s-1 + 1 I s-2] and _^ s _ 3 _i (s a/4-1 .\)(s i 2) + ^ -s/2 e" " _ s ri i +2 (s-l)(*-2) _ 3 i i i Then and «(*-!). 288 .Exercises 8. Exercises 8. 4 s 21 + 5s + 2 4 s = 7 T5-nr l)( s 2 + n + iv. +6 2 V6 5VSs 2 + 6" 5s 2 + 6 l y/6 5\/6s2 5 1 5s 2 + IT .cos on mi and + £2(^2 — sin 5\/6 and x 2 represent displacements positions. mi = m2 = = 1. + k 2 (x 2 — x\) -^2(^2 xi(0) = ~ 0. Taking the Laplace transform of the system gives -2%{x } 1 + (s + 2)X{x 2 } = 2 s so that s 2 <fif s4 + 2s + 2 _ + 7s 2 + 6 2 s 5 s2 + 1 1 2 5 s + 1 and [0[ . Since the net forces acting respectively. In this system xi sin f + 5 —kixi f 5V6 mi and of masses m2 from their equilibrium m 2 are — and k 2 (x 2 — x\) — k$x 2l Newton's second law of motion gives mix" — — k\x\ 7712x2 Using = sin v 6 v61 5 £1) V6 1^2^- . x 2 (0) = 0. *i) ~ zi(0) k3 x 2 = . s = .cos t + 14. and .2 5s 2 + l 6) 3 r„2 2 (s 2 5s 2 +6 1 Then 11 = % cos * + 5 \ sin ( — ? cos s/6 5 and ^2 = 4 . and x'2 (0) = 1.2 13. The system is = Z" + -3ii -xi) 2(x 2 = -2{xi-xi) H (0) =0 *'. 2 - 5 ~ H 5 ki=kt = ks = 1. -1.(0) = 1 *a(0) = 1 4(0) = 0. Kirchoff's second law. (2 + a 2 )X{x }-X{x 2 } = -l 1 ^{x }-(2 + S 2 )^{z 2 } = -l 1 so that Then xi 15. (a) By Kirchoff's - E(t) first = — -^sinVSt law we have Rij + Lii'2 and E(t) = ii and + 83. = (c) ii (a) By i2 + i3 = Kirchoff's have E(t) i3 — = tf = = 80 -3- - 9 20 c - . + £q so that q is + Rii 2 = + fi2i3 + ^i 3 = 290 E(t) 0. 20 .000 = 400)i?{i 3 } s so that vr 1 _ ~ _ 8.01i 2 + 5i 2 + 5z 3 = 100 0.»-™«»3 0. 12 = .9001 first Li[ 80 . By 22 = Jfci+Z^ = 12 -^sini/St.2 we obtain taking the Laplace transform of the system.0125i 3 + 5i 2 4. law we have ii Rii 2 and E(t) = = i2 Lij + CR\i'2 — CR-iH so that the system Li'2 -flii 2 + 13- + Li'3 By R213 Kirchoff's second law.0025i 3 - 13 = 100 — 9 100 _»„.Exercises 8. . and . = or Lii'2 +Ri 2 +Ri 3 E(t) and on each loop we have L 2 i'3 +Ri 2 +Ri 3 = (b) Taking the Laplace transform of the system 0.20e. on each loop we = CR\i 2 - CR2 i3- Then .™ _BMt — g + — . 90m e 9 — .000 s2 80 + 900s 1 80 _ 9 3 1 9 s + 900 ' Then 13 16.5i 3 = 100 gives + 500) #{i 2 } + 5ooy{i 3 } = (s 400y{i 2 } + + (s . 17.2 (b) Taking the Laplace transform of the system i' 2 -10i'2 + + i' z = 10i 2 120 - 120 + - 2) + 5i'3 + 5i 3 = gives (s °U(t -10s2{i 2 } ™ + sX{i 3 } = 2{i 2 } 10) + 5(s + 1) %{i 3 } (l .-*)- (i + 10 s 240 + 5/3 s +2 (i-. Then + 48e.240e. t2 and ^-T3 e + 375 e I469 + 145 113 COSt+ 85 Slll( Tl3 145 s — + TTT7 ^r-^r + 15 113 s + 2 1 85 113 s 2 1 + 1 .-).e" 2s ) - so that 120(s (3s 2 + + lls 1) + 48 (i-„-)- 10)s s 60 + 5/3 s +2 + 12 s and 240 3s 2 + lis 240 -. Taking the Laplace transform of the system i'<2 + 613 = 50sint + 612 + 613 = 50 sin f + 13 lli2 gives ( S +ll)if{i 2 } 6^{i2} + (s 50 + 6^{i3} = + s2 l 50 + 6)^{i 3 } = s2 + l so that 20 50s (s + 2)(s + 15)(s a + 1) — 1 = -tz -+ 13 3 + 2 375 1 1469 s + Then 20 _.2 " 3 [l2 . -2 >] * (* - 2).5(t .5 " _2 ^3 .Exercises 8.2) and (c) ij = i2 + 13 = 12 + - 240e.a 288e" 5 '/ 3 - - 300e [240 e -2 ' - .300 e .2 < ( 2).60e" 2£ - *3 = 12 i3 = 240e" 5i /3 [l2 + 48 e ..5(i~ 2V3 - 60e" 2 < l -2 V(t >] .2 «~ 2 >] °ll(t - + 288e.6 "3 . i2 280 cos t IT§ Ri 2 + Li' - + — 810 .000 s(s 2 + 200s 6 + 20. KirchofF's second law.005i 2 60 + i2 .Exercises 8.h = gives -200if{i. + } + 200) (s — 120 + 100^{t 2 } = «J?{il} = ^{13} so that * 12 ' _ " 24. ( law we have Rt2 and E(t) = — ii Li\ 30 - 11 V2 = + i2 + 250 _.000) 1 6 "5s 5 3 (s + + 100 100) 2 + 100 6 100 2 5 (a + 100) 2 + Then h= 5 _ 6. mot. on each loop = CRi 2 so that q _ 15t ii = i3 1 q = we 19.100t cos 100t. Taking the Laplace transform of the system + 50i 2 = 0. Taking the Laplace transform of the system 2i[ 0.005i'2 + i2 = I .-iow cog 1M( _ 6 e _ 100i 5 5 1Q0( 5 and ti = 0.|e. ( e +j^-9 e By '3- + CRi'^ Then - = E{t) - = 0.005i 2 + + 50i 2 = i2 60 -h = gives 2si?{*i} + 50^{i 2 } = — s -200^{ii} + (s + 200) %{i 2 } = so that Xin} = s{s 2 + 200s 6 1 6 5 s 5 + 5. 20.000) 5 (a + + 100) 2 100 6\/2 .2 I3 18.5i[ 0.(50^2 } 2 100) 2 havt CRi'2 so that system ' 100 2 ' is .(50V^) 2 292 5 (3 + 50V2 . By Kirchoff's E(t) = Li[ + = first 25 TS m 1 --* 2 . Using across the shorter loop gives 1 ifl.s 2 + 1 and = 50e _1 e" (t_1) sin(i - 1) 293 = 50e _l sin(* - l)<3f(t - 1).^e. and taking the Laplace transform of the system we obtain oil ( S +l)J£{i 3 }-^{9} = 0. so that 2{l} = 50e (a + _1 l) e.Exercises 8.005i 2 Using KirchofF's first + j 2 = | . 1 . (1) so that Then dq 1 „. E(t) = 50e-"W(t - 0.2 Then 5 5 5 and h = (a) 0. dq/dt we have ii — ^3 = dqjdt.100*cosh50\/2t - law we write Kirchoff's second law = ii 12 + ^3.hRi + hR\ + i or £§ + *2i3-^ = (b) Using L = R\ = R 2 = C = 1. q(0) = i 3 (0) = 0. and Summing the voltage drops across the longer loop gives E<S)=i l R l + Combining this with (1) L^ + Rii*. bOe^e-^^t - = 1) 1). Since E{t) E(t) = = ii and summing the voltage drops ^e- 100 'sinh50v^(. iRi + -5. . we obtain + R2 3 . 6x2 + + 10y 10xi 294 + f2 + + 2 + 1 1. X2 x2 x2 - = = y'.=( + = -^'-20 2 3 . Taking the Laplace transform of the system + 02 + 80i = 40? + 0? + 202 = ' 2 gives 2 4 (s + 2) +s } 2 + s i?{0.} (s 2 2 i?{02 } = 3s + 2)^{02} = O so that (3s 2 + 4) or Then 02 S ( 2 + 4)^{02 } = Is 3 s 2 s2 2 s2 +4' 12 3 2 2 + 4/3 = -cos-7=t--cos2t %/3 = 0i . n Let xi = = = y. Let 3. 4 V3 Exercises 8.2 22. y'. so that .cos 4 0" and 12— so that -3s 3 . and y" and y" x3 = = = y".4j/ + sin3t 3j/' —2y' and x\ = xi x'2 = 3x2 + j/"' so that — + sin 3t. Let xi 2. j.3 1.1 Exercises 8. 4tj |y so that = x\ = x'2 = -2x 2 + 3y" x'j = x2 x'2 = £3 - xi Qy' x 3 = 3x3 . y. y'.cos 2f. .. -ix 1 + 4x + x4 = 7. Xi y"' = = -\y + |e' so that x[ = x2 x'2 = £3 = y"r and j/ 4 ' . x2 = j/. Let ii 5. x4 = X3 — 3:4 x\ = 2X3 and j/ x\ = x2 x'2 = Xi x's = X4 j/". x3 = y". Let xi =y.Exercises 8. 4x2 - Prom x' + 4x - u' = 7t 10 so that . y so that (-l/t)j/ + [(4 = = - X2 ( 2 2 )/( ]y so that X2 4~f 2 1 9. and y".X2 = = y. Let ii = y. = 4> — Xi+t.X2 = y*. y'. and y" = + Ay + 10.3 4. and y" = Let xi = y. = x\ - 1y" Ay' -y+1 so that i2 4 = ^3 6. x2 = j/'. x2 — y'. -\y'" 1 x[ 8. x3 x3 = = y". Let xi = j/. + sint — sint. is x' = -2x + y + 5t y' = 2x + y- a degenerate system. + y" = and 24 3x" = 2cost 3j/' = cost y'. 13. we obtain Dx = 2 t + 5£ . Then xi = x2 x2 = -cost* x3 = X4 = -cost < 2 2. + -sint„ 1 1 1 / Xi - we obtain cost = 5f — -sm(.2 Dy = -x + 12. Since Dx . Let yi = is xi. Adding equations. i2 = smt = i'. 2 . w — x\. y3 = £2. This 11.3 we obtain 2x' + 4x - 2y = lOf 2y' . and x" £3 = y. From x" — 2y" Let xi = x. It. and j/4 = y'\ x 3 so that = Wl mi ' 3/3 = 3/4 T712 296 mi .Exercises 8.4x - 2y = -At SO that 10.Dy = the system 14.-x and Si - degenerate. Z3 = 10i 2 -42 2 + 3^5 - and y.3 + lx 2 -lx I -4 ^ -^ "5T ' 1 4 2 X2 - 3 - + 3/25' . and = z'i =^ 25 so that 22 A = 23 = 23 - 4zi 323 + 425 4 = 25 4= 16. 22 = Dx. The system = 50X1 + 4a:i = 25e-'/ 25 - 25e _3t/25 . Taking the Laplace transform of the system gives so that „. Let 2] = x. Let z\ = x. is I1 = ' X2 = 2 . z4 = y.Exercises 8.6t 2 + 4t - 10 =-y + 6f 2 + U + 10 so that f 21 = 22 £*3 = £>23 ^3 - 3f 2 + 2t - 5 = -^23 + 6t 2 + 4i + 10. 23 = x". + 2) + 25 (25s+l)(25s 2 3) 25 1 a + 1/25 1 2 3 Then 2 2 and x2 18. z2 = x'. 2£> 2 x 2Dy = y . 17.3 15. 625(25s . y y = — 1c\e~ and (a) (b) a4 A+B = B-A= 4-2 5+6 + 9-10 -6 8 — a2i3 obtain y 8 +6 8. A-B = +4 2 6 -10-9 f-2-3 4-0 1-2 7 11 -1 f" \u X 2A + 3B- 2 6-5 ^-2-4 ^ 3 +- 2 y 19 -6 18 > 24 -30.Exercises 8. Let a\Dx + a^Dy = b\X + foy + aiDy = b3 x + b\y a 3 Dx so that if then the system Dx + Dy = —x . + l\ (a) = l ^ (c) «3 Dx + Dy = —\y we Exercises 1. and C.4. Since and ai 5 f -4 i\ i 5. is 21. X2.2 "Too 12 100 1 5- = . and #3 be the amounts of salt in tanks A.10. so that x[ 1 = 1 2- 100 4= *3 6 3 T^ x l -" 50 50 1 + X3 -160*. a2 = 0. degenerate.4 . B. -5 ( + 2 -1-0^ B-A = 0-4 2-1 [-4-7 — 2 — 3 > = 28 G 1\ --1 4 11 3 f (b) 0.7^X 2 6 1 100 100 = 13 Too ~ X2 h> h X3 12 + X3 ibo - 20. 298 — —2x and Dx = —a:.3 19. Let xi. respectively. Then x = c\e~ l . 15+ 4 ) 27 -4 -1 M9 6 3 22 \ 6-12 -4 + 4 + -32 -6 + 12\ _ 18 -3 + 1 5.4 -2\ (2 (c) 2(A + B) = 2| 4 \3 3.6-10 -9 + §J + -6-12 19 -18 15+16 -30 31 4 15 -10 .20 1 + + 24^1 18 6 + 12 9 ~ 30 - 48 -36 -4 + 30-24 .Exercises 8. 1/ 12 AB = -20+ 10 BA = 6 3 .5 (b) BA = (c) A2 = (d) B2 = (a) (a) BC- 6 2) -6\ -30 + 12 -11 6 17 -22 8) . A(B + C) = / (b) BA 8 ( -6 -16 6 5\ -4 -5 5 5J 8 10 = (-16) 3\ 5^ 5 12 0) .-32 (b) 8 '-2 -30 / 4. (a) AB = '-2-9 + . 3 -3 + -15 30 8 -24 + -10 ./ 2 8 2 -2 24' 3 (b) A(BC) = (c) C(BA) = (d) 3 /0 0^ U o.-20 24 J -16 + 60-36\ 4 16 - 30 + -6 / > + 20 24. (a) A + Br = 10\ 8 16 20 20 25 / ( 6^ + 12 \5/ G :) 8 \ 10 (2\ 10/ 4 V-5/ /-2 5 -1 7 3 7 5 11 \ -1 4 -( ::) = ) 9. not defined. (a) (b) (AB) T /l 2^ 3 U 4y (.4 18 21 24 28 AB is x 1 and 1 C 7. (a) — ( 4 = 104 72 -18 (AB)C (180) -10 /2\ T (b) B B = / 4 4 (2 5) = V5/ 4\ (c) A+ BT = 8. 16 / V 300 54 {-26 4 \ -10) 8 u 2 3 x 3 the product is / AT A u 78 f -1 1 -1} 6 (d) Since 2 (1 ^ is 99 \ 132 -33. .Exercises 8.3 7 10\ .38 75 J ' = (A 1 + B) T = -14 -u 20 -11 '-4' -7 T :*)( 2 1 -3 -6 5 BT A T = 10 (b) "(1 -6 -14 9 1 7 38 10 75 2 -11 20 8 2 -11 20 8 + 11. Since det A = 27. Since det A = 0. Since det A = 3.12 -10* \ 8 1 . A is I A= 2.4 -t 6f 12. A is singular. Since det A= 4.-1 Au = 4 -4 22 = -l A 32 = 4 nonsingular. A is 4 2 -10 nonsingular. 5 .-6* {.433 = -5. | 2 3f + | -* a -a. + l\ +t 3t-3 et ^ ( = 2* | V-iot/ / 2 +t- 10* - i 3 -38 -2 +3 -5 \-6f + 4 /-9* 14. A is nonsingular. -2 7 cof&ctors are -411=0 Ai2 A21 = -1 A 22 ^31 = ^32 1 =2 =2 = -2 A u = -4 = -3 A33 = 5. 1/-5 -8 4 18.-I 19. 16. Since det 3 The nonsingular. 2 13* + . \ -1 ( 2 -4 Xi3 n 2-2 -3 = 22 4 23 = -19 . A is The cofactors are A u = -1 421 = 7 A Z . Since det A = —6. 13* ( + 4/ + 1^ . ^23 Then -1 2 -4^ -3 1 -2 5 / 2 / 20. A-i .Exercises 8. 14/ 1/4-5 3 17.-(>) 15. l-l 2 A is nonsingular. 1 IF -2 -5/ A" = ±<T* 26.-1 22.10sin2t / + 8e. A is 24.3 ' <iX 28. 23. A is singular.4 Then -1 22 7 -1 .ie"» — e* cos t ! + 8sin2t \ : X= -6cos2(.. Since det A{t) = 2t 2e # 0. The f-\ \ 27 21.\ / Exercises 8. Since det A= —9. -e 4t 3e 4i _( e* sin t . Since det 8\ T -13 7 -I = 3t 2e ^ 0. A- 5 -1 -4e 27. An = -2 A 2l = -2 . -13 nonsingular. / 2e sin t -1\ 7 5y . dt »«t 3t cos 3i 4e 4t 2 + sin 3t ^ — irsinirA 6f J 302 2e' cos f ( IT cos2( 2e-' A is nonsingular.12e. Since det A{f ) 25. A is -5 -1> -19 \22 / -5 J cofactors are = -13 Ai3 = S =5 A 23 = -1 = -1 A i2 = 7 A 33 = -5.4. -2 A = 0.3t \ . -2e 2( + 4e" 3t / 2e* so that dX —— — A 4e 2( - 24e" 3i -4e 2t .-1 4 nonsingular.3! 7 4-14 -19 Ai2 Ai-i Then 1-1 -S S . 3 . y = 1.0 2 + 1) .3 In 2 + ( .4 (=2 -i t=o s=t (c) £m*)**= =2 -2(/(t 2 30. + - . 2 11 -55 ) .0 14 /l | 1 Thus x = 3. and 2 = -5.Exercises 8.3 s° l) 2 s J ( s=0 ( 3 - f 3 (a) ~dt 2f ri 1 /tan" 1 .6 3 1 4 0-3 5 6 1 .(5/2) -2 1 1 2 J 31. 1-1 t-1 a *t \ * E 3 1 / i 1 3 t=0 / 3t 2 /2 i=2 2( ( 6s/(s 2 + /3t " /I 1 l) 3 ln(t 2 (3/2) t V -2 + /l 4 9 2 In 2 6 +3 2/(s 2 -1 . .0 * - - 4 - jr/2 2 -1/3 14/3 \ -5/3 28/3 | 1 . + l) ^ + 12s 2 3 4( + 2 tan' 1 2t fl -28 - 1 3 l) 1 3^ j 1 I -5. = and x -3 1 4 2 --2 7 /l 6 - 11 -5/2 \ 5/2 ) j 34. 7/2.0 | /l Thus x -14. -2 4 1 1 un 1 l\ 1 6/7 fl j 9 1 9\ 1 -7 -1 -7 -1 -35/ ijt. -16 4 5 -10 1 -5 1 1 1 -5 j | . -5 -4 fl 7 \ 2\ | 33.4 /5 32. 1 5\ | 1 - . 3 we t find -5 j/ = 5 — 7^ | -1 /l 0/ VO |t. 1 -1/2. y 2 - 2 1 -2 = 3/2. . 5. | = -5 — Letting 2 (1 /l 6^ 1 f. 304 11. 8^ and z HO 2 2 -4 4 - 5 Letting z -6 — -1/2^ 7/2 J H 8 \ | -4 we \ find y 5 -10 = — 2 + 2(.0 1-5/ 1 \ 9 ij -45 J 9 0. Exercises 8. and x =4— | -35 1/7 1 j 4^ 5 | I V4 -3 = Letting a fl -1 VO 1. 1 —8. and ^ — —3. 10 -10 and 2 = \0 8 — 2(. /i 1 -8 j ^0 1/2 -2 -1 2 -2 6 -3 -7 -5 8j 4 | \0 fi 36. j V2 3/7 1 3/2 1 = 4 1 10 = -21 3 | fl Thus x 2/7 (1 | | k 2 \ 1/2 | 10 3.. + 4f 2 -17 17/2 1 j \3 4/ 1 --2 . 8\ 2 | 1 -2 -2 | j 6 t 22/7 . = y -3.0 = 2 1 ( 35. . =!• i 2 2 1 37. 1 -2 2 -1 6 5 5 I V4 -7 7 9 >o 5/ ^0 5. find X2 /I 2^ 4 4 j j 39. 2 4 3 1 .4 n -1 1 n -1 -1 1 -1\ -1 -1 /l i 1 1 _1 \4 _1 o —2 2 1 \o 3 2 -2 1 4 1 —1 A <j 2 9 4 -1 5 -2/ n u 4 I 1 4^ 5 \0 | /l - o 1\ 1 1\ /l 1 [ 1 1 1 1 j 1 2 1 2 1 | U ii = n 3 1 2 1 1 7 1 3 Thus x2 1. I There 41.0 1 0. -1 1 1\ 3 /l 1 -1 3 1 -1 -4 1 -1 -4 3 -3 -12 n 1\ 1\ 7 I -1 1 -4 1 -4 -1 I 40.A 2 -7 8-A = (A-6)(A- 6 we have -7 2 1 I -7 2 -2/7 I 1) - 0. o\ 1 2/5 1 -5 -1 -20 -4 0/ 1 1 1 1 o\ 1/5 f Letting 13 /l 2 I \Q 07 = we t.Exercises 8. solve d et (A . We is no solution.—5* and 2 —5 1 \Q 0/ y J = x\ 2^ (\ -2/5 ^ 2 —3 3/5 1 j I VI -1 2 -5 7/ I \0 5 7 7 I There /I is no solution.AI) For Ai = —2 =*. 6 i3 = n o\ 1 1 o) I and 14 2. 1 - 0. 38. 3 = . 1\ | i = -1 . Exercises 8. We solve det(A - AI) 2- = A 2 For Ai = 1 1 - = A(A-3) = A we have 2 1/2 1 1 j 2 so that ki = -\ki.4 = %k2 so that ki For A2 = 1 .If Aa = 7 I then 1 K2 = 42. We solve det(A For Ai = A2 = —4 we - AI) = —8 — A -1 16 -A 0\ f\ 0/ U -1 I 16 = -3*2- If (A + 4) 2 = have -4 so that k\ = Jfc2 = 4 1/4 I 4 then Ki = -1 306 I .If £2 = -2 then 1 K2 = 43. If k2 = 7 then we have -2 2 1 -1 j I -7 so that ft] = fo. If k2 = 10 I 2 then Ki For A2 = 3 j = -1 we have -1 1 1 -1 j I 2 so that fci = ^2. = 2&2.25 If A2 = 4 then fl -I 0\ -1 (I 01 j I 0-9 9 1 -1 ] I \5 -1 -4 0/ 0/ .A —5-A 9 = A(4-A)(A + 4) 4. If £3 -9/5 1 -1 0\ -9/25 (\ o) = 0/ 25 then 9\ K. = 1 -2)0 1 | -1/2 1/4 A2 (A A 3/2 we have -1/2 If = - then 1/2 then 1/2 1 2 1 | 1/4 1/2 | I I so that k\ = — 2&2- = If Afc 1 then -2 K2 = 45.4 44. We solve 5-A det(A - AI) = -5-A 9 -1 -A 5 If Ai = 4.A -1 4. = 45 I .Exercises 8.A -1 -A -1 then /5 -1 -5 s5 so that k\ = and k% = 01 9 §£3.AI) — 1 A 1 1/4 For Xi = 1 1 so that k\ = - 3/2)(A - 1/2} = 0. If ho. We solve = det(A . 0. We 3- AI) A = 2 - A 4 Ai = 1 = A)(2 - -A = and kt = 0. If &3 = n 0\ | 0\ | 1 | j 0. . .4 = so that kj and k^ — k$.4 so that &i 0\ | then K2 = 308 o o o.Exercises 8.4 A2 - then 1 so that k\ (3 1 (2 If w solve det(A If 9 1 then 2 then /l o = and k% = 0. If — (\ 0\ 1 j \o j 1 | 1 -1 . = k$ If 1 then K2 = If A3 = -4 1 then /9 U so that k\ k$ and k 2 — 9&3 - 1 - 1 9 - 1 4 *3 If ON 1 -1 l\ 0\ -9 1 0/ o) then /1\ K3 = 46. 4 If A3 = 3 then /O 0-1 1 md = it i *2 = *3 j o \0 ~2 0. If = *a -A 1 0\ | V = 0. We solve 1 det{A - AI) -A 1 6 2- = A 2 For A = 3 6 A 3 1 1 - - -A 1 A we have j -2 6 | 0^ .0 I (3 - A}(1 - A) 2 = . If 0\ I | \4 so that -1/2 fl fl\ I | = 2 then K3 = 47. If A3 — = 1 1 ^ 0> —1 j oy 1 2- I 1 i -A = —3 (1 j o 3 n . that ki = 3*3 and k 2 = Ag.-2 we 2 4 -1 -2 | -2fe.A Exercises 8. We solve det(A - = AI) -A -1 4 = -A -4 -2 For A! = = A2 A3 . have / so that ki + 2) 3 = -(A 0J | and fc3 = then 1 /-2 48. £3 If k\ = (0 0\ 1 1 1 1 1 1 oj ^0 0. 1 — 3t —Si we have -1 +3i 2 -5-1*1° 1 J -5 so that = fci (5 + |i) h- If k2 = 1 + 3i I 5 then K2 = 50.A Exercises 8. We l + 3i solve det(A . We solve det(A For Ai = 3z - AI) -1 .0 so that k2 = and =0. For A2 = 0.AI) = 2-A -1 5 2-A 1 = -A 3 + 6A 2 - 4 13A + 10 = (A - 2)(-A 2= (A-2)(A-(2 + i))(A-(2-i)) = For Ai = 2 we have to -1 0\ 4/5 (\ 4 5 0^ I I 1 j I V0 a 1 I \0 0/ 310 0/ 0- + .A - 2 -5 -A 1 = + 9=(A-3i)(A + 3i) = A2 we have -1-3* 2 -{1/5) 1 + (3/5)i I I -5 = so that ki (5 - If |i) — 1 = fc 2 3* ! I 5 then = K. then 1 /1\ K2 = / 49.4 For A 2 = = A3 1 we have 0\ 6 . 4x2 + a'4 X2 J 0. i 0/ then K3 = 1 51.If = fca * then K2 = For A3 — 2—i we have —10 /i so that fe] = -1 ~ik<i and A:2 = 5 i 4 .0 1 i -i&3- If A3 = / 0) 0\ i i 1 {0 0.4 AB = I.4x2 J ( + a'i%i + &2X2 + a 2 x.If £3 5 then K. Let A= an 112 tt2i a 22 Then x\\ d f a\Xi :)( x2j dt\ (13X1 ' >>*«»^C a-i 52. = 5/ For A2 =2+ i we have -1 f-i /l 0> -i j 0> I 5 -i 4 1 -i —* 1 I I \ oy Vo / I so that fcj = ife = and &2 *&3.4 so that kj = -1*3 and k2 = = 0. so that 'an ai 2 \ /fen hi a2 i 022/ \ ^1 ^2 ( 1 1 .2 \ \ a3x\ + a'3 x\ + 0.Exercises 8. Assume det A^ and K + a2%2\ _ + 0. B = (° t. Then IB = IC and 54. LetX= [ 1. Then \ Then X . Let X= I 2. No. "22 A .A _1 AC. Since (AB)(B _I A _l ) = A(BB _1 )A _1 = ALA -1 = AA" = I 1 and (B _1 A _1 )(AB) . 55. det . 021 022 det A 6l2 _a 2i . consider A= . Since A / 022 \ -a21 -ai2 an J A is nonsingular. . = -au deTA = an det A Thus B= A-_j = „ 1 1 det 53. 312 B= C.5 1 1.Exercises 8.B _1 (A _1 A)B = B'lB = B _1 B = I we have (abj-^b-'a- 1 . AB = AC implies A _1 AB . 1 °\ Vo 0/ f andj ll Exercises 8. .4 Then Oll6ll + Oi2&21 = 1 ttll6l2 + 112&22 — 121612 + 021622 = and 021611 and by Cramer's + ^21621 = 1 rule _ " 611 = . Then o / X' 6.5 3. dt a. dt _t. Let X= . + y + 8sin + t (2f + l)e 41 11. LetX = 1. = X+ 4 3 dx _ ~ 4x + 2y + e = -x + 3y ^ at ( . ~kt dx 8.Ay + z + 2e (t- 4)e* ~ = -2y + 3z + . ± c t t. = -f 3i . -3t 2 Then x'=r5 ' X 2 1 -1 t . di dx 9.9z ~_ x . = 7x + 5y . Let X= -9\ f-Z 4 6 -1 10 4 X /x\ 4. r f Ue-'cos2W) l ^-=4x + y + z + 2e5 dt 10. LetX = y . W Then -1 1 X' = 0\ 5.Exercises 8.y + 2z + e dx _ ~ 3i - 7i/ e 9/) + I e x+ - dv Zt. Since X' we = -5 -10 e _5t -5t and see that 3 -4 21 dt dt at + 4sinf + 2t e '.Se' dz — = ~2x + 5y + 6z + 2e~' + 5t ^= ( . X2.X2. . Yes. Since cost — 5 cos ( — cos t — sin ( J i sin ^ we cost \ X= and * \ see that f X' = i o 1 1 -2 19. Yes. Since X' we 4 5 = 2 -4 -1 and te* 1 -1 X= 4 -4 -1 see that 2 1 15. and X3 are linearly dependent on —00 < t < 00. No.X3) = 17. -1 16. — 5 cos t — cos t — sin t J ^ sin t IVfXi.Exercises 8. Since 3/2 X' = we -3 .5 12. Since X' / 5 cos t = ^ 2 cos f we — — 5 sin t \ . since W(Xi..(5 4 2 cost — 5 sin A — 4sint / ( see that 13. since W(Xi. Since we 1 2 6 -1 Ul -2 / (°\ x' = and 1 \ (0 X= -1.0 see that i 1 6-10 = X' -1 -2 X. n X. since 18. and e< 4 sin f / -2 5 -2 cos t X .X2) = 8e ^ and Xi. -1 / = -2e~ sl / and Xi and X2 are linearly independent on -00 < t < 00 2t and Xi and X2 are linearly independent on —00 < t < 00.Xa) 314 .n ^y-a -3J/2 see that 14. 3/ 25. and X3 are linearly independent 00. since —oo < t < ^ 1 and Xi.-6 1 1 /-1\ 3^ xp + oj 4 I sin 3t. and A= 6 1 Vi 0\ 1 1 0. Since 2 ^ e* + * I ) 2 1 3 4 and te* = e ie 1 thatt /n 1 K = c l 24.X3) = —Me 20.X2. on .5 W(Xi. 1 / 3\ / Xp + 4 [ sin 3t 3^ -3 sin 3i / 3 V 3 cos = see that 1 2 -4 2 . Let / X = 1 /2\ 6 -1 V-5 I e-\ Xa = 1 V 1/ 1 VI/ /O e 3t . Since we 2 1 1 -1 -5 X„ + see that 23.Exercises 8. Since 2 -1 we see that *-G i) 22. Since 3cos3t\ / f we — 3sin3t / 2 -4 2 V-6 1 and . Yes. 21. X2. set for X' = AX on 00.0. 26.5 Then /-6\ xi = p 1 — AX.AX. _t ^ \ 6 ( -2 ) . p" 2t V-2J 3i F = and W(Xi. Let Xi = -1- y/2 X2 = x„ = 1 I U2 + 2 )t+ ( 1 .X2. = x2 = set on —00 < / e t 2t = < 2^/2 ^ AX lt -2-y/2j 2\ and W(X].e "'\ Xi and X2 form a fundamental . X3) —00 < t < 20 ^ = AX 3 . Xa. e 7t\ j /3 e " 316 . and -1 -1 -1 1 Then x.V2 /-2\ so that l\ Xp is a particular solution and 00.Exercises 8. and X3 form a fundamental so that Xi.X2) e^' = -2. 2 [ so that and which give cj = 3/5.sinf 2sint\ ' 2cosi have x(*)= CI „7t . 5 \. =C}e C2 = 1/5. e 5 and M+ c\ = —1/5. c2 Vl = = 2/5.fi„7t 32.3 = 7l 6 a 2t. 1 -. c 2 Vl = 1 = /-I 1/2.3 so 2„2(. ( 5' [ -fe< 29.Exercises 8. *(() " ! e 2t \ "\ # -1 (t) and + sint — 3sin( + cost / fe — -3e* - 1 2 (— 3sint + cost \ -3cosf . c2 /I W e 5t = and 1/2. #(t) 30. Then V2 = 1 £ /-l e 5t . #(f) e 1 28.r 2 1/2. Then ^d V 2 = -i 7* * ( ] e» + I ] e «.5 -e' 3te 3e' We - -e 5t /3te -e' 1 " l e } K -2sint 2 cost / \ 3 cost 31. We have X(t) = Cj e ' + c2 I j e 5t 1 so that X(0) = ci + C2 X(0) - ci + C2 and c\ = and which give c\ = -1/2. I . = +e 'Sie' ' 34. 3 cos i \ -2sinf / + + sin t J c2 \ cos ( — 3 sin t so that and which give ci = —3/2. (to)Xo and follows that ci —5 cos t .«(()] Xo = 0. X= S^)*" 1 *(t)X we ^ - = see that 0. and Xo.Exercises 8. c 2 = —3. C2 = —1/2.(t\ 33. We —1. 318 [*(t)*~ l (to) . Since Xo . Since is = *(to)X = IXo X = *(t)* arbitrary it -1 = = C= 0. Then + sinf 2cos( 3 cos ( + sin t ff-^toJXo and X(t) = = AX X= we know that *(«)* -1 (to)Xo. and ci = J c% 0. 36. Then 1 fe' + e' -3ie ( 9te< have X(i) 2cosi = ci . 4"(t)Xo solves X' = AX. Since the column vectors of 9(t) solve X' X(to) 1. We i fi -t — 2 f _I eP -t _l + 2 lJ>t + ^2 Ie P 5t 2 have 3 / 3 V V-l / so that ' X{0) =d X(0) = and ' Cl 3 — which give —1. Since X(t D ) = »(to)C = Xo we = ci '— 3cost = have 37. and *(*) 35.5 H. => 0/ For Aj /l = 4 we obtain -1/2 I ON so that Vo j Ki 0/ I For Aj = —4 we obtain 'A 2 I 0\ /l 0/ Vo 1/2 I so that 4 K2 — 0.Hi. The system is and det(A . = For Ai ON /l 0/ V0 5 we obtain -1/2 I so that For A2 = —1 Ki 0.AI) = (A . e a e 5! .. The system ^ I -t M + c.AI) = (A - + 5)(A '-4 2 4 -2 1) I = 0. is and det(A .Exercises 8. we obtain '2 2 4 4 0\ /l 10/ V0 I 1 I so that K2 = ' [ Then X = Cl 2.4) (A + 4) = /-4 2 8 —4 \ I 0\ 0. Then 3. The system is 2J V-5/2 and det(A - AI) = (A - 1)(A -5 -5/2 + 3) = 2 I 0\ 10/ 0. For Ai /-5 = 2 we obtain 1 ( | so that V 0/ Ki .6 1. The system is x'=[320 2 |x .Exercises 8. . The system is 9 1/2 x'=[ IX 2. For Ai 0\ /I 07 Vo = 7/2 we obtain -3 I 0' so that -1/2 1/2 For A2 = —1 Kr 0/ we obtain '3/2 9 1/2 3 I 0\ fl 07 V0 6 I so that K2 — Then 5.. Then x=o m.' Then 6.1 4. + c ! r]. For Ai 0\ (\ 07 V0 0\ /l 07 V0 -12 J 8 V - = 8 -5/2 we obtain I I so that Kj so that Kq 0/ we obtain I -1/4 I I 0.6 For A2 = —3 we obtain -1 2 -5/2 5 I 0\ /-l 07 V 2 I so that Ko 0. The system is and det(A For A2 AI) = -10 = (A - 8)(A + '2 -5 I 8 -20 I '20 -5 I 8 -2 10) = 0.1/2 and det(A - AI) = (A - 7/2)(A -3 9 + I 1) = 0. 2) = For Ai 2. and A3 (~ 7 ) ) . K3 = . . K2 = 3 and .2/ 8.AI) = (A - - 1){2 A)(A + 1) = 0.-3 1 | For A 2 = —5 we | obtain 2 I 0\ /l I Oj V0 —2 I so that -3 6 K2 — .6 and det(A - AI) = + 5) = A(A '—6 2 = For Ai 0. A2 = 5.Exercises 8. The system is /l .-5 J / X = ci f-7\ 5 I so that 4\ / _ 7\ 3 f-7 e 5t + c3 5 = 5/ 7 we obtain . and A3 we obtain /1\ (2) ) Ki = = -1 3 K3 = and .AI) = (2 - A)(A ( - 5)(A 4 7) 12 -1 0\ 5 10 4 ^0 5 = = 0. VI 0/ Then 7. = 2. The system is X' and det(A . ON /l o) \Q we obtain -1/3 j Ki = | so that . 1 f OX 2 Vo and det(A -V 1 = X' -1/ 1 = For Ai A2 1.V so that X = (A (') e* ci + c2 /1\ 3 e 2t + c3 . K3 = U/ so that / X 11. and A 3 2 \ -2 I 11/ so that /1\ X = ci 1 W ( e 3t + c2 M -1 2' e- 5( + c3 -2 11 322 „6f = 6 we obtain . and K3 = 0/ ^ -5. 1-1/ \ 2 1-1/ 5^ I 4 f K3 = and . = A2 ^ ( 1 V-i/ 5/ \ 5)(6 2 ( .0/ For Ai = 3. that 4\ X = ci j * + We have det(A - AI) = (A - 3) (A 1 ( + 6 c2 -1/ 12. A 2 = -1/2. ^ c3 = -1. We have det(A - AI) = -(A + = /0\ 1^ + C2 ci + 1)(A 1/2)(A 1 f + e' 1 + 3/2) = For Ai 0. -1 .3 3/ \ so that /-1> X = ci 4 We have det(A - AI) U/ iy I 10. = 0.Exercises 8. = \2 = and A 3 1. -i . /0\ 2 we obtain fl\ 1 W and .6 = -(AH -3)(A + = 2) /l K = L K2 = . i\ -1 / = -A(A - 1)(A / 1^ - = 2) For Aj 0. and A 3 = -3/2 we obtain 4 ( f-12^ ^ K[ = K = 6 2 . 4\ / e \ - A) = 0. if Ki is chosen to be _ | cost X = c\ 2 cos 15. and C3 = = 3 -1/2. /2\ ^ f = K3 = . in Problem 15. In Problems 15-28 the form of the answer will vary according to the choice of eigenvector. = XI) (2 - - A)(A 5 3)(A + 1) -3 2. C2 = 5/2. e sine 4t + c2 2 sin f For Ai =4+ i t — e 4t . and A3 = -1 we obtain -2\ and . cos t we obtain so that x = 1 2 +i l ' 5 \ J e 6 <4+0* = /2eost-8int\ \ 5cos( J / cosi 4 + 2 sin i l \ 5sin( .t Exercises 8. = -1/2 and K2 = and A2 = 1/2 we obtain f j J that X = C! „t/2 If = X(0) then 14. We have det(A the solution has the form | .1/ so that l-T X = ci -3 e 2t + c2 e | 3t + c3 |e-» 2/ \ If X(0) then ci = -1. -2t. We ci = 2 and have det(A C2 - = 3. | 1/ .6 13. *) { = For Ai 0. - = AI) A2 - 8A + t + sin = 17 0. A2 = 3. We - have det(A = XI) (A + 1/2)(A - = 1/2) r Kj= For Xi 0. For example. We have det(A - AI) = . we obtain -3T '1 so that X 1 = 1-3 [ 2 ^J e <5+30< = /«»3t + 3Bin3t\ a I 2cos3( /'sin3(-3cos . 3i we obtain '4 + 3i' 324 — 3cos3t^ 2cos3t 5 '. + cosf\ /2sint .6 Then ..cost siRt ( ) e 2cosi I /-sint-cosf « 2sin* \ J Then X = ci 18. 2cos3t Then X = ci cos 3t I V 19.. 4t e . + i we 4 obtain '-l-i' 2 so that X. U . We have det(A - AI) = A — 2 8A + 17 = / - For Ai 0.Exercises 8. We have det(A - AI) = A2 +9= + 3 sin 31 ^ }e ' 2cos3t 0- For A t ot 5( + c2 J = ( siti3t \ I. We have det(A - = AI) A2 + 1 = + c2 = . Ki-. 4( e / For X\ 0. 1 = f'" 1 "'^^'^ 2 y .' sin X 2 ... / I. 4t . 5sint V we obtain i '-l-i Kl 2 so that —1 — I i\ 8 1 \ 2 — /sini it \ J cos*\ 2cos( / + j — — cost H sin* 2sini Then (sin f — cos f \ / — cos ( — sin t \ 2sin( +C2 2cos( 17.'2 X = ci — cost sint\ 5 cos* 16. 4i + C2 = For A] — sint — { cosf\ e 5 + 3i .10A — t cost \ + 34-0. )e S( ' . Exercises 8. / ' t j For Ai n — cos t + c3 -sini COS 22. 2 cos 2t i \ 1 cost . We have det(A - AI) = \2 + 3 cos3( / 4 sin3f. \ Then X = Cj . = -3 sin ( .6 so that 4cos3f .3sin3t\ . + c2 + 2A + 5 _ 1+M)( = = 0- 5sin3i \ I = -1 + For Ai 2i we obtain so that Xi „/2 + 2t\ ( 2 cos 2f - 2 sin 2t e ' + cos2t 1 Then X = ci 2cos2( - 2sin2i _t e We have det(A - AI) = -A (A 2 + = l) 2cos2t + c2 + 2sin2f sin2t 0. + 2sin2C sin 2t cos 2f 21.3sin3t\ .'4cos3i 5cos3( 20. we obtain -t . /4sin3i + 3cos3t ( 1 5cos3( / 5sin3r. - +i j cos ( cosi sint \ ) Then / (1) X = cj + c2 We have det(A - AI) = (A + 3)(A 2 - 2A + 5) = 0. = For A] we obtain /1\ \°/ For Aj =i we obtain f-i\ K2 = i so that x2 = — sinf i / sini \ / cos t sini ( . 6)(A 2 .2A + 2) = Ki -cos2i .2sin2( For Ai = = 1 we obtain 2 U/ For A2 =1+ i we obtain m so that x2 = e< i = 1 cos f ' 1+i >< faint \ — sint ( — sin f e ( +i .2sin2f > -3 cos It .8A + e 20) = + For Ai 0. We have det(A - AI) = -(A . Vo/ For A2 [ we obtain K2 = V 326 2^ cost c3 — sin f / 1 = 4 + 2i / sint \ > sint \ cos = 6 2 sin 2t 2sin2t \ 0.\ Exercises 8.2 sin 2t I l +i I \ 2 sin 2t + sin2£ \ I Then X=a I 0\ -2 e" 3( + -2cos2f c2 I -2 cos 21 + We have det(A - AI) = -3cos2t 2 cos 2f 1/ 23. cost s cos t j Then cost { X = ci 2 e' + - c2 V 24. - t / we obtain .6 For A2 = 1 + 2i we obtain -2-i ( K2 = - -2 3i 2 V that ( X2 = -2cos2( + sin2t \ -2 cos 2t + 3sin2t e 2 cos 2t y -cos2t. e +c 3 3 sin 2t (1 - A)(A 2 . For Ai t = K.6 so that sin 2* ^ / . /4cos3f .2cos2i / ( — cos 2t y X3 = 2/ \ 2 sin 2i Then '-cos 2t' sin2f \ X = Cl e 1 6£ + c2 e ] 2 cos 25.Exercises 8. obtain \ -5 1 \25j For A2 = — 2 + 3t we obtain /4 K2 = + 3A -5 ) so that X2 = p (-2+3i)t -5 = McosSt-Ssin 3^ — 5cos3t f4sin3i + 3cos3t\ -5sin3( e" / / Then X = c.3sin3t^ . 7 28 K = = For Ai 2 we 4t . / 28 \ -5 e 2t + c2 V 25 / 26. For A 2 = 2i + + 3cos 3^ ) \ = -2 --2t —5 sin 3t c3 / V -(A /4sin3( 2( we obtain 0\ -1 we obtain (~2-2i\ K = 1 2 1 / so that f-2-2i\ X 2 = '-2cos 2( \ 1 J /-2cos2(-2sin 2^ + 2sin 2t\ cos 2t cos 2t + i sin 2i sin 2t 2t . . We have det(A - AI) = -5 cos 3f e" + 2)(A 2 + 4) = 0. We have det(A - = AI) (2 - A)(A 2 + 4A + = 13) 2sin2f 2*7 0. . 2 cos 2* 2 cos 2t „ 1 e 5i . . X = ci -7 e l + c2 6) .2t + + 25) = = 1 sin2f \ . sin 2t we obtain 25 ^ K. We = C2 = —1 have det(A - and AI) C3 = = 6. Then s /25 1 cos 5t — 5 sin5t N ' sin 5t . + C3 cos 5t cos 5t K + 5 cos 5t sin 5t sin 5i „ .6 Then ( X=a We have det(A - = XI) - A)(A 2 1 /'-2cos2t-2sin + C3 cos 2( s (1 + 2sin2f\ cos 2t Oz I) ." . For Ai 0. Ki 5 + 2i we obtain 1 = 1 -2i that 1 1 e . '-2cos2( °^ -1 e. . 27.( cos5( = — 5sin5i^ +i cos 5( J cos 5t 4 sin5f ' + 5cos5f s sin5t ) sin 5t ( .2ij X = ci Uos2( + 2sin2f cos2t 1 . cos 2t + \ 2 sin 2M ct e +c 3 = 328 -2 sin2t\_ +i sin 2( / - sin2t V sin 2t If X(0) 5! . If 4\ ( X(0) = 6 V-7/ then ci 28. = -7 | 6 / For A2 = 5i we obtain K = 1 3 V J 1 so that ( 1 + 5A x2 = ' e 1 I 1 5. - A2 10A + 29 = = For Ai 0. Exercises 8. then 29. We ci = —2 and have det(A - c-i XI) — 5. = A2 = For Ai 0. = we obtain '1 K= A solution of (A - AiI)P K = is P= so that x = Cl 30. We have det(A - AI) = (A + l) 2 = 0. ( 3 i + ca = -1 we For Ai obtain K 1 A solution of K is (A - AiI)P = P= 1/5 so that X = c, le-' | + ca te-* + x 31. We have det(A - AI) = (A - 2 2) = 0. = For Ai 2 we obtain K 1 A solution of (A - A,I)P = K is -1/3 P= so that X = ci )c [ a+ c2 (e 2' + 1 32. We have det(A - AI) = (A - 6) 2 = 0. For A] = 6 we obtain KA solution of (A - AiI)P = K is _ /l/2\ Exercises 8.6 so that X = ci 33. We have det(A - AI) = (1 - A)(A " 1 e I - 6( = 2 2) + c2 0- te , 6t 2/ For A] = 1 + 1/2 I ] e we obtain /1\ Ki = \1/ For X2 = 2 we obtain K2 = K3 = and VI/ 1 V0 m Then X = ci /1\ 1 + c2 e' e 2t + c3 1 W uJ 34. We have det(A - AI) = (A - 8)(A + = 2 I) 0. = For Aj we obtain 8 (2\ 1 \2/ For X2 = —1 we obtain / K2 = 0\ -2 1 V 1\ ( K3 = and -2 V 0/ Then X = c\ ^\ 1 ( e Si U, 35. We have det(A - AI) = -A(5 - A) 2 = + c2 ®\ -2 For Ai = we obtain -5 \ = + c3 V f-4\ Ki = For A2 ! 1/ I 0. f e 2/ 5 we obtain /-2\ K= V 330 1/ 1\ -2 0/ A solution of (A - AiI)P = K is P 2 T„ so that '-2' X-d -5)+^^ Je j 36. We have det(A - = AI) (1 - A)(A - if = . 5( U + C3 For A, = l 5l + w obtain '1' K. = For \ 2 = 2 I we obtain 0\ K=|-l 1/ A solution of (A - A 2 I)p = K is / P= 0\ -1 0, ao that '1^ X = ci ( e' I + c2 °^ -1 °' e + c? -1 We have det(A - AI) = —(A - 1)3 =a For Al = 1 te 1/ ^ 37. j ) ( 2t we obtain -(!) Solutions of (A - AiI)P = K and (A - AiI)Q = P are /0\ 1 I \0/ (1/2" and Q= I 2c + Exercises 8.6 so that (Q\ /0\ 1 1 X=q 38. We have det(A - AI) = (A 0< f l i - 4) 3 = 0- 2 + r ay w 1 t 1 = For Ai we obtain 4 /1\ K Solutions of (A - AiI)P = = K and (A - AjIJQ = P are /0\ /0\ P= Q= and 1 \0/ so that \1/ m X = ci (0\ e 4t + c3 r. e 4t + 1 1 / 39. We have det(A - AI) — (A — 4) a = 0. For Ai = 4 we obtain K= A solution of (A - AiI)P = K is F.1, so that X = Cl | ( e 4t + c2 T; i te,,+ ,e ' ii If then 40. We ci = —7 and have det{A - ci AI) — X(0) = = For Ai 13, = -(A + 1)(A - l) 2 0. K, = V 332 i = -1 we obtain te 4t + Exercises 8.6 For A2 = 1 we obtain /0\ K2 = K3 — and 1 so that /1\ (-1} e" 1 + c2 e + c3 ( 1 uJ If = X(0) then ci 41. We = 2, C2 have det(A = 3, - AI) C3 — (A - and = 2 W 2. 5){A + 5) K = 1 = 0. = For Ai n K2 and = -5 we 5 and A 2 obtain =(~l so that *(*) = 3e 5t 5t E _ e -5t^ ___ 3e- 5 ' __ * _1 (t)=io _ and ] 1 / 3e -5t e« g -5( 5t 3e Then x = *( 42. We have det(A - AI) = (A = )*- (o)X(o) l f + 3/25) (A + 1/25) = 0. ^ e5[ For Aj + ^_ 5[ = j. -3/25 and A 2 = -1/25 we obtain so that /_.-31/25 p -t/25\ / 1 9 P -t/25 Then 25 X = *(t)*" 43. Using 1 (0)X(0)= 2 e -3t/25 ow™ ( X = x K in t i ' 1 tX =l l 333 IX 5 , 25 2 -(/25 .»«,. „-(/25 \ , Exercises 8.6 we obtain For nonzero solutions K we must have (A — 4)(A — = 2) 0. If Ai = 2 then Ai = 3 and A3 '3' = Ki and A2 if = 4 then K2 = so that X = at* 44. Using X = x K in t we obtain 2-X -2N For nonzero solutions K (0 7-X 2 we must have (A - 3)(A - K,H and \0 6) = 0. If K =( 2 ^ so that Exercises 8.7 1. Solving 2- A -1 we obtain eigenvalues Ai = —1 3 -2 and A2 Ki= = -A = 1. A 2 -1 = (A-1)(A + 1) = Corresponding eigenvectors are and | J J Thus 334 K2 = ( J = 6 then Substituting into the system yields 2ai + 3h = - ai - — from which we obtain ai 2. —1 and b\ — 9 = 26i = 7 -5, Then 3. Solving 5 A -1 we obtain the eigenvalue A = 8. A = = (A corresponding eigenvector is -A 11 \ 2 - 16A + 64 - 2 8) - '3' K= Solving {A - 8I)P =K we obtain Thus Substituting into the system yields from which we obtain ai — 1/2 and X(t) = ei bi ]e | B 5oj + 9&i -ai + llfr 1 = = -2 = -6, —1/2. Then '+c 2 + 1 3. Solving 1 3 A 3 1 — = A A2 -2A-8 = 335 (A-4)(A + 2) 7 Exercises we obtain 8. eigenvalues Ai = —2 = and A2 Corresponding eigenvectors are |_M = K, 4. K 2 =(| and Thus Substituting into the system yields 03 + 363 = 2 (12 + b3 = 3a 3 from which we obtain 03 = ^3 = 1 - X 3&2 + b2 + 3a 2 —1/4, + 3/4, 02 1 = 2a3 = 2& 3 = 1/4, 63 = A + ai 3o! = + fci —1/4, 01 = 02 +5= 62 3f>i = —2, and (>i = 4. Solving -4 1 — A and A 2 = 1 4 we obtain eigenvalues Ai = 1 + Ai Ki= (J) — , + 2A — 17 = Corresponding eigenvectors are Ai. K2 = and (~* Thus cos 4f + [ ^ I 1 sin 4f e + c2 cos 4f — — sin 4£ = d| .. cos 4t +c 2 |« / V sin At \l ^ J cos 4t — sin At Substituting into the system yields as 4<i3 — 4&3 = — 4 + 63 = 1 <i2 — 4fcj = (13 4d2 + 62 = i>3 336 — 5ai — 46i = —9 — 56i = —1 4ai 3/4. Then Solving 4 - A 1/3 9 6 we obtain the eigenvalues Ai = — = - 2 - 10A = 7. Corresponding eigenvectors are A + 21 - (A - 3)(A 7) = A 3 and A2 K '=L Ka= and 3) (g Thus xe =c 1( e _. 62 1. 01 + ( ) and 1. Then Solving -1 . + ^-' mi ')e- — sin At cos4( = 1/17.Exercises from which we obtain 03 ' X(*) = = = 0. = 55/36 and 61 = + 56j 3 = -10 —19/4. 63 — sin 4( e' ci — aa .A 5 -1 we obtain the eigenvalues Ai = and A2 2i Ki = 1 = -A ) 1 A +4= a —2i. t+ f»i i/17 _) ( = 1. . Corresponding eigenvectors are 5 I ' = + 2i J and K2 = 5 [ \l - 2i Thus X = Ci 5cos2t c cos2( - 2sin2f \ 5sin2t / + ^2 / \ 2 cos 2f Substituting ^ = C) cost+ {Z ]sint + sin 2t 8.) -+ ri^. c2 Substituting into the system yields + ^1 = 3«i 9ai from which we obtain ai 6.4/17. Then + (]) C - 1/17 5. sint. -A and A3 = Corresponding eigenvectors are 5.-2 ci from which we obtain = cj —2. . and ai = —3/2. 61 1 X(t) =d f — —7/2. -3 \ cost .2.Exercises 8. into the system yields — 02 + 562 — - 2 = + 56i + a 2 + 1 = -oa -oi + 62 X(t) = = 5cos2f / = -3. — -2/3. and 5sin2i / +C2 ci Vcos2(-2sin2t/ 7. Thus /1\ X c = ci e' + c2 f 1\ 2 1 W -St Substituting x„ = e 61 41 \ c l/ into the system yields -3ai +bi + ci = -1 + 3c! = -26i . \-2/3 J + /-l/3\ ' \ 1/3 Solving 1 -A 1 1 2- A we obtain the eigenvalues Ai = 1. Then fl\ ^ e ( + c2 1 e 2l 1 + c3 m 2 /-3/2\ e 5< + I 338 „4( -7/2 -2 . (1} Ki = and 1 K3 m = 2 . ai 61 + bi + b2 = -ai from which we obtain 02 - = aj \ + sin 2t / V 2 cos 2t bi — / + Then 1/3. A2 = = (1-A)(2-A)(5-A) = 3 5 2. 62 \ -1/3. 5t f-8\ + u 9. Solving -A 5 - 5 = -(A-5) 2 (A + 5) =0 A -A 5 we obtain the eigenvalues Ai = 5. and A3 = —5.-5 56i = 5ai = -40 10 01 /I X(t) = 5( e Cl + C2 1 e 5t + C3 e . and 5ci .7 A Exercises 8. .= [_M - (A 1)(A and K = 3 * ( -4 Ci -1 2) = Corresponding eigenvectors are Thus Xc = - e' + C2 339 „2i 8.3A + 2 = 2. and 1 1 K3 = U) \ Thus /1\ ( e 0l + c2 1 \ e 5t 1 + c3 I e~ 5t Substituting bi \ c i/ into the system yields from which we obtain ci = = -1 2. Corresponding eigenvectors are 1 / Ki = I K2 = . 2 V-l/ Solving -1-A -2 we obtain the eigenvalues Aj = 4- 3 = 1 and A2 = A 2 . = Aj 5. K. 4c2 - c\ c\ = and 13 c% = = -4 + 602 + 6 = -c\ Then 9 5.-2 -5 1 60 and X = ci[ c If XD = ( ^j then = is + 13.)^ + ^(t)et U -1/ so that 340 6t + \E/L 2 . we obtain ] -(Ri + R2)/L 2 ) \i 3 ) {-ft/Li so that '. .7 Exercises 8.9 6 ) 6 Setting we obtain . (a) By KirchofFs E= ij Ri first + Let I = t and second (on each + L2 i' dt \i 3 j (b) = 13^_ ^e + 2 i 3 loop) laws. 1 I (_ i = -3 + 46i = -3 3ai from which we obtain a\ 2f>i )e« 4 + ra (- e » + (. E = so that 12 ( i Xp = ^ ^ ) 2 . Substituting into the system yields -ai - — —9 and X«)^ 61 = Then 6. 2 so 1 X(() 10. ~12e.Exercises For I{0) (c) = h{t) = i2 ^ (t) we j find c\ = -12 + i 3 (() .2* -3/2 -5/2 Cl -3/2 1 —3/2 and ai = -5/2. 8.( and C2 = lSe"61 + 30.2A - A(A = 2} Corresponding eigenvectors are 2. K 2 -= and ( Thus 2t Substituting into the system yields From the If we first let b\ ~ system we have 1. then 01 = 02 - = 11 - &1 + 2 = (X 2 -a 2 + b'i = -01 + 61 - 5 = 62 «2 = 62- 0. .2 — &2 1 and - The second system becomes oj — 61 = a2 ai — 61 = —02 a\ = —02 — — 4. 11. Solving _ \ 1-A _i -1 i -1 we obtain the eigenvalues Aj = 1 = -A and A2 = = M Ki I = 2 A . 2 — 5. -6. This gives 02 = (>2 = Thus X(t) = . 8.'1 3e'\ . 8. )*-( i and Xp = 3. Prom .Exercises 8.8 Exercises 1. and . X' = j'2 3 -n X + /o\ -2 It we obtain Then 2 C so that -2(e~^ (2te~ t t u=/*.F*=/(-^. *u=[*) +(_° 4 t FVom 3/4 -\j we obtain 342 \-l 2te* + 2e- 2e l . From we obtain Then *= . 1 = l 2e j (-2 3 U"' -e" ( so that and 2. e 21 (l Then 2e ( e 2t and so that and # 1 = _ e -2< 2e -2( 2 '.8 Then / 2e t/2v 10e 3t/2 J _l -3t/2 "2 e e -3t/2 _3 e .4 we sin2t -Mx + f\2cos2f/ ) 2 obtain \2sm2t) V2cos2f/ Then / -e 2t sin 2t Ue 2 'cos2t e 2t cos2i \ 2^3^12(7 ~ 2t -if ~2 " e ^ ~V le 2f 3 e_a 008 2t cos 2* £ B -»aiii2t 9111 2( so that U 5 cos 4t ^ Fdf = ^^ ^ ^ sin 4t ^ and X P = *U=( ^ cos 2t sin 4! 5. — S sin2icos4f - I cos2f cos4i \ 8 e 5 sin 2t cos 4t / From we obtain Xc = et cl (l) + C2 .Exercises 8.( /2 5 5 so that U Fdt = 13 / *- \z t | and 4. . From x' = [' 2 . From - 3 A H 1 we obtain *— (!)«" + «-("iV- Then so that "-/•-"-/( )*-( 61e 3. and 12 x' = * u ='"o 8. 344 „ f" 2 ^ ) .( 4 /-2e-* + Je-** -ee \ at _ 2„-5t + r-)*-( ' u -/*" ip *-/(-^«" and e" 3£ 1 e 20 7. From we obtain Then so that 2e. 4 . 2(e 3.3t .Exercises 8._i.+ ) (: 4/3r From we obtain 4^ .8 6. Prom we obtain -1 e' 1 + c2 te* -1 + 1/2 Then $ and -e ( Ae'-te' 1 e = _t - 2ie~' -2(e 2e"' _( 2e~* so that U= Fdt = e _i . ±e< From X' 3 2 -2 -1 = X+i " \--t we obtain Xe = Ci| _ Je l + c2 te' -1 1/2 Then e *- l te* * and 1 e~* = 2e 2ie" 1 -2te _l _t 2e -' so that PA - U = di 6e -2t and 1/2 -2 10.8 Then 4> 4e - -2eat 3' 36 6 *-!=f and „-3( „3i . so that U= /#- 1 6 Frf(= 3 24 = df / t c lit' fi^^ T3 + e S 12 te 2I and - -te* 9.4te _t df 2e 345 - 3e-' + 4ie _( \ -2e -i e .Exercises 8. 3 cost/ -3 sin f + 3 cost 3 cos H* = and 13.8 and 11. ='i /sect x+ o we obtain . 1 / J V t \ / 3 cos t + 3 sin t \ + 3 sint J V 3 sint . t Then — sin t cos t \ ( .'cosi\ / sin t \ I sint — cos t Then #= cos i sin £ \ t # _1 = . e' cost * _1 = . and | —cost! sint / cos ( sin \sint —cost t so that U= 1 / *-'Fd£= /( )dt t = lnj sectl and X„ = .'t *U = s 12. cost + sintln sin — cos t m | t i sectl I sec | t] From we obtain — sint \ .'0 . / . e~* \ cost sint / so that U= J t / *" F(ft= r ( . From x -1\ . From X' = 1 — 1\ 11/ X 346 / cost \ sin t . cos t J cost V sin i .Exercises 8. and sint / / — sin t cos t \ . te" 2' 1/2 +| | e -2t U/2 2/ Then 1 2t+ 1 1"\ e 2 4f . „-2( Prom X--I Mx + — 10/ f \ sec t tan t we obtain cos t \ . i i Then — *- sin cos t t l e cos t sin * and _] / — ' — sinf cos t cosi sin t -t f so that and .sin t / / sin ( cos t V Then cos t sin t \ -sini * _1 = . t \ sin.2t .2tln! 4t + 3lnf-4tlnt 2t 15.Exercises 8. ( i t From we obtain .'-sint\ cos t /cost'.-i ^ -2t * = ( 2t + -1 2 so that u=y *- i Ptft= j 2 In t dt -lnt and + lnt . lj 2t+n . t and cost/ / cos t Vsint 347 — sin t cost J . . cost sin 14. . .8 we obtain . t cos£\ / + . . \-±smt) 348 In sint + /-2cost\. le* } * -1 = . — cot esc £ — cot In esc t £| £| I From 2\ 1 -1/2 1 /csct _ \sect J we obtain X = C1 2sint \ i ./ \ In csci \csc £/ | — cot t\ j and ' sin t | \ cos £ In 17. In sect \ sin£ ' . cos(\ #U = . Then *= / 2 sin £ \ cos ( 2 cos £ \ — sin t . . In sect| J | and X p = #U= /3sint\ . J — tan t\ [I — tan 2 £\ /( \ tant J Vln|9eci|7 J and X„ = 16. e~< so that U= J/#" 1 Fd£= /f. From -1 Wotf we obtain X = ci cos t \ . ] \cost. and cos £ \ / k sin £ ? \ j cos £ — sin £ / .sin Je* £ C2 / . . - \ £ In sintj J j . 1 \te |cos£y . .) . * J \%cost-tant W(l.K + C cos £ / 2cos£ V . . = V sin I — sint /cost I cos £ so that J .8 so that / .— sinf — sinf / £ | + \ sin / /sinf. c — sin / sin t +c 2 . \ £ / cos t Then sin£\ cost i *= I — * and sin cost t _.( Exercises 8. ?*. \ tant/ £ + ln|sec£|. . . 2i and 3> 1 — -2( lp e 2 o 5e e. Then I *= 1 e -1 e 2t 0\ 4 .1 Exercises 8.cos 2 ( sin 2 — 1 2 sin cos 2 f 2 — * + .3* e*/ so that /ie u = j * _1 Fd£ = ^ l„-2t c -|e 2 '\ dt j i = ¥ / and xp = *u = -e + \e 2t + ( 2 3t ht e \te 2t > . .cos! - In —2 cos f — cos t | secf + tani| \ sin t and 3 sin (cost X p = *U = 19. From we obtain - cos t .o we obtain ^ ( -1 + C2 H W 1 (0\ e 2t + c3 3t e .cost) In (sini cos ((In [sect | sec t + tanf| + tan(|) From / x' = / 1 1 e X+ 1 e ( \ 2t 3^ .3t / . Xc-Cl sin / cos \ f + C2 1 | + sin t \ £ 1 I cosf sint / Then *= cos ( — sin cos f t + sin ( \ cos t sin ( * and ) = 1 — sin ( cos £ + sin f \ cos t sin f — cos t J I I \ J so that ^ _ y ^-lp^ _ ^2cosi y" + sini - 2 sin t sect ^ 2 sin ^_ ^ — t .8 18. e Exercises 8. From we obtain _ e 4t c e 4( £ 2t 2t i J and X = *$- 1 (0)X{0)+*^'*.2t v 2"' \ -2f so that /-te. 2 f + ) f 2 \ 4 2 2 ie +* ^Jj e.Fds = *1 -2 „4f .e -» e. fe 22.t -e~ t -2e~ (ft + 2t\ = and -3/4 \ -1/2 ^ X B = *U = -1 -1 "1/2^ 21.4.ai + 2i . -3/4. Prom MI :l) x+ i/t we obtain i i i +t -t i t 350 i + -l t ^ e 2t - + 2t- . From /S -1 1 1 = X' -1\ I o \ -1 t 2e'7 we obtain i\\ Xc = 1 ci e< + c2 1 W e 2( + c3 a) Then e* e 2* e 2( -( \ and $ = 1 a -2t — .8 20. --.-+ I2/) 5t Prom -2\_. we obtain L\i'2 ( \sint . and X = *X(tt/2) + * h/2 f 9~ l Fda = * ° ( ) + *- \0J l)-{l) 25. t 2ot+6e 5 6 *-^=*.( 23. w .J Exercises 8. (a) By and Kirchoff's first E= i\R\ + d_ + L 2 i[ fh\ dt(i 2 j (b) Let I = V *2 ) 3cos( -{°i) i — 12+13.fl. 3 /2 we obtain *= /sin ( — —5 cost ^ 2cosf 3 cost sinf + \ 3cosi/ /sint . * > = + 5 cost \ — 2cosi 3cost sinf — \ 3 cost/ . /-ll 3\ 351 ZlOOsinA (E/L 2 { - \ 1 - E = iiR\ +13/^2+ £2''i> so that 11 f ~ 2c03t - COSt and second (on each loop) laws.f^-5/u*.8 and x . 8[Rt R2 /L2 f-(Ri + Ra)/L 2 { \ -R2/L )\i 2 ) Ri/Lj l so that / .G) ( ' - (l) + (!) ^ From X' = /4 1\ « s) /50e 7t \ X+ o ( ) we obtain and x=*x(o w)+ */A 24.60t-12e:. + .) ./r'F* - ln +• (-j) • . A*k '.5. . For A= .. r.<« .A... 4. ant J \ 12t &e (12sin( and T =*Ti= ^ sin *-I cos( ™ fat I I coat so that If 1(0) = ( „ ] then ci = 2 and c 2 Exercises 1. fc = 352 2. J J \3Qe I * . 12t = ... ^ | we have Aa = '0 ° Ji i a3 = aa = 2 ) = i .o | ^ = ^..Exercises 8.3. *=1.I.8 and Then *= I - .9. 6. 8. >-= ra : i 1 (A 2 = i = 1 i (o 2 A4 = s : . In general . i 1 1 1 ' i I A 5 = AA 4 = AI = A and so on. = a - cost) J .. For A= 1 ( ^ _ coshi { -sinhfA \ — sinh t cosh(-t) / cosh t J 0' ^ | we have A2 = I) a 3 = aa 2 = A 4 = AA 3 = u 1 = i (o ir "Uf'J (i 4 ° °) { 1 " ' Wi ^ dG 8 for A=l ° 16 and so on..Exercises 8. = I + At+ilt 2 + |A* 3 + --= I = I cosh (l ^+^+ + t *... In general A*=[J 2°J ( 2.) / cosh t sinh \ sinh cosh f / + A sinh t = t ( and /cosh(-I) sinh(-f) \ V sinh(-t) 2. 1 0\ /l } and e" 2t ' 353 .9 Thus A (A A A3 2 2 . _( et . 1 0\ 1 (l 1/1 0\ 0\ +^+i+ .3 Thus. + l.) +A ( t+ .. + . cosh s 1 1 solve X' — sinh s sinh s — + c2 t cosh a V . To solve 1 = X' x+ 1 we identify to = = e X(() 0. Using the result of Problem 1 (cosh ( sinh 4. / I cosh c\ 6.sinh 3 sinh t sinh f t (cosht sinh t + cosht ci t sinh f + C2sinh( + c\ sinh f + C2 cosh c\ cosh + cosh t + C2 sinh £ ci sinh t + c2 cosh t ci f' (cosh t cosht + C2 sinh t + ci sinht + C2 cosht c\ To (3) in the text.9 3.sinh t t 2 + 1 cosh s sinh t cosh 1\ = sinh s . and equation 1 C + e tA /' e-* A F{s)ds ci = Problem results of /cosht 1 sinh t . tA = F(s) and use the t sinh t \ sinh t cosh t + + cjsinhf sinh t + C2 cosh t cosh t sinht sinh t cosh t sinh cosh / cosh t t cosh t (cosht sinht / sinh \ sinh cosh 2 ( 2 1 sinht 1 cosh 1 / ' t t \ sinh — a — - 'o cosh 2 J 354 cosh s — sinh s cosh s cosh s + sinh s — cosh t . t sinh _ f cosh f cosh t\ I sinh sinht / \ cosht Using the result of Problem 2 X= e" I = + «=2 Ci e 5.Exercises 8.cosh 1 4. and use the Problem results of 1 and equation (3) in the text. t \ /s A ( i /cosh sinht/ t \ A sinht/ 2 A /sinht*\ /cosh \cosht/ \ sinht/ solve At 2 we 1 identify fo = X(t) 0. 1 _ - e - e"' e 2 -ie"* e 1 3 -se e' e' c2e .2i C2e + -t I„4( 2( 2 .4s 1 + e e £ l„2t 355 2 + 1 ds (3) in the text. To sinh t da + C2 cosh ( / i I. .9 we identify tn = 0.2t se + C2e cie' e 2f + it + „2s 2t cie e 2t . j e" sA F(s) ds Jtn (coshf sraht\/ci\ /coshf sinhf\ sinht coshi/ \C2/ \ sinht cosh(/ Jo \-sinhs rt ( coshs — sinhs coshs\ coshs sinhs / ds /cosht sinht \ c\ sinh t + C2sinh ^ + C2 cosh f / \ sinht coshi/JO \0 I c\ cosh t + C2 sinh t\ I cosh i sinh ci sinh f + C2 cosh t / \ sinh t cosh t / \ ( + C2 sinh t \ ^ f c°sh * sinh f ^ ( t + C2 cosh t J \ sinh t cosh t J \ + C2 sinht \ /i cosh /cicoshi = { c\ cosh V ci sinh (ci cosh t ci 7. = F(s) e (A e = M C + e tA and use the results of I' /' e~ sA Problem 2 and equation F(s) ds l e cie 2i o 1 .' Exercises 8. F(s) = ( coshf \ | \ sinh X(() = e tA C + e tA f* ( . 2i lis -e -2s -3e~* c\e -1 -3e.1/2 PDP" = 3 V 1/2/' V -1/2 3 1 3/2 2 1 -3 6 1 and D=f\0 10.A Exercises 8. Corresponding eigenvectors are K.4A +3= 356 (A - 1)(A - 3) = ° 5 (3) in the text.9 8.2t l .a C2e C2C „-2s \ t + Cie' 2t ( tr 1 C2 E Problem 2 and equation results of ' I e / c\ \ \ 2* ) +1 2 «3 e' +Q e 2< -3 + l e 2 Solving 2- A -3 we find eigenvalues Ai = 1 6 — 3 and A2 - A2 - 8A + 15 = (A - 3)(A - 6) = A = 5. ) ^ ( V 1 C + e tA f e~ sA F( S e' e 21 C1Z ds e 2t JJo e 2t \ e21 + 2 e e -3 _1 2 2 -3 - 3e* + = 5 9. Ka = and u u Then P= 1 1 P"l=f . -2( IP e . Solving 2 -A 1 1 2- = A 2 . To solve x-i we identify to = 0. . = F{s) X(t) = e tA and use the . -1 = e ePDP.9 we = find eigenvalues Aj 1 and A2 = Corresponding eigenvectors are 3.= j + ((PD p-l) + I f 2( PD p-l)2 + i. (2) in the text n e D p-l)3 + A2 • + . . e Xnt j 357 + Xn t + ^{X n tf + . P -i = p e 'D P -i. £D + i(iD) 2 + i((D) 3 + 12. Ki = M I K3 = and ' J Then so PDP 11. \ - + 1 \ • xl) x\ 7 2!< K) I • I /A? \ A| + + 1" VO /l A^y + Ai( + ^(Ait) 2 + l + A 2 ( + ^(A 2 i) 2 + 1 > • e < - A2t .0 - 1.) Exercises 8. From equation e (2) in iA 1 = 2 1 1 2 the text .(3(p 2! = P H. From equation lD - 1 = . since X' = X' + XJ. Prom Problems 9. True. and nonzero solutions of det(A — AI) AK = AK then A(cK) = cAK = c(AK) 10. and B = C. False. 6. 4.8.C J e 1 Z)(. Prom Problems e _ 1 Q 5I 3t_| e 5( _I e 3' 4. AB = AC. A _1 AB = A _! AC. True. True. A -1 = -A 4 -2N/-2 -3 1 j ~ \3/2 1 -1/2. since 9. by the definition of an eigenvector. 11. 11. by 7. . _l e 3( + Ig5t 3 e 5 t/ y C2 10-12 and equation (1) in the text X = e tA C = Pe iD p. False. True. True. \-l 2 358 = = A(cK). 0. consider A= V . 5. True. and 12. they are the zero 8. True 3.Exercises 8.o \ e 3t J U 4* )• Chapter 8 Review Exercises 1. since : Theorem if 8. and equation the text (1) in X = e fA C = Pe^P^C 3„3t 3 14.9 13. since complex roots occur in conjugate pairs. = AXi + AX 2 + F = A(Xj + X 2 )F = AX + F. Chapter 8 Review Exercises 12. False; n i 2\ i (1 1 3 i 1 -1\ 3 1 J U 13. Prom (D - 2)x + {D - 2)y = Dx = 3 - (2D- l)y. Then = y Substituting into 14. From (D - 2)x (D — 2)x + C2e - 21 c + (P — -y=t—2 and Dx + (2D — and 1 cie 0, 1 = 2)y -3a; and 3 x * ! = -c 2 e - + (D - 4)y = = 1 gives C3 x = l)y - -C2e l -cje 2t and + 03. 5/2 so that 3 -cje = = -6 we obtain (D - 1){D - 2)y 3 ft zt + 5 - . -At we obtain (D - 1)(D - 5)x = 9 - St. Then 3 and y 15. From (D-2)x-y = = (D - -3x -e* and -t+ 2)x + x = cie y = (£> = sin2f and 1 (£) 2 = -cie* - 4)?/ = + 3c 2 e 5( + _ + -7e' we obtain (£> —*. - 1)(D - 5)x = -4e* so that + c2 e 5 + ie' ' Then 16. From (D4-2)a: + {D + l)y - 2)x + e* = + -cje' 5x+(D+3)j/ = 3c2 e 51 cos 2t - te* + 2e ( . we obtain (D 2 +5)y = Then y = 2 ci cos t + ci sin ( — - cos 2i + J 7 - sin 2i o and x= = -i(Z> + 3)j/ /13\ — - \5 ci - 5 c2 + ^cos2( sin i / / 1 V 5 + — -C2 — 3 \ -ci cos 5 / I i 17. Taking the Laplace transform of the system gives S y{x}+^{y} = 12{x} + s2{y} = 359 1 ^+ 2 5 1 3 3 — - sin 2f — - cos 2t. l 2 cos 2i - 7 sin 2t. Chapter 8 Review Exercises that ^{x} = s - 2 1111 + +1 + 2) 2s s{s-2)(s 8s-2 + ~4s~ 9 1 +2 8 s Then x = -] + \e 2t + 4 o %" 2t and = 2/ 9 +1 = -x' 1 2t 4 o 18. Taking the Laplace transform of the system gives s 2 X{x} + 2 S X{x} + 2 s 2'{y} s 2 s(s-2) 2 and s-2 %{y} = - so that 2{x} = 11 11 2 s 2 + s-2 -s-2 = _3 1_lJ_ + 2 - 2 3 2 {s 4 s 2) 2 s 1 s-2 1 (s-2) 3 1 - 4 3 2 {s - 2) 2 Then x = l-le 2t + te 2t 2 2 and y = -\ 4 2 3t ( 19. (a) X= | , 20. Let xi -f -( + 2 + 12f 2 + 8t + 3 4f 3 =y, x 2 = y', x3 = y", (b) 1 \t X' = and 14 = 2 ^e 2t 2 - y'" so that Dx 3 = x 4 21. Let xi = x, X2 = y, 2:3 = £>£, - ^3 - and X4 = Dy Dxi = X3 DX2 = X4 Dx 3 = x4 £>X4 - + 2e' - |f. so that 2x 3 = -X3 - 3xi Xi - 2x! - lnt + + 5* 360 2. te 2t 1 + 24f + 8 } Dx\ = X2 £>X4 - + 6t + 5 -Zt 2 12( } + lOf -4 . ' Chapter 8 Review Exercises 22. If x= c\e l + C2te + sint and t y = c\e + c2 (te + e') +cos( l l then = x' cie* + + e') + cos t = c 2 (te' y and j/' = cie' = ~(cie* + C2(ie + 2e - sin t l ) + c 2 te' + sini) + = —x + 2y — 23. We have det(A - = AI) (A f - l) 2(cie' + c 2 (ie + e*) l - +cosi) 2cosi 2 cost. 2 = K= and A solution to (A - AI)P = j. ^ so that r/ n K /o . 1 24. We have det{A - = AI) (A + 6) (A + 2) = x 25. We have det(A - = AI) A 2 - 2A + so that -«(-i)^ + «(0 e 5 = 0. For A = 1 + 2i "" we obtain Ki 1 = ^ Xl -fV +a0f \i J j and ^WJ^U -(\-sin2tJ \cos2t) Then v X = c, coa2(\ t , e' + c2 26. We have det(A - = AI) A 2 - 2A +2= 0. For A = /sin2(\ , e*. \cos2t/ V,-sin2t/ 1 +i we 3 Ki = obtain * ^ ^ ^3 1 " ( 1+f j ( _ ^3 cost + sint ^ ' V 2cost j ( ^ ^ - cost + 3 sint 2 sint \ + sint\ 2cost , / / V 361 — cost + 3 sin (A 2sinf J ^ j Then (3cosf j , and t is P= ^ Chapter 8 Review Exercises 27. We have det(A - AI) = A 2 (3 - A) = so that /1\ 1 1) \ 28. We have det(A - AI) = -(A - 2)(A - + 3) = 4)(A f~2\ X = ,1, so that (o\ 3 Cl e 2' + c2 1 We 7 \ f e 4t + c3 -3t 12 {-is) u, 29. ,31 1 have Then e 2t 4g 4( e _4 e -2t ;~2( 4t and 2e~ 2t - 64te~ 2t \ — dt Wte -it ( 15 e - 2< + 32te~ 2t -e- 4i - Me -4t so that p 30. We -1 - 1 4t have Xc = /2cosf\ ci e \ — $\tit} , 1 + c2 /2sin(\ { \ cost J Then *= / 2cosi 2sint A cos t sin f — sin t I sin V-si cos I t ; cos t and cos t — sec t sin f sin = dt J t - In sec V | t — cos f so that — 2 cos (In \ 31. We | — 1 + sint In sect | + tant| \ + tan(| / sect have / cos t ^""H + sin t \ 2cos ( I sin t — cos f +t ) 362 *{ 2sint + tan i| \ e Chapter 8 Review Exercises ' cos t + sin t sin — t cos sin t -cosi — 2 cost 5 cos t + 5 esc — Asini - A cos t + A. ^ sinf + ^ sinf + 5 In — 2t te 2i sin )e + e 2t 2t + c2 \ * , ,2t ,2£ esc t xP = *u=| f| — t| I — j sin i + cos ' cot In esc 2t 1/2 1 /2 — t cot f|. „2i + ( -te -2t -te -« _ -21 ^ -1- U=^ t / I te = csci / f -I . t \ -te u = /*-^ ss — cot I sint -1 e ( 1 1" csc( -1 X c = c,| t t \ A,cos(+ A,sint / — 2 sin —5 cost — j cos t ' 2sin( 2 cost 1 )tV' + ( „-2t i2 - t e -2t 9 Numerical Methods for Ordinary Differential Equations Exercises 5. +4= Setting x 6. Setting + 2x 7. Setting x 2 8. Setting y we obtain x c = ?/ —y = — x2 9. Setting \Jx 2 — c c we obtain a family we obtain y + y 2 + 2y + 1 = a family c = x2 c; of vertical lines. a family of lines with slope -2. of hyperbolas. + c; a family of parabolas with veritces on the we obtain x 2 + + (y l) 2 = c 2 ; j/-axis. a family of circles centered at (0, — 1). + y 2 )~ = c we obtain £ 2 + 2 = 1/c; a family of circles centered at the origin. y(x + y) = c we obtain 2 + xy — c = or y = — ^x ± ^Vx 2 + 4c; a family of hyperbolas. 10. Setting (x 2 11. Setting = c — 4, we obtain y = -2x + c 2 9.1 12. Setting y l j/ j/ 1 +e = c we obtain y — 13. Setting (y — l)j{x — 14. Setting (x - y)j{x + y) = 2} = c c c — e x \ a family of exponential curves. we obtain y = cx + we obtain y = (1 - 1 — 2c; c)x/(l origin. 364 a family of + c); lines. a family of lines passing through the 1 Exercises 15. Setting = x we c form a family of see that the isoclines vertical lines. y \ + y = ewe obtain the isoclines 16. Setting x y 9. = —x + c. V, / / \ .X. ** \ V 41 V \ 17. Setting y = / s —x/y = c we obtain the isoclines 18. Setting l/y = c we obtain the 20. Setting xe y — c we isoclines y = -x/c. y 19. Setting 0.2x 2 +y =c we obtain the y 21. Setting y isoclines — cos |x = cwe y = cos ^x + c. obtain the = In c 22. Setting y = (1 — 1 - obtain the isoclines In x. — y/x = c c)x. V we obtain the isoclines 1/c. Exercises 9. 23. Solving ax + By T~ = yx + oy for y we obtain cy a family of 24. y = cx is lines through the a solution of the differential equation and only if _ _ ' and only oix + jx + 5cx and only 2 and only Bex + (7 - 0}c - a]x = if 6c if if if [6c if —a origin. ^ if c 2 + (7 - -a= 0)c if 8-7±^{3- 1 )l + 4a6 26 if and only if 25. The isoclines of y' = 3i + 2y are 3x 2 - {0 + 2y = + 4aS > f) c or 3 If 26. we choose The c = —3/2 then 3k isoclines of y' = 6i - 2y + 2y — —3/2 are 6x - 2y = is 27. we choose c The = isoclines of 3 then 6i 1 y = — 2y = 3 is 2x/y are 2x/y c a solution of the differential equs c or » If 0. = -3* + |. a solution of the differential equation . — cor y = 2 ~x. c Setting 2/c 28. The = c = we obtain c isoclines of y' = 2yj{x ±y/2. Thus y + y) are 2yj{x = ±\/2 £ are solutions of the + y) = y y = c or c 2-c 366 x. differ* EULER h=0.4823 367 .40 0.8840 .7596 3 . Then y EXACT 00 .1105 IMPROVED EULER h=0. — x are and y solutions of the differential .1230 2 2 3 3 .0574 3 . + are (5x a Setting (4c 3 - 5 = (3c - = 5) of the differential equation 0. 1 = 1 — x + tan(i + c) j/ 2. The — c) = c we obtain c = or c = 1 — Thus y . 3c - 10 3c 2 - 14c 5x are solutions Thus = c 4 and c = -1 and y = Ax = + 3y) + l)(c c or +5 4c + 0.4525 2 .Exercises 9. .5142 .4 = (c-4)(c + l) = are solutions of the differential equation isoclines of 1 y = (5x + l(h/)/(-4:r + 3j/) and — c 5)/(3c 5 and 3/ - 10) = — ^x = and = 3/ = we obtain c c 10j/)/(-4a. respectively. All tables in this chapter were constructed in Consequently. 0. 2 2 2 2 . .8845 3.5958 0650 . .0000 2 .2 Setting c/(2 equation 29.3075 .5931 3.8254 2 . = isoclines of y' (4i + 3j/)/j/ are (4x + 3y)/y = c or 4 c - Setting 4/(c-3) — —x and y 30.2440 2 . so J.1 h=0. . x n and yn will be indicated as x(n) and y(n).2 a spreadsheet program which does not support subscripts.9082 . Exercises 9. Thus c = -1/3 .3c.20 0. — du — dx + and j/(0) = 1 — x + tan(x + — = 2 gives c = 7r/4. 1.0000 2.10 .2261 2 .1 h=0.50 2 .30 0. .2727 .0000 2 1000 2 . — Let u x+y— so that y' 1 = (x +y— l) 2 becomes u2 and tan -1 u=x+ c.S858 3 0378 3.05 2 0000 2 .05 2 0000 2 .1228 2.3085 . 1220 3049 2 2. The c we obtain c 2 — . 1 x(n) vln) 2 0000 0.10 1. isy bU j . 5384 368 . 4870 0.0557 1.0500 0. 05 vfnJ 3 UUUU 4 4UUU .4260 .40 1. 2 2 2 1 .3200 0.25 1.05 Vln) 0.3647 2.5315 .45 1.1950 2.1943 0. i UU in 1.8207 xln) UU UD 1 1U 1 1 c 1.30 1 1360 1 .20 0.3070 0.35 1.0000 0.40 1.1 xln) 0. 6000 .05 0. 3683 0.50 1 .9800 . .50 .40 0.9830 0.1506 0.30 0.1001 0.1b 1.3050 0.00 0.2010 0.0460 . 0000 10 . vln) D uuuu .00 2 0000 .45 0.15 0.2538 0.20 1.9424 h = 0.00 .20 1 .1000 0. 1 . 0.05 1 .2018 0.05 xln) Wn> .4183 0. 0914 0.40 0.20 1.50 .30 0.25 0.0288 0.2V 1 .2715 0. 2 . J . 3617 0.2 n = xln) 1 .40 0. h = 0.30 1 .6300 0.Exercises 9.5702 2.35 0.0582 .10 1.4143 0.1349 0.8000 0.50 .40 1:50 1 .30 1.50 h = 0. h = xln) . .35 1. / C 1 1 "J 1 Zzs 1. 4770 0.10 0.20 0.00 YW 0.30 1.45 1. 4 lis . 15 1.0623 . 05 V(n> 0.2360 .0000 0.2676 0.0375 . x(n) h = x(n) 0.25 0-30 0. . 00 10 0.2278 0.35 0.20 0.3753 1. v(n) 1.0526 0.30 0.1 x(n) 0.25 0.2 h = x(n) .40 0.05 0.35 0.20 0.5070 6267 .15 0.30 0.00 0.0075 . 5735 1.0000 0.40 0.3492 0. .00 0.3427 .35 0.50 h - . 1863 0.45 0. .0025 .20 0.10 0.45 0.1 x(n) 0.45 0.00 V(n) 0000 0. 0.40 0. 1 0.0905 0.1905 0.1134 . 0100 0.10 0.30 0.7670 1. .15 0.4124 h = 0.2220 1.0500 0.50 A f\ f\ A 1 1.05 xln) 0.50 YW 0000 0.1000 0.30 0.10 .3058 .1429 0.50 . .0250 0.0703 0.40 0.25 0.05 0. uu 0.0500 1053 1668 1 1 1 1 .20 0.30 0.15 0.3144 1.20 0.50 v(n) 0.05 0.00 0.8371 . 0300 0.9332 1 .50 h = 0.0976 0.10 0.40 0.0150 .3782 .Exercises 9.0000 1 10UO 1.4039 1.0000 .0601 0.2731 0.1005 05 V(n) .0000 0.10 0.20 .4198 h = 0.40 0. 1 Exercises 9.2159 0.00 0.40 1.20 0.15 0.00 1.7219 h = 0. 30 1 35 1 1.05 1.1231 1.10 1.10 15 0. 00 1.40 0.30 0.00 u . 50 0.6217 1.0000 1. 0000 0.30 0. 1 .1 vln) 1 . 50 1 1.50 h = x(n) 05 Vin) .1538 .40 0.0000 1.50 .5000 5125 0.35 0.25 1 1.20 0.35 . 1U 0.10 1.00 0.10 1.2696 .2057 .40 0. 40 1.30 1 .5565 0. 0000 1.5293 1.20 1.45 0.2921 1.0000 1 .53 95 0.5000 1CA U.20 0. 5613 0.2194 0. 50 h = 0.4440 1 .c bzbu 0. 5548 f\ .1115 .20 1 1.10 0.5559 0.05 .0298 . .2245 1.45 0.5527 .6902 h = .1 xln) Y(n} 0. 1 1 .20 1.0147 .2 h xln) 0. vln) 0.0500 1 .15 1 Vln) 1.3651 1.00 .0775 . 05 xln) 40 0.5639 h xln) 0.5322 0.5232 0. vln) 1.0000 1.0506 .0588 1. 1000 .05 xlni 1.1619 1. 5431 0.30 1.3505 1.5072 0.30 0.25 . .0049 1 . 50 1 .45 1 1.25 0.5452 5496 0.0191 1.05 0.1039 1.5547 0. . 5465 .10 0.3451 1 .4475 1.30 2 .) Exercises 9.00 . 1512 .2 h = 12.2342 1.45 2.15 3.4230 0. h = 0.05 0.40 1 50 x in " n> < 1 00 05 in 15 20 1 ?5 .05 x(n) V(n) 1.0000 0.10 0.20 0.35 2 . 05 4 .4234 .9900 1.6231 1.40 2 3226 1.00 5.6228 .30 1 .5747 0.50 2 .20 0. 1 .20 - h 1 i 1 .3652 0.4952 1 .50 2 0592 .45 1.0801 fl .3095 0. j ^^^n ^^75 u J *± ^ ^ M ~> 5 i _> 5746 5868 0.1 xln) V(n) .5751 1.30 .3098 0.40 1. (b) h = 0.0501 0.30 2.35 0.00 .00 5.5989 0.50 13. . (a) h .40 2 .5991 0.2028 .0000 1.7236 1.2030 0. v(n) 5000 5250 .40 .5470 h = 0. 5000 n 1. 6109 . 1 0.25 2 .05 x(n) .2546 1.9763 1 .4832 0.30 0.20 3.0000 0. 1004 0.10 3.45 .35 .1005 0.9452 1.5499 0. 1.20 3. .25 0.40 0.1 X V 1.15 0. 10 1 1 -1- .50 .00 1 .0000 1.0.2554 0.1786 1. ^0 V -J 1.7009 1.10 3 .50 V(n) 0. 13 97 1823 h = xln) 1. 50 f\ fl u xln) 0.35 r\ . .5490 0.2622 0. 10 1.0984 1 1389 1 1895 1.5492 0. 05 0. 00 Uo ±U 0. 372 .2231 .40 .20 1. uysj 0.30 0.20 0.15 a ^ ZU 0. 10 0.20 h = 0.25 1.20 0.2 h = x In) .4053 0.05 1.0663 1 .0414 1. 00 1.5362 0. .10 . .50 V(n) 1.40 1. U .00 0. 5000 0.30 0.4054 0.1 1. 5214 0.20 1.25 0.5294 0.30 1. 0000 1. 0095 1.2526 1. 1. .00 1.35 1.40 0.5444 .1 Exercises 9.0000 U U400 /I 0.15 0.40 0.3260 . 0404 1.30 1 .3715 0.5469 .5000 0.40 0.3363 0. 05 Y(n> 0000 0024 0100 1. 0.15 1.45 .25 0.05 Vln) 0.50 h = 0.5449 0.50 h = 0.5408 . 1.5215 0.5503 0.45 x(n) ^ 0. n 00 in 0.10 1.05 .3315 1 .5359 .3364 0.35 0.5495 0.0967 1 .1866 1.40 1 1 . 1.2623 3001 0. . vln) xln) 0.45 .30 0. U . 00 0. 5116 0.50 .30 0.0000 1822 0.0228 .50 h = 0.1 xln) vln) 0.5484 . 50 h a A fl \J 11 1 1 .2515 10 .0805 .45 0.2541 . .1110 1.2 h = xln) . 0.00 > on n os 10 U.20 . .20 ?5 X 0.40 0. V . 1499 1.8598 2.00 . 1 1.40 h xln) 0. U \> Vu VJ 10 15 0. . 0200 0. 5556 1.1122 0.J 2 0.40 50 .0615 .0000 0.20 ns fl 1 v In ' X £ J^ u 1 7 IRQ X J J oo X o/ 1.0488 n 7 i 11 ' 0000 1. . 05 10 15 1.1217 1.0619 0.30 .50 x In v In 1 1. . 1 _L fl \t 0.0050 0. .0451 0.1266 n illili on n O - .25 .45 0.1022 0. 00 .35 0.35 1. 6939 ?s . I OS (1 x In —J 7 1 0.1 U .45 0.35 0.0200 0313 U V J XJ 0451 0.40 0.6600 1.4172 1.1266 .50 0. - 0.30 1 40 SO .6880 2 X & *± ** J. . 1056 000(1 1 o X - YW 1.20 . dO 1.0526 1 1113 1 1 77S — X X .20 0. . 3? 1 1 .Exercises 9. 1 ? .4361 .30 0.1916 1. 00 10 30 (1 0.1564 1. y ) .0000 0. 0805 0. 373 .1 vln) 0.50 0050 0113 0. 9060 1. Ob v(n) 1 .0000 1.5866 0.0000 x(n) .1 V(n) 1 0000 1 1079 1 2337 1 3806 1 5529 1 7557 en 5U 00 05 10 15 20 25 30 35 40 45 50 = 0.9711 2.4938 1.1 Yin) aa i 5000 1. y) dx so that y(*n+l - y(zn) = f{Xn.0519 1 1079 1 1 C A 1. 10 .6224 IMPROVED EULER EULER 1.6106 0.2469 1. cTc f\ r\ U f\ a /4 *i .2000 1. 0. 5986 1 1. 374 = hf{x n . 1 J .yn){Xn+l ~ %n) or Vn+i =Vn + hf{x n . 5744 1 1 4a 0.4634 1. Integrating y' = f(x.10 5250 i nn 5498 1.yn ).Exercises 9. . 6668 2. a U x(n> Vin) 0000 1.40 16. h = 0. OU / 1. 00 1.50A 0.5622 CI A A A . a <in n .7988 we obtain y / Jin dx= I Jx„ f(x.5530 1 .30 1.5498 0. y) ft 7> .2 12 A — AA U 10 A 1A 0.1684 1. JU 0.yn ) .2337 1 AA O 1 3043 h = xl 15. 1. 6224 (e) 1 1 1 1 1 1 1 1 >i IT 1 1 YW 1.6503 1.5125 U .5987 .20 A TA .20 1. 00 05 10 15 20 25 30 35 40 45 50 1 .7560 AC A 3UOU 0. h=0 x(n) rtc .6427 8. 1. y'dx^ I we obtain Vn+i)] ^Tl+l n+l 1.9452 1.50 2.yn ).30 0.8x„ + 4)-h 2 .05 x(n) vln) 1.2342 1.50 1. + f{Xn+l.05 x(ni v(n) 00 2. 3.40 1.3226 1. 0000 4. h = 0.9900 .35 2.00 5.20 1. = 4y - 8x 2t/) +4 so that J/n+l =Vn + (ten - 2yn + l)k + (4yn 0.Vn+l)].1217 0.4172 1.2(4a.2541 1.3451 2 .5237 0. We = yn + (2x n - 3yn + l)h + {9yn use .1916 0.30 0.45 1.45 2.40 1.3 17.8150 0.50 1. FVom y1 = f(x.4952 1.4475 1. We h = 0.50 3y' = 2-3(2x-3y+l) = 9y-6x-l Vlnl 5 0000 3 . .Exercises 9.05 1. y) m \[f{x„. n+> { + f(x n +l \[f(x n} yn ) + f{x n +i.10 1. 1564 1.2482 0.3 1.6x n .00 1.7009 1. 00 0.20 3 .1786 1.9763 1.yn+ i)\ or Vn+l - Vn + ^[f{x n . y) and f(x. yn ) .y„) + f{x n+ i.0801 so that lfn+i 2.1056 .l)^h 2 .10 3 .0000 1. .1 use y" = 2 - xln) 1.3212 0.35 1. . h = 0. 10 Vin) 2 0000 1. so that -y{x n ) « y(x„+i (x n+i - x n )^[f{x„. 6600 0.0592 h = 0.25 2.yn} Vn+l = yn + hf(x n .7236 2. where Exercises 9.1 y" =4- V xln) = 4 .20 0.40 0.40 2.5751 1. 05 10 1 .6571 15 1.1499 0.yn+1 )]dx •'In 'it.30 2.4124 1.2546 2 .25 1. 15 3 .20 1. . 1122 .30 1. 4053 .40 .35 so that v(n 0000 0.10 0.45 so that yn+i = Vn + (4 + Vl)h + use y = e.0952 0. 376 1 V .20 0.1000 .0000 1.30 0. 50 .2999 0.2617 0.2622 0.2551 T090 0.y ) =— -2y (2x n + 2 2x n yn + 2yl)\h 2 0.4310 1.4372 1.0000 0.20 0.05 0.2519 1.y (e.40 .3357 50 0.0525 1. h = 0.20 .3647 0.1111 1.05 .3378 1.20 .00 .20 1818 .40 2 !/ + 2y 3 2y x In \ni 0.25 . 0.2020 0. 05 .30 0.30 0. h = 0.20 0.0561 h = 0. We h — h = 1 ft use x.1770 1.Exercises 9.0500 1003 .1100 .3082 00 05 10 V 15 X -J 0.40 0. 1822 0.50 2 .50 .2230 0.25 .4046 .3 3.05 xln) Vln) 0.40 yn+l =y n + e-y^h-{e-^)\h i 0.35 so that - yn+l 4.5438 0.00 0.0950 .00 1 .4211 0.0000 0.0300 use y" 5.15 0.45 . 50 o noon 0.30 0.3714 0. 75 .2025 0.15 .2490 0. = y" = = 2yy' + 2y(l 0.4825 0.8557 2.6910 1.1 xln) Vln) .35 0. We + (l+yl)h+ yn (2y n + 2yl)~h 2 = 2x + 2yy' = 2x + = 2x 2 2y(x + 2x 2 y +y 2 + 2y .45 0.0000 0.6783 .10 0. 1 J- 510 J -L U 0.10 1.0488 0.5456 0.10 0.5535 1 .4223 0.10 0.3363 0.40 0.00 .05 vtn) 1.50 ) 3 We h = xln) .30 1. .1397 0.1 xln) vln) 1. 20 .1264 0.20 = = x{xy + y/y) +y+ x 2 y+ -Xy/y + y+ xy + 1 1 0. We n use — n ^f^J y" = v n U + 2yy' 1 = l + 2y(x + y 2 .7551 0.0200 0.25 . 1078 1.0519 1.5000 0. / ri 1 UUUU UU1J nn £ n Ju U .00 0. 0804 0.45 0.10 1.1265 . 50 0.20 0.30 0.35 0. 05 0.30 0. h = x(n) . 0.2327 .35 0.5524 1.20 0.5482 use x (n . U^UU U .50 that = Ifh+i 7.10 .5213 . U .45 0. 00 1.5480 . A\J U .5293 .40 ) nh —— 1 . u \j = 1 + 2xy + 2y 3 * .30 0. .3 6.45 0.1682 1. 10 .05 0.5116 0. + 2yl)-h 2 y" = - 2(x y)(l - y') = 2(x-y)[l-(x-y) 2 } = 2(x~y)-2{x~yf so that 8. VJ 1 A AAC1 0.5357 .40 0.05 0.35 0. We h = 0.5475 0. 0.7522 h = 0.1 x(n) V(n) 0.4628 1.0805 0. We yn + (in + + (1 + 2a:„jfr.5490 YW . U11J U .3039 1.05 x(n) 00 0. uu UD .50 V(n) 0000 .0451 0. - h = 0.0000 1.1021 0.40 0. ±U U ISc 1 -wr U u n .25 0.0615 0.20 0. YAHl uuuu u UUJU uu n n J.40 0.5504 .50 0.0451 0.5000 0.3802 1.J Exercises 9.00 0.5466 0.1 use y" = xy' x(n) 0.5487 0.50 0. + Xi\/V» + ^ Vn + U U U > .40 .15 0.3790 .5441 0.50 2^y 1 1 - that 1/n+l =yn + (XnVn + \/V n )h + { x nV«.2334 1. .30 0.6495 1.30 y/y .40 0. 15 0.10 +y+ ^ V(n) 1 .5406 0. ZD U .5438 0.1075 .5213 0.5355 0. 3 -TERM TAYLOR V(n) 1.13 87 1.10 0.10 0. 3226 use y" = +y - 2xyy' -V xy 2 = 2xy (xy 2 ~~)+P X) X \ = 2zV-2y 2 + 1 J- X 1 1 1 1 X* % 10. Using the Taylor .1857 1.50 0. l use y" = = = 2x 2x 2x 10 X V J.30 0.fv = 0. 00 1.2y(y .3301 h = 0. 1.4938 Vln) . 10 1.40 1. J + 3y 2 y' + 3y 2 (x 2 + y 3 ) + 3xV + 3y 5 1 J.6225 h = 0.20 1. 5988 1.40 0.30 1. \J\J KJKJ J.3 9.y 2 ) = y-3y 2 + 2y 3 so that yn+i 11.2469 1.5498 1. y) = 1 yn + = ax + 0y [*t + vt)h + so that fx (2x n = + 3a£i£ + a. » \J i 1 £*£aJ nd t t.40 0. We M + y- _ £) h + 2xlvl . 50 1.Zyn2 + 2y 3 )-h 2 .2vl = ri- YW = y-y 2 .45 0.5745 1.20 0.7988 . —YSRl ir/n x(n) h=0 . 0969 1.4600 5.05 0. 0000 1 . 6107 1.+ Exercises 9.30 1.5622 1. 1891 1.15 1 . and 378 3i£)£ft all a .9711 .6226 W Uj If) { use y" . ncci y ddj VjOJ 1.2518 1.35 0. 1.0100 1 20 1. x2 \J .2400 1.25 0.2000 1 . .00 0.30 1.0000 1.1 x(n) EULER 1. 0410 1.9060 1 2 IMPROVED EULER Vln) 0000 1.6353 so that yn+\ 12.5125 1.0000 1.00 1. » Uua j i K on m j.10 1. Let f(x.1 y<"> 1 00 1.5000 1. higher derivatives are 0.1 x(n) 0.45 1.40 1.5866 1.50 0.50 so that = fTi 1 . 2 nn i .6345 2 .5000 1.5374 1. * mm 1 J.5745 1.05 x(n> WnJ 1. We =Vn + (v» - vl)h + 2(y„ .5250 1.5987 1.20 0.6668 2 6427 8.5499 0. We — — h 11 h = 0.35 1.40 1. in J.5250 0. and y From = 5 we find c = 6e~ x = ce 1 - x. 5.3 series expansion we have for f(x.7992 7.5509 8.5300 6.50 {x„ + yn - l)h + {x n + yn )-h 2 .yn ) .5454 8.y n ). y) .0000 3 -TERM TAYLOR V(n) 5. Since f(x n yn ) .40 1. so that y = &e x.5454 8 .5310 6 . 6 5300 1262 6.fn) \h[2y'n = yn + = ft.1 - x.1262 6.7954 7.f(x n . For Taylor's method we use y" h=0.3847 . To solve the initial-value x integrating factor e~ we have + ^M/fan.1 x(n) = l + y' 1. IMPROVED EULER V(n) 5 .yn ) + ah + @hf{x n .7954 7. + V* + ^» fn+1 )] + fi^Vn) + ah + f}flf(x n + h{a + 0y'n . 5.30 1.Exercises 9.yn )hf(x n .3847 EXACT V(n) 0000 5.0000 5. 00 =\+x+y -\ = x = yn 1 +y so that I/n+i + .f(x n: yn ) +fxixn. 10 1.1284 6.20 1. = + 0y'n = y£ n w»d a y' Vn+l - J/n = 13.yn)h + fy (x n . y„)] )\ 2 i£- problem analytically we note that the differential equation linear with is Then . !/n) + + ^[/(^n.3923 . 50 V(n) 00 10 20 30 40 50 V(n) 0.2465 1.4 X x{n) 1 1 1 .0200 0.4110 1.1037 1 V(n) .1266 V(n) .10 .7561 n) Y(n) 00 10 20 30 40 50 0.3365 V(n) X 00 10 20 30 40 50 X 1.1115 1. .30 0.0989 .1480 1 .0000 1. 5744 0.5463 xln) .3163 .40 0.30 0.40 0.0000 1.5250 0.00 0.30 0.0000 1 1 .5531 1.10 0.0417 .0000 1 .5000 0. 50 0. 6562 n) 00 10 20 30 40 0.0000 1003 .4397 1. 5987 0.10 1.00 .20 0.20 1.6225 .00 0.4055 x(n) 0.5498 0.2624 . 5358 0.3333 1 1 380 .1079 1 .2337 1 1.Exercises 9. 0.50 5.5213 0.4228 0.40 1.0000 3. 50 0.30 1.0050 .40 1 . x(n) V(n) .20 0.0533 2 Y(n) 0000 0953 1823 0.20 0.2284 2 6945 2.2530 1 .10 .00 .4 Exercises 9.0805 0. Yin) 2 .40 0.2027 .9724 3. 00 1.6961 2.0101 1.5000 0.0670 X (n) .10 0.30 1.1905 1.3093 . 5482 0.20 1.5493 x(n) 1. 5443 0.50 0. 0.3807 1 . 0451 0.50 X 00 10 20 30 40 50 0.0000 . 0432.\n(a- In - dt dA = dt (3 A) A)} =t+c a(t a-pA A a.0A — A= 1 + c) . . ( 1 ^ ^^ + 1 dv + v \ - h W5( and v{5) a= -1) a 35.0 32 .128 and = we obtain 0.025v o 2 = 1 2 3 4 5 f{t.a(t+c) e ae a < i+c > . 2.0 34 .0A -[\iiA.7678. Separating variables and using partial fractions 1/1 ^v^- and iD 2^32^0.VO025 v\) = t + 13. OOOC 2570 9390 9772 5503 7128 .4 11. Separating variables dA A(a - a\A a. See the table in the following problem.Exercises 9. /3 + e" ac e c. v).pAt a ^ + 0e a(t+c )} A = ae al t+cl < Thus A(t) a Q = 1 + /?e a t+c < > + e-*(t+0 = dt . OA 12.0 Vfr025 we obtain Solving for v <n\ lill o we have + VomBv 2^32 .025 Since v(0) = we find c — 0.0 35. 14.0 35 . Let 25 . Write the equation in the form 1 n) — = 32 -0. 00 49.2128 ' A (observed) t 15.02 5 v(n) 1.128 A{t) 0.8235 and 2. + 1 1 1. ) .0432 1 2 3 2. 78 13.4 From A(0) = 0. + -h] n + k).95 12.32 49. .50 12.0282 16.93 1. 6934 2. Simpson's rule on [i n > r in = is x„+h Jx n For f(x.y) + h\ f(x n )+Af[x n O + -h)+f(xn + h) f(x) the Runge-Kutta method gives ki = k2 = hf[x n + -h fc3 = hf[ fc4 = hf(x n hf(x n ).9425 903 .24 = + e- 8. 0000 (n) 00 10 1 20 1 30 1 40 8.Exercises 9. 40 47.64 36.24 so that e~ ac = « - a/0. 50 49.30 36. 53 A approximated A (exact) 1.23 47.46 36. and f(x n ) + Af(x n + 382 + f{x n + ft)] .24 we obtain 0.8235e.2511 1.63 (dava) ( X ' 4 47. 22210646. = we 1.8) 2. « 1. Comparing exact values with approximation! 1.5 1. REM ADAMS-BASHFORTH/ADAMS-MOULTON REM METHOD TO SOLVE Y'=FNF(X. y(0.Y 240 PRINT "X". H 160 INPUT "NUMBER OF STEPS = = 1.X "Y0 =".09181796.02140276 compared to yi 1.Y) 290 K2=H*FNF(X+H/2.2) = e 1. so that and j/ From y(Q) = 1 we find c obtained in Example 1.Y+Kl/2) K3=H*FNF(X+H/2.Exercises 9. For — y' y = x — 1 an integrating factor = is e e x . fts = e^-ze" 1 + c) = -x + = —x + and y 1 find y(0. anc .09182470 compared to y 2 j/(0.N 160 INPUT INPUT PRINT 180 "X0 =".5 Exercises 9.22211880 compared to y3 100 PRINT 150 INPUT "STEP SIZE=".42554093 compared to y4 .02140000.Y+K3) Y=Y+(Kl+2*K2+2*K3+K4)/6 330 Z(I+1)+Y 1."Y" PRINT REM COMPUTE 3 ITERATES USING RUNGE-KUTTA 250 DIM 220 230 Z(4) 260 Z(1)=Y 270 FOR 1=1 TO 3 280 K1=H*FNF(X.Y) HERE 130 REM GET INPUTS 110 140 170 IF N<4 GOTO 200 210 REM SET UP TABLE 190 (AT LEAST 4)=".6) = x ce*.42552788.Y) 120 REM DEFINE FNF(X.Y+K2/2) 300 310 320 K4=H*FNF(X+H. y(0A) a 1. 2109 1.60 1 0.5569 x(a) 0.80 1.1039 initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector Yin) 00 80 Yin) 2.1003 .00 0.00 0.00 .20 initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector predictor corrector 0.00 I vinl 0000 1 .6841 1 0234 1.Z(3))+37*FNF(X-2*H.65B5 0.4228 0.Z(3))+FNF(X-2*H.4112 1. 0.2027 3093 10 20 30 4227 40 4228 50 54 62 5463 6S40 6842 8420 60 70 .80 1.Z(2}))/24 410 420 X=X+H PRINT X.Z(4))-5*FNF(X-H.20 0.2049 0000 . x(n) 0.BO 5.5 350 X=X+H PRINT X.0000 0.YP)+19*FNF(X. 7332 0.Y 360 NEXT 340 I 370 REM COMPUTE REMAINING X AND Y VALUES 380 FOR 1=4 TO N 390 YP=Y+H*(55*FNF(X.40 O.8423 1 1 1 90 1 00 384 .6461 0.0297 1.Exercises 9.40 0.60 0. .5376 1.40 xfn) Yin) 0.Z(2)) -9*FNF(X-3*H.20 0.GO O.Y 430 Z(1)=Z(2} 440 Z(2)=Z(3) 450 Z{3)=Z(4) 460 Z(4)=Y 470 480 3.1483 0292 0297 2552 1 2603 1 5555 1 5576 initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector .2027 0.Z(l)))/24 400 Y=Y+H*(9*FNF(X+H.7232 4.0000 1.Z(4))-59*FNF(X-H.7328 0. NEXT END x(a) 0. 3433 3.1837 00 .60 0.0201 0. WnJ x(n) 1 0000 3 2 .34B6 4.0000 1 .2102 00 xfn) initial condition Runge-Kutta Runge-Kutta Runge-Kutta predi c t or corrector 0000 10 0003 0026 0087 0201 0200 0379 0379 30 40 50 60 0630 0629 0956 70 .9603 3 3486 3 3 3486 7703 7703 4 2280 3 0.5 Wn) x{n) 00 . 4414 1 .6028 3.2385 .60 .2230 Runge-Kutta pr edict or corrector predictor corrector 00 10 20 1 30 1 1 40 50 80 9719 1 90 1 . 3 1 J.00 1 . .Exercises 9.0026 0.2276 4.80 1.0956 B0 1359 1360 1837 90 .1362 1360 0.20 Aft f\ 0.20 0.00 0.40 0. 9719 2 .4414 6949 1 60 x(n> 0. 2 2 740 70 v(n) .23 79 0.2384 2384 1 predictor- corrector predictor corrector predictor corrector predictor corrector y(n) 00 20 predictor corrector i . 80 1.0000 .00 .2280 initial condition Runge-Kutta Runge-Kutta Runge-Kutta predi c t or corrector predictor corrector predictor corrector initial condition Runge-Kutta Runge-Kutta Runge-Kutta predi ctor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector predictor corrector .2740 2 6028 2 .6028 2 9603 2 .0630 0. 7561 1.0291 3.5207 initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector predictor corrector predictor corrector predi c t or corrector predictor corrector predi c t or corrector predi c or corrector V<n) 0.60 0.t Exercises 9.90 3 .2.30 0.6214 3.5208 WnJ x(a) 0. 0499 1.20 0.3807 1 1.80 .1079 1.5531 1. V "'^3\ 2.2337 1.00 WnJ 1. the Taylor polynomial through k = 2.9961 2. Since the three-term Taylor formula error is x n<c<x n+ thus i.0000 1. the formula agrees with is = The 2.00 0.00 0.5531 1.9960 1. Wher6 is is y"'(c) 3.1) =s yi 386 xn — <c< x n +i.40 initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector 1. the . = 4.20 0.0918 Exercises 1.0289 0.40 0.5203 1. 0918 1.00 3 .5 x(n) 0.10 0.6.0052 1.40 0.0000 1.2811 0.2337 .2312 6211 2.0214 1. 7560 1. 9.30 initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector predictor corrector 1. Since the improved Euler method the Taylor polynomial through k a second-order Runge-Kutta method.5151 3 .50 0.20 0. local truncation error (a) Using the Euler (5| ( c) method we obtain where j/(0.0000 1.70 2 2.6213 3.60 0.80 1.6180 2. Since the fourth-order Runge-Kutta formula agrees with the Taylor polynomial through k local truncation error is y 4. the local truncation -gj- where xn < c < x n+1 .5530 1. 1.10 0.9961 2.00 0. 0. 05 we obtain y(0. Thus an upper bound 0.Exercises 9.02e 1.2214 = 0. Thus an upper bound 0.1. method we obtain y"'(c) Since e 2* e halved we expect the error for h (a) Using the improved Euler 1 (b) Using y " < 2 = Comparing 0. the actual error - is j/(0. global truncation error 0{h 2 ).221403.1) ^6 = 8e = ' e 2 = e 20 2c < 0.1) s= 0.221403 - (a) With 0.05 to be one-fourth the error for see that this Using the three-term Taylor method we obtain y(0.6 (b) Using y" = Ae 21 we see that the local truncation error 4^^=0. is 2c 2c < 0. (d) h the case.221025 — Using the three-term Taylor method with h = 0.001403 which is less than 0.0114 with 0. is see that the local truncation error y "\c) y(0. = 2' 02 1. when « y\ — = 0.221025.1 = ' = e ' 2 < for c < 0.22.0114.21 = for is = 2 ' 01 ' = 2 e we obtain y(0.000378.22.1) = j/i 0. 6 e ' 2' 2 .001628. is less = when the when h = 0.001403 we 6.05 - 1.221403.l) (d) Using the Euler method with h (e) e is 2c < for c < 0.05 to be one-half the error = y(0.001333e ^6 = 8e M! 6 e c With global truncation 0. see that this an increasing function. is for the local truncation error (c) = Since y(0. 1. than . e error 0(h).1) = (c) Since y{0.1) = 2 e ^y2 — < 0.1) (b) Using y"' (c) we obtain 0.1.0244.21.05 halved we expect the error for h Comparing 0.221025. for the local truncation error is Since y(0.1) ?w y\ = 2 e than 0.214 we 5. the actual error isy(O. = = Se 21 we Since e 2x is an increasing function.001333e Thus an upper bound 0.1. is = -yi Se 21 we see that the local truncation error the local truncation error < for 0.1) as y2 = 1.1) - yl = 0.1) 2 e = is The error in (d) step size is is 0.001628.001628.0244. 1. is = COOl^e 2 ". is the case.2214. 1.001628. 2e 1.0214.000378 with 0. the actual error is y(0.02^. s/'(c)y= Since e 2x an increasing function. which = 0.1.1. 0. = 1.001333e 3/2 = 1. 1. (d) Using the improved Euler method with h (e) The error in (d) the step size is is 1.001403 which is less 0. 2c is .221403 = 1.025.1) (d) Using the Euler method with h (e) The (a) _2c = 0. when h error for 8. is see that the local truncation error pW( c J£. - — 5e _2x 2x is we is 0.0234 we 9.2214. e~ the local truncation error is 2c 0.1. see that the local truncation error 10e 0.221402571 Using the Euler method we obtain Using y" 0. for the local truncation error (c) = Since y(0.Exercises 9.000003257.0109 with 0.6 (e) The error in (d) the step size is - 1.000000187.8125 < . 3c < for 1 - is t/(0. halved we expect the error is = yi -acM! = a decreasing function.1.1) (b) Using y'" = global truncation error 0.05 to be one-fourth the error for method we obtain (a) Using the fourth-order Runge-Kutta 5 = halved we expect the error for h is With 0.1) 02 e = see that this is e 2c 2e < e 2 ( 01 0.000002667e = e 02 = y\ 1.8.05 1.025(1) Since y(0.1) With 0.000002758 we the local truncation error step size we obtain halved we expect the error for h is 5e Since e~ = 1. 0.221025 Comparing 0. which y(0.= Since e 21 global truncation error 0(/i 2 ).1) we obtain = for h yi = < c 0.221402758 the step size = 0.221402571. Thus an upper bound 0. error in (d) 4 — (0^L = 0.8125. global truncation error O(h). Thus an upper bound = 0.1) With 0.1)«tti (e) The 0(/i ). 1.1. 0. « y\ = 0. see that this -2c see that this is « = y% is less for than 0. 0. < c 0. the case.001403 we 7.8234. 388 < c < 0. (b) = Using j/ '(x) 32e 2ir we 32e ) is an increasing function. is = = when the case. Using the improved Euler method we obtain y(0.05 to be one-half the error is when when h = the 0.000000187 with 0.1.025e- = e° = -10e -21 we is the case.825.05 0.0109.001667e. Comparing 0.001667(1) < = e° = 1 for 0. (d) Using the fourth-order Runge-Kutta formula with h = y(0.1.8234 - = 0. the actual error is y(0.000002758 which is less than 0.000378 with 0. the actual error (c) error in (d) is 0. 0. j/(0.0234.l) - yi = 0. (a) (b) is 1.05 to be one-sixteenth the 0.1) = h— 0. e = ~ see that the local truncation error Comparing 0.025. Thus an upper bound for .221402758. < for 2c .000378. o Since e~ 2x is a decreasing function.001667.1.000002667e ) 02 y(0.000003257. 000003225. = — 38e is |0. is - 0. |y(0.1) is = = 0. is 0. = (c) (a) (b) = the step size 0. (d) Using the fourth-order Runge-Kutta method with h y(0.000003225 we see that _3 ^ I_1 ' we " y see that the local truncation error {c ) % = 38e-^-') % = is this is the case. see that the local truncation error 40e for global truncation error 0(/i 2 ). halved we expect the error for h Comparing 0.823413627| = 0.000305.001587 we y'" Using = — 10e we 0.1) -21 = we obtain 0.l) — 40e -2T Thus an upper bound 0.2c M! = is less « y2 = than 0.1) (d) Using the three-term Taylor method with h (e) The error in (d) h (a) (b) = is the step size 0.000305 with 0.1) (d) Using the improved Euler method with h (e) The error in (d) when h 10.001587 we 0.001587.000003333.2c 6 Since e _2x is a decreasing function. y\ see that the local truncation error l 0e g/(0. the actual error Since j/(0. halved we expect the error Comparing 0.05 = 0. e 0.823413 is is - y(0. (c) when 11. the actual error is = < 1 for < c 0. 0.6 0.823781.1)m y 2 (e) The 0(h (a) ).000003333(1) = a decreasing function.823416667.000000185 with 0. 0. the actual error = Since y(0.001667e. global truncation error 0(h 2 ).05 to be one-fourth the error see that this when the case.825.823413 is - 1 < for c < is j/(0. e~ 2c < the local truncation error 0. is 0. 0.000003333. which we obtain for h = y(0.001667. which — With y2 is less = than 0.823781. |0.823413441. 0. error in (d) 4 error 12.8237181| -yi = when the case.001667.1) is _2c < e° 0.823413627.823413. |0.05 we obtain 0. which is less than 0. .1.001667(1) 0.1. is = r.05 0. = is = e° 0.1.1) Using y^(x) = we .05 to be one-fourth the error =.823413441 step size 0.000305 with 0. is -^M!=0.000305.823413.1) 0. halved we expect the error for h — With global truncation error 0.001587. Thus an upper bound for 0.1.8237181| — j/i = = 0.1. when the when h = Using y" = 0.Exercises 9.05 to be one-sixteenth the Comparing 0. 0.000000185. With Using the fourth-order Runge-Kutta method we obtain y(0.001667.000003333.1) - = yi\ 0.1) as see that this Using the three-terra Taylor method we obtain ^(0. 120 Since e _2x is the local truncation error (c) = Since y(0. h We = 0.i/£ 13. = have Uo.5) ^).Exercises 9.9424.5.2325.005950.l/^0.080108. the error for h With 0.19.i/£b.3 ( c -'». ' we see that the local truncation error h3 (b) Since 0. « y(1. " 3 < c_1) < e _3(1_1) = 1 for 1 < c < 1.05 = 3 19(0.5) = 0. y(1.q$ ) fa 4.05 y'" e we obtain i?ci.(1.059166. we obtain 0. < e" 3 ' 1 "" 1 ' = 1 for 1 < c < 1. 0.05 Em = 0.10. the error for h have .l/^o.1109.l/£b.05 is see that the local truncation error h3 y"'{c) (b) Since e" = y(1. 1.5) ^" 1 ' is So.52. We actually 4. 2.5. the error for h With 0.05 = = -3 * 1-1 ' is = IHe" 3 ^" a decreasing function ^6 = 1 ' < x< for 1 19ft 1.05 2.1 is global truncation error we obtain ^ 0(h 2 ) we expect = i 14e mV -3(-i>^ 6 3 2.005950. (d) Since y(l. e 3 is e.05 5:3 We 2.053216.059166.$) (d) Since y(1. e -3 < c-1 > < e -3 ' 1-1 ) = < 1 for 1 c < 1.080108.5 and */"(c)y <19(0.1 0.5) w 2.05 h = 0.05 is actually have 2.6 (b) Since e^ -1 is ' a decreasing function for < 1 x < 1.5 and y"'(c) (c) ~o < 3 19(0.053216.5 and 3 v'"(c) (c) h - < Using the improved Euler method with h we obtain — = 2. (a) Using = 2.05 0.019.8207.05 is actually .1 £ is = .1) (1) Using the three-term Taylor method with h we obtain y(\.5) =a 2. (a) Using y'" = 0.1 0.5) £fl.05 = = With = 0.1) (1) a decreasing function for 1 < x < 1. while the error for h = 0.026892.5) #0.5) « = Eo. With k — 0. (d) Since y(l.026892. we obtain y(1. With h = 0.l = With h 1.05 * 4. while the error for we expect global truncation error 0(h) = = — lHe -3 ^ -1 . 390 Eq A 0(h = 0.1 is global truncation error 4 -52. while the error for 2 we expect Eq.1 = 0.05 = = w 0. >. e is 3 (c- -3 ^ -1 * .0532. = — 114e _3 ^ -1 we 14.\/Eq.1) 2 (1) (c) Using the Euler method with h 1.5.019. = 0.Eb. the error for With h = 0. 3 -.5 . 0.l/£fl. (5) = is — ( c+ l)2 * < ~ . (a) (c) for —< 8.05 is ^ 2. 3 < - . 3/(1. 1) „ 3 = 1 for < c < 0.000122595. £b.5) With h = 0. = 0.We actually have 2.05 we obtain 0.i/£fl. y"' (ar+l) 3 we see that the local truncation error is h3 1 ' 6 * (b) Since (l + is a decreasing function for (c + l) <x< 3 0. 0. (a) Using j/ 5 3<. y(0. + If we see that the local truncation error 2 2" 0. (x = 1 ' < e^ 3 ! 1 we obtain 0.5.Exercises 9. while the error for global truncation error 0(h 4 ) we expect £b.05 = j/(1. e = 5 8.06.4055.5.05 w (a) Using = 0.55fe < 1 B < x e- 3 <c.000006757.i/£b.i/Eom = actually have -3 2.55(0.5) 2.4124.0S « 18-14.05 = 0. -.053216232.i 0.05 We is we obtain £b.6 15.5) = = (1)^^ = 0.053338827.005. .0143.5) f» 2. the error for h = >.1) (1) = 0.5.1 we obtain 0.5) 0.5 1)2 and < y"(c) (c) Using the Euler method with k y(0.5) (d) Since £0.1 £ is = .05 0.000333.4198.5 = y(1. j/{0.i = * 1' = 1 for 1 < c < 1.0000855.05 17.1 = 0.x = -1026e ' we see that the local truncation error w 120 y 3 (b) Since e~ < 1_1 a decreasing function is > and y (c) (d) Since h = 16.1 is £b.0069. — (c l) + 1) and ft m i) 3 y"'(c)~< (1) 3 = 0.„ (0 * + .„ (0 + = 1 for < c < 0. (c+1) is (x+l) a decreasing function <x< for is h2 1 (b) Since ^ 1 =- Using y" With Eo.053222989. while the error for h we expect global truncation error 0(h) i?o. 0.1 1. Using the fourth-order Runge-Kutta method with h With h 16. 000333.1 is B Using the fourth-order Runge-Kutta method with With (d) Since ft = 16.05 is actually we see that the local truncation error is I) 1 = _J_ !t >» -.5) With - have Eo.5) = 0.405419.6 (c) Using the improved Euler method with we obtain (d) Since E 05 = j/(0. A ft = 0.1 we obtain y{0. + (x —+—^ 3 (x is is y(0.05 = = 0.405465111.5.405465.5) 0. is actually 4.05 & 0. 0.— Exercises 9. while the error for we expect Eo.05 We is = we obtain y(0.5 1) and y"'{c) (c) t/(0. = ft global truncation error 0(ft v (b) Since 0.i/Eom = ft We 4.5) as 0.l/£b.i = 0. With 0.o5 = —+ (x 1) ^ 5 0.\/Eqx>5 0.405281. have £*.405465108.05 0.64.05 ) We 4. 392 we expect £b.405465. r (c 5 1) < —-ttt = t: (1 + 5 1 for < c < 0.98.000000003.i/Eom -tj -. 2 — 18.05 « . (a) Using j/ 5 ' (b) Since ^ ' -.22. actually have Eo. -.5) « 0.405465168. the error for 0.J = + < 1 for c < 0..5) ^0. t /£t». while the error for global truncation error 0(ft 4 ) 1T. With = ft 0. With ft = h = 0.i 0(h = 0.000002. 0. the error for h = 0.5) Since = (1) Using the three-term Taylor method with we obtain (d) ^6 < = = 0.405270.5) = « y(0. ft = 0. while the error for 2 we expect £o.1 0.000823. j/(0.404643.000046.5) « £0. = ft ic) a decreasing function < for x < 0. 0.000184. +\) 5 is ft We See t ^ at ^ e '° Ca V W 120 a decreasing function ' truncat ^ on error ' s ' (c+1) 5 < for x < 5 —+ ^—tt 0.5 1) and y (c) i5) (c) —< (1) ~^ = 5 ft = 0.05 0. the error for 19.1 we obtain 3.000000060. 5 With = = . —^—^ < (c+ 1) 1) J (0 V.5. 0.05 = Boa 2 ) 0.1 is global truncation error Eb.i/£b.1 we obtain yf0. (a) Using y'" 0.05 j/(0.000195. 1(4u find c\ and 1 The substitution y The initial c2 = — 5.2000 m2 4400 y.9 y is — + 8) = 0.9.I1/0 -4 + 0.2) = + C2X 2 FVom .9 + 0.20 -1 .00 -2 0000 1 0000 0.2443 m2 1600 3298 initial leads to the system y' corresponding to Thus y 5. the 6.ltii - -2 + 0.00 -2 0000 1 0000 0. The solution of the Cauchy-Euler differential equation we find a — —1 substitution y' — u and ci = Using formulas (5) ml 0.1(1) = -4y 0.2) - = -2e 2x + 5xe 2x Thus y ).10 -1 S321 2 4427 0.20 -1 4919 4 4753 u .1(9) = U1 = + 0.9 + 0. The initial The we 2.2 -1. Exercises 9. and — (6) in the text w.1(2(9) - 2(4)] = 10 m = yi + O.1 (Ju . general solution of the differential equation = -2 =u n + h(4u n .2) ss 1.5360 4 8064 X y u 0.4yn u n+1 c\e 2x and 2.1(4 + 0.4918. -1. y corresponding to x.1(10) = 5. to the iteration formulas = = 4 and uo Then 9. with x corresponding to f.Imi = 4.1000 0. 2/1 =i/D + O. u' = 4u — 4y.4928 4 4731 Ruiwe -KuttB method wi th h= 0. conditions are yo — —2 and uq = yi = w + 0. + C2xe 2x from .9 u leads conditions are j/o ) = + 1 -1.68.1 ml 0.7 Exercises 9. y = c\x and y(l. the initial conditions y(0. The substitution y' — u leads to the iteration formulas =Vn+ hu„. Then 1. = — x + 5x 2 is ml 1710 3444 m4 2452 4487 k2 kl 1 1 2000 7099 1 2 4200 0031 k3 *4 1 4520 1 2 0446 2 7124 3900 X y u 0.1(2.^w) = uo 9 4.2800 k3 3 k4 .1«o = ui = ii + V2 = tfi + O.7. and we obtain m3 5280 9072 k2 kl n\4 2 4000 3 . 1. The general conditions 3. 20 1.0000 1.9983 9996 9997 y u 1.0000 5. y corresponding to x.2 m2 m3 2.1 !tl3 m4 .1973 k3 k4 1.0000 1.10 4 9500 1. 5 0000 -5 0000 22 3750 -10 6250 29 07 81 -16 0156 10 6309 -22 5488 55 9856 -31 2024 75 394 k4 5000 6875 .0000 2.3516 3076 9B21 00 10 20 30 40 50 i2 il t 0000 7500 2 5 7813 2 0703 7 4023 6104 9 1919 -1 .2000 .0165 m4 2.20 6.4640 2 2 0000 6594 Runqe-Kutta method with h= 0. 9443 -55 1758 -12 7344 6 31.9865 1.0002 Runge-Kutta method with h-0.00 4.0000 0. 0998 (5) and 1 .1 0.9500 1. and 2 2 x xl the text with x corresponding to (6) in (.l ^- 10 0000 8 .20 1. u' = 2u — 2y + e' cost.4660 m4 5320 kl 6000 k2 k3 6599 k4 6599 y t 0.0017 ml 1.3298 0.0000 11.2315 2150 2480 2157 .0023 1. Using formulas (5) corresponding to y.7500 -2 10 1563 -4 13 2617 -6 17 9712 -8 0000 5000 3750 3672 S867 kl 5000 -20 0000 13 .3507 -75 9326 -17 7734 8 12 .0000 2.0000 11 .9950 1.0000 1. obtain ml m2 .0000 0.0500 9501 0501 0.8000 kl k2 2.00 1. 0703 -40 0000 -8 7500 5 22.4000 4600 ml m2 m3 .00 4 0000 1.20 6 0000 0000 10 .9979 0.0000 9. u corresponding to the text with y corresponding to % and y. substitution y' m4 = u 0.9000 1.4640 2 2 2 0000 3150 6594 Runge-Kutta method with h^Q. The 9.4375 -28 7500 0000 -5 0000 4 17.10 1.7 Exercises 4. and u we obtain Runqe-Ku^ta method with h=0.0001 9.5619 11 4877 2 0000 SOOO 8125 3 we . The m3 m2 ml 0.0019 X k4 k3 0.9998 1. leads to the system y' Using formulas k2 kl (6) in = u.3587 u 0.7170 u 0.00 1.2487 2315 2659 ml m2 m3 !t>4 kl 3000 3299 k2 k3 k4 3150 3446 3150 3444 k2 k3 y t 0.2155 0. 1. substitution y' = u leads to the system . 0000 m3 m2 ml 2 2800 2 3160 mi kZ kl 2 6408 1 2000 1 .20 2.4086 -0.7730 .0785 1.3846 -0.10 -3 4790 -0.3055 y 2. .7648 .20 8.3200 .6524 3.Exercises Runoe-Kutta method with h=0.0000 0.2682 1 8218 3692 k4 1 1 0195 7996 X t 0.00 6.00 5000 0.8176 -0.6535 .10 7.20 1 2 5000 0207 1904 y 2000 0115 2 3592 1 9.2 m2 ml 9400 .6500 .4112 0.4571 -0.4006 0.3055 y 2.1 m2 ml m4 ia3 k2 kl k3 k4 X t 0.0600 2 k4 3940 3 7212 t x y 0.7120 -0.6000 m3 1 1060 m4 1 7788 1 4000 2 .10 1.6862 1 6600 0117 k3 k2 kl 1 .3579 -0.0845 y 1.3502 Runge-Kutta method with h-0.0000 3.10 0.20 2 1589 y 2 2000 3279 Runge -Kutta me thod with h= 0.4800 -0.0000 0.0000 3.4328 -0.8216 0.0000 .8963 3.0000 2.2 m2 ml 6400 1 27 60 in4 ra3 1 7028 3 .00 0.1494 1 1 0700 2289 m4 m3 ra2 1 1 0745 .9200 -0.5000 -0.0000 0.3300 -0.4785 -0.0731 0.Kutta method with h= 0.00 1.0000 2.4571 -0.1 m2 ml .3579 -0.1 ml 1 1 .20 8. 9080 -0.2 2 .00 -3 0000 5 0000 -1.4199 Runae -Kutta method with h=0.1 m2 171] 3000 5193 m4 m3 3850 6582 4058 6925 m2 m3 k2 kl 5219 8828 1 7000 1291 1 8650 4024 k3 1 k4 9068 4711 1 1 1343 8474 x t 0.20 -3 9123 y 5 0000 4 6707 4 2857 Runoe-Kutta method with h=0.3290 -0.7200 -0.4000 k3 1 k4 .7816 1 .4280 1 6632 x t 0.3558 k2 kl - 1 3200 1 7720 k3 2 k4 1620 3 5794 X t O.2 ml m4 k2 kl k3 k4 X t y 0.00 6.7736 1 4790 0862 m3 5324 1 1929 m4 .00 -3 0000 .00 0.3382 Runae -Kutta method with h=0.3000 -0.0000 1.3858 -0.6000 -0.20 1.4342 -0.2340 1 1 1496 3193 k3 k2 kl k4 6000 7073 .0000 -0.4199 Runae -Kutta method with h=0.7688 kl k2 k3 7075 8307 x t 0.20 -3 9123 4 2857 Runqe. 3226 -5.2167 identify P(x) = -10.0 4.0000 -0. is 0.75 0.2259 -0.0000 and h 1.3751 0. = 0.0 1.0 We equation 1.2w + i The finite difference solution of the corresponding linear system gives 0.3356 -0.0000 0)/6 = 0. = f(x) 0.0000 1. is 0.6 o. 1. 00 . Then the finite .8yi_i .4 3.3333yt _i f{x) xje 2".3308 -0.4000 = 4.Exercises 9.2.0000 0. Then the finite difference Then the finite difference + 0. identify equation 0.2355 0.6448 0. Q(x) = 1 0-25.0000 difference equation The - (1 - 2.5.0 0. Q(x) = -1.0625a:.0000 3. Then the finite difference 2 0.5000 3.1667.6 1. f(x) = 0. - (2 is yi+l +0.0316 -0.3333 3. We = P(x) 2.8.5 6.8333 2.0000 2.0400- 0. Q(x) = 25.0625yi -Vi 5. and h = (1 - - = 0. 2. f{x) 1.0 0.8889yi + 1. = - (1 = 0)/5 0. .6667yi+i X y 0)/4 0.6306 0.6774 -2.9600 identify P(x) = 0.a 7. is + 2Vi-l = 0-04.25 We x 2 and h .5807 W+i - The = 0)/4 solution of the corresponding linear system gives 4.0000 = (1 + 1.1667 3.8 -0.0000 + Xi)e 2x '.SB00 -4.0172 -0. solution of the corresponding linear system gives 0.6667 3.0000 1. Q(x) 0. 0.4 0.00 We Then the is 0.0 0.2 0.25^ The = and h 0.2778(1 solution of the corresponding linear system gives 0. 0324 Y 0.7200 15. = solution of the corresponding linear system gives equation The = + ^1 =0. = Q{x) 9.8 Exercises 9. 1.5 identify X P(x) = 0. = and 5x.0 0. We identify equation = P{x) 0.0.0000 0.2 1.1411 396 1.2. ft = (1 - 0)/5 = 0. - 1.96?/. 50 1. f(x) 3. and . 000 1. finite is 0.4167jfc+i The = f(x) 0.625 1.6 1.4168 -0.3 .0000 1 9. solution of the corresponding linear system gives Y 1.0625\ / 1.125 1 .7109 -2.250 . Then [-2 + O. = 0.1875\ J -2- Vi+i - = f{x) .2778(4^).8992 -1. .6510 -0. - = l)/6 0.2 W + 0. and 0.1988 -0.750 1.3353 0.1667 1.0000 We identify X .0156\ "2„ + /. (2 1. and = ft (1 .05a:i)!K-i = O.1.05(1 .125.1875\ 0.0469 + Vi x% \ 1 ( tt-i = o.4 9471 a:.8842 P(x) = difference equation 1.0)/10 = 0.01561a a*. 1.8 6.Exercises 9.5 146 5 .1594 -1.7 0.750 0.9640 2 = 1.0 identify equation = x.250 0.000 1. + 0. .Wtf.05(1 = 0. 0. = (2- ft — 0. Then the is (1 The Q{x) solution of the corresponding linear system gives 0.S918 -1.73 57 1.0000 We identify 1.0000 ft = (1 - 0)/10 = 0.Xj)l»-i = 0.0 1. W+ 0. x. 1 the finite difference 0-Olxi. = ft Y difference equation 8. Q(x) —) K+l + ^1 The = Q(x) 3/a:.0000 -0.3070 -1.9 1.6667 1.875 2. ft = 1.5000 1.2660 P{x) = . is [1 10.2704 -1.1541 -1. finite difference .5833yi_! = 0.000 0.500 1.Olarj. f{x) .375 1 500 1 625 1.875 0.1.000 5.8 0. Then the finite .2913 - (2 = l)/8 2.6430 x~ 2 f(x) = lnx/x 2 and . Ay/x is 1.2 5097 Q(x) f(x) = x.0000 We identify P(x) = is + The X — 3/x 2 0.1626 -1.3333 1.Xi)]yi+ + i 0.0000 Then the 0.0625N — ^-J W~l + (l = 0.0000 -0. J solution of the corresponding linear system gives y 1.125 3.0681 1.1 0. equation P(x) The 0.4304 -1. and 0.5149 1.0000 We = l-x.1667. Xii / ( 0.05xi)yi+i - l.OliilK + [1 - . We identify = P(x) difference equation = Q{x) 5. 0. + 0.2064 -1/x.6855 1. + (1 - solution of the corresponding linear system gives 0.5826 1.375 2.0000 1.8474 2. solution of the corresponding linear system gives X 7.8333 2. 5 3.5 4. Q(r) = 0.000 5. have from .0000 Then the 0. . solution of the corresponding linear system gives 0. We 0. finite difference is +1 -2u.0 3. finite difference J «_ l = 0.250 0.5 50.5. we .0000 72. 0.0000 Then the 0. .125 0.6233 2/r.625 2 .7 .875 3.375 1.0000 0.2225 0.4 0.750 .1 0.0 0. (a) The difference equation 1 is + + yt+ i (-2 + ^ h 2 Qi) yi «_! = h 2 fi The equations the same as the one derived on page 530 in the text. + h 2 Qo)ya + (l - \p V-i \ = 2 ft /o — 2ft. Y P(x) 0.0 2. 1 solution of the corresponding linear system gives 1.2118 = 0.6615 identify equation The X 12. yi + 2»i (-2 + + h2 Q ("2 ) yo + + h?Qo)yo = 398 h a /o Jp ) (in + 2h~ ya.42 96 is 0.0 1.8 - (1 1347 . allow jfo-i- i to range from Since yn (b) Identifying j/o n— 1 we obtain n equations one of the given boundary conditions. not an unknown.3 1.6 0. 0.5440 = 0.3492 identify P(r) The = Q(x) 0.0000 13.77 89 0.000 0000 We = 0. are the same because the derivation was based only on the differential equation. f(x) equation 0.8490 2 - (4 = l)/6 1.9 0601 1.3333 90. and h = 0)/8 tt + i-2.+ fl-^U_ =0.2 0.2222 83.500 1. is = to y(0).6190 100.0 2. y~i — y(0 — ft).y-i = (5) in 2ft. with y-i (l or + ~Po) = yi + ^Pbj + yi (-2 i = 0.0 0. not the boundary conditions. ('l + ^U f(r) = and h 0.8 11.3216 0.1363 0.4444 97. 0.Exercises 9. 2ft The difference equation corresponding to 1 becomes.0625»j + = -4.9386 0.7202 0. - Pq.0000 94.125. 2ft) = h?fo the text If j/i. 1 = ~ V-i] [yi and yi = if'(0) = 1 y(0 in the it is + h) we or j/i n+ 1 unknowns y-i.5 0. 0. - 0.8929 0. . 2462 2 .0735 2 .2 0.15 1.1197 2/i> - - - .00 1 . .0.30 1 1 . 2.5695 2 6944 2.8269 2. 2. RUNGE KUTTA 0000 0736 2 1556 2.4532 2.1 X (n) 1 .5672 2 .0000 .40 1.3454 2. = 2h to the yo> we obtain 0.4 O.3450 2 .4527 2 2 2 2 2.1157 0000 1556 2 3446 2 5680 2 8255 3 1167 2 .8589 -1.0 0.50 EULER 2 0000 2 2 2 2 2 2 2 2 2 3 0693 1469 2328 3272 4299 5409 6604 7883 9245 0690 IMPROVED 3 -TERM EULER TAYLOR 0000 1549 2.6 1.1197 .3275 -1.40 .30 1.35 1. 2 2 .5689 2. 2 IMPROVED 3 -TERM EULER TAYLOR .2459 2 2 2 2 2.0000 Chapter 9 Review Exercises h.20 1.05 x(n> 1.10 1 .3439 2 .1556 .45 1.00 1 . n obtaining (c) Using n = 5 + we may simply add 1 equations in the the equation 3/1— y-i n + 1 unknowns y_i .0 -2.25 1.9686 2 2 1187 3 3 . list of n difference equations I/n-l- .5695 2 8278 3.10 1 .B 0.0755 -1.20 1 .8278 2 9696 .6937 2.8246 3 .3454 2 . 3 .2755 -2.Chapter 9 Review Exercises Alternatively.05 1.1554 2 . 2 2 0000 0735 1555 2460 3451 4528 5690 6938 8271 9688 1188 RUNGE KUTTA 2 2 0000 . . 2 .50 EULER 0000 1386 3097 5136 7504 0201 2 2 2 2 2 3 h=0.6126 -1. 50 .2027 .6610 0.3087 0.00 EULER 0000 0500 1001 1505 2017 2537 3067 3610 4167 4739 5327 . .7194 .5512 5513 0.5376 RUNGE KUTTA 0.45 0. 0. 0.5000 5000 0. x(n) n uu nn u .1163 IMPROVED 3 -TERM EULER TAYLOR 0.70 .80 .0500 0. O— u J. h=0 . EULER 0000 1000 2010 3049 4135 5279 .1511 0.4202 0.15 0.75 .0000 .05 0.90 1 00 5000 6000 7095 8283 9559 0921 .30 0.4201 0.3088 0. .3638 0.40 50 . IMPROVED 3 -TERM TAYLOR EULER RUNGE KUTTA 5000 6050 7194 8429 9754 1 1166 .0452 1.50 60 .1169 RUNGE KUTTA 5000 0. .95 00 .20 0. h=0. 0000 0. 400 .2030 0.30 0. 5377 0. 60 65 .4202 4782 0.3637 0. 50 55 0.4201 0.2026 0.3087 0. 9757 1.3638 0.6049 0.5000 .25 . 9083 0.7194 0.4207 0.7191 0.3088 .9757 1.5512 0. 1 1 8430 9082 9756 0451 1168 0.40 .1512 . 8431 0.9082 0.9752 1.6048 . .4781 0. 1 1 6024 6573 7144 7739 8356 8996 9657 0340 1044 FTTT CTJ ft ft ft ft 3 -TERM TAYLOR n 0000 1000 2025 3087 4202 5377 0.1003 0. 5376 .20 0.6609 6610 7194 0.5000 .1168 0.35 .2552 0.70 80 .10 0.05 x(ni 0.6049 . 05 x(nl 0.9755 1.0500 .8427 0.8431 0. 1 EULER 5000 5500 .2551 Iff ft iqupa n ftft 1001 0.5382 IMPROVED 3 -TERM EULER TAYLOR 0000 0501 1004 . X fW EULER . 0451 1 .2551 .90 . 1ft ft 0.Chapter 9 Review Exercises 4.2027 0.4782 5378 . .7193 7801 0.3092 0.85 1 .7801 0. 1169 0. 1003 0.6049 6049 0. 1511 .7800 8430 0.2026 0. .00 .10 1 1. 8446 3 .1095 1.2183 1.1497 5.30 1.0000 1. EULER In) ILL- 1 00 1 10 1 . Yin.25 1.7689 7. 6197 3.23.5994 1.2405 1.20 1.2360 4 1528 . 6036 2 1909 1 IMPROVED 3 -TERM EULER TAYLOR EULER 1 .30 1 .1524 .4734 3 17E1 4. 1 0000 2000 4760 8710 4643 4165 .0783 .5910 2 h=0. + hu = y + 0.0000 2.Chapter 9 Review Exercises 6.1789 2 6182 Using = + hu n Vn = we obtain (when h = = 0.9988 1.20 0.45 1 7. 1. 1 1 2 1 .5300 1. 5 RUNGE KUTTA 2.05 1.1 we have xx = xa + 10 - yo + h(x - h{x Q + yQ = 1+0.35 1. L-< ! 0000 1.30 0.3) 3.40 1.2.8338 1 .1(1+2) ) y ) = 2 + 0.7567 3.2340 .2380 .2) Vl=IfO xlnl 0. .8515 .3 1.1 = 1 + 0.10 0.1(2xo + VZ = pi . .50 3 1 1 3 5 1. hj£ TMPROVRD 3 -TERM EULER TAYLOR . RTTWHR XViyi VJ Using xq + h(2x„ + = l)y„.1253 .2415 1.2510 1 KUTTA 0000 2350 5866 1453 1329 2208 1 1 1 1.4029 1 .6001 1 .2363 5.8523 1 2.7389 1. 3.40 1 . 15 2 .1911 2.2745 . + O.3296 4.2)1 = yo = 3 u = 1 = 3.4010 1.2401 1.1458 .6036 1 8586 2. 3 . yo initial condition Runge-Kutta Runge-Kutta Runge-Kutta predictor corrector = 2. 1.lwi - + + (0.3925 6.50 2 3 5 0000 1.3284 2. .1088 1.1 we have .20 1 .2755 4.1(1.6401 3.2415 1.1(1)3 = 0.9 1.1091 1.05 x(n) 0000 .6350 4 .40 9.3595 1.3 h = 0.1(1 When -2) = 1.1uo = 3 ui = u + 0. — 1. 0000 1000 . 5 1 Vn+\ 8. 2.1799 2.1 3 + (0.2) yi y(0.0000 1.00 0. and h = 0.4004 1.6404 .1)l l)yo 3. 8 0.55(1 + x).9 9.4083 0.3-1.3 0.84.1(1.0000 t.1 0. solution of the corresponding linear system gives 0.Chapter 9 Review Exercises and ' Thus.0655xj - 1.2) = Q(x) « ) 1. i(0. 6.9) = 1.3 + V2 = Vi + /i{x! . 0. 3/(0.7770 14.9 + and 0.1(1.1049 0. and h = (1 = 0. Then is K+i + [-2 + 0.001 or y i+ i The + (0.0 0.2 4.5 0. f(x) = 1.7038 14.9345)^ + yi_i = 0. .62 = P{x) difference equation x2 = xi + h(xi + yi) = 1.9) = 1-62 0.0655(1 + Xi)]vi + yi-i = 0.5396 0.0 0000 .19B7 8.yi - 1.4 11.9450 402 0.001.2) 10.7 12.84. - 0)/10 1.3640 13.3 + 0.2847 4. We identify « 1.6 1.1. Documents Similar To Differential-Equations-by-Zill-3rd-Edition-Solutions-Manual(ciitbooks.blogspot.com)_text.pdfSkip carouselcarousel previouscarousel nextEquações Diferenciais Vol. 2 3a Ed. - Dennis g. Zill e Michael r. 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