Engineering Mathematics II 15MAT21QUESTION PAPER SOLUTION MODULE - 1 DIFFERENTIAL EQUATIONS – I d4y d3y d2y dy 1) Solve 4 4 4 3 23 2 12 36 0 dx dx dx dx (July 2015) Sol : The given equation is written as 4D4 4 D 3 23D 2 12 D 36 0 A.E is 4m 4 4m3 23m 2 12m 36 0 solving we get m 2, 2, 3 / 2, 3 / 2 yc c1 c2 x e 2 x c3 c4 x e3/2 d3y d2y dy 2) Solve 3 6 2 11 6y ex 1 July 2015 dx dx dx Sol : The given equation is written as D 3 6D 2 11D 6 y 0 A.E is m3 6m 2 11m 6 0 m3 m 2 5m 2 5m 6m 6 0 m2 m 1 5m(m 1) 6(m 1) 0 m 2 5m 6 0, m 1 0 m 1, 2, 3 x 2x 3x yc c1e c2 e c3e ex 1 ex e0 x .1 yp D3 6 D 2 11D 6 D 3 6 D 2 11D 6 D3 6 D 2 11D 6 ex 1 24 6 x 2x 3x ex 1 GS is y c1e c2e c3e 24 6 Department of Mathematics, SJBIT Page 1 Engineering Mathematics II 15MAT21 d 2 y dy x 3) Solve 2 2 2 e sin x July 2015 dx dx Sol : The given equation is written as 2D 2 2D y e x sin x A.E is m2 2m 0 m(m 2) 0 m 0, 2 yc c1 c2e 2 x let PI as y e x ( A cos x b sin x) Differentiating w.r.t x y ' e x ( A sin x b cos x) e x ( A cos x b sin x ) y" e x A B cos x ( A B)sin x)} e x {( A B)sin x ( A B) cos x Sustituting thesevalues inthe given equations and solving we get 1 A 0, B 1/ 2 yp ex sin x 2 1 G S is y yc y p c1 c2e 2 x e x sin x 2 4) Solve : y 4y 12 y e2 x 3sin 2 x Jan 2016 Sol : We have D 2 4 D 12 y e 2 x 3 sin 2 x AE : m 2 4m 12 0 m 2, 6 yc c1e 2 x c2 e 6x e2 x 3sin 2 x yp 2 2 D 4 D 12 D 4 D 12 e2 x 3sin 2 x x D 4 4 D 4 e2 x 3 D 4 sin 2 x x 8 4 D 2 16 xe2 x 3 2 cos x 4sin 2 x 8 80 General solution 2x 6x xe2 x 3 2 cos x 4sin 2 x y yc yp y c1e c2e 8 80 Department of Mathematics, SJBIT Page 2 Engineering Mathematics II 15MAT21 5) By the method of undetermined coefficients solve y y 2 cos x (Jan 2016) we have D 2 1 y 2 cos x Sol: AE : m 2 1 0 m i yc c1 cos x c2 sin x Assume y p x A cos x B sin x yp x A sin x B cos x A cos x B sin x yp x A cos x B sin x 2 A sin x 2 B cos x substituting in given equation and comparing coefficients we get B 1 and A 0 y p x sin x G.S is given by y c1 cos x c2 sin x x sin x 6) By the method of variation of parameters solve y 4y tan 2 x (Jan 2016) D2 4 y tan 2 x Sol: AE : m 2 4 0 m 2i yc c1 cos 2 x c2 sin 2 x y A cos 2 x B sin 2 x cos 2 x sin 2 x W 2 2 sin 2 x 2 cos 2 x sin 2 x tan 2 x sin 2 x log sec 2 x tan 2 x A dx k1 2 4 4 cos 2 x tan 2 x cos 2 x B dx k2 2 4 G.S is given by sin 2 x log sec 2 x tan 2 x cos 2 x y k1 cos 2 x k 2 sin 2 x 4 4 4 Department of Mathematics, SJBIT Page 3 D 2 2 a a a a i i 2 2 2 2 The general solution is given by ax ax 2 a a 2 a a y e c1 cos x c2 sin x e c1 cos x c2 sin x 2 2 2 2 8) Solve ( D 4 m 4 ) y 0 ( Jan 2016) 2 2 Sol : A. Engineering Mathematics II 15MAT21 7) Solve ( D 4 m 4 ) y 0 ( Jan 2016) 2 2 solution : A. 4 3x 4x yc c1e c2e Department of Mathematics. SJBIT Page 4 .E D 2 m2 0 ( D 2 m2 )2 2 D 2 m2 0 D 2 m2 2 Dm D 2 m 2 2 Dm 0 2a 2ai 2a 2ai D . D 2 2 a a a a i i 2 2 2 2 The generalsolution is given by ax ax 2 a a 2 a a y=e c1cos x+c2sin x +e c1 cos x c2 sin x 2 2 2 2 ( D 2 7 D 12) y cosh x 9) Solve (Jan 2016) 2 ex e x Sol: We have ( D 7 D 12) y 2 AE : m 2 7 m 12 0 m 3.E D 2 m2 0 ( D 2 m2 )2 2 D 2 m2 0 D 2 m2 2 Dm D 2 m 2 2 Dm 0 2a 2ai 2a 2ai D . 2 2 1 x 3 x x y c1e c2 e 2 x c3e 2 c1e 2 Department of Mathematics. Engineering Mathematics II 15MAT21 1 ex e x yp 2 D 2 7 D 12 D2 7 D 12 1 ex e x 2 20 6 General solution y yc y p 3x 4x 1 ex e x y c1e c2e 2 20 6 y y x sin x 10) By the method of variation of parameters solve (Jan 2016) Sol : D 2 1 y x sin x AE : m 2 1 0 m i yc c1 cos x c2 sin x y A cos x B sin x cos x sin x W 1 sin x cos x x sin 2 x 1 x2 x sin x cos 2 x A dx k1 2 2 2 2 4 x cos 2 x sin 2 x B x sin x cos xdx k2 4 8 G.S is given by 1 x2 x sin x cos 2 x x cos 2 x sin 2 x y k1 cos x k2 sin x 2 2 2 4 4 8 11) Solve (4 D 4 8D3 7 D 2 11D 6) y 0 (June 2015) Solution: The AE is 4m 4 8m 3 7m 2 11m 6 0 By inspection method m=-1 and m=2 are two roots 1 3 then we get 4m 2 4m 3 0 m . SJBIT Page 5 . ..2 0 x 4 4 5 4 2 5 1 2 1 e x y c1 cos 2 x c2 sin 2 x x 4 2 5 13) Solve ( D 2 2 D 2) y e x tan x using method of var iation of parameters...e x tan x B dx dx sin xdx cos x k2 w e2 x Hence general slolution is y e x A cos x B sin x y e x c1 cos x c2 sin x e x cos x log sec x tan x Department of Mathematics. (June 2015) Solution: The AE is m 2 2m 2 0 m 1 i yc e x c1 cos x c2 sin x u e x cos x v e x sin x u v e x cos x e x sin x w x e2 x u v e sin x cos x e x sin x cos x vf ( x) e x sin x. SJBIT Page 6 . Engineering Mathematics II 15MAT21 12) Solve ( D 2 4) y x2 e x (June 2015) Solution: The AE is m 2 4 0 m 2i yc c1 cos 2 x c2 sin 2 x 1 1 yp 2 x2 2 e x D 4 D 4 2 1 1 D 2 e x 1 x 4 4 5 2 1 D2 D2 2 e x 1 ... x 4 4 4 5 1 2 1 e x 1 2 1 e x x .e x tan x sin 2 x A dx dx dx w e2 x cos x sec x cos x dx sin x log sec x tan x k1 uf ( x) e x cos x... I xe x 2sin x The solution is y = C.1 1 D x 2 xe 2sin x 2 P. +P. SJBIT Page 7 .F c1 c2 e c3e x 2e x 4 cos x 2e x 4 cos x P.I f ( D) D3 D 2e x 4 cos x D D D3 D 3 2e x 4 cos x 1 1 D D x 2 xe 4 cos x 2 3D 1 2D x 2 xe 2 cos x 3. A cos 2 x B sin 2 x.F x y = c1 c2e c3e x xe x 2sin x 15) Solve: ( D2 2 )y= x 2e3 x e x cos 2 x ( Jan 2015) 2 Solution: L( D) D 2.I .1 C. which leads to C. Also. x 2e3 x e x cos 2 x e x e cos 2 x L( D) L( D) L( D) Department of Mathematics.F c1e ox c2 e x c3e x x C.F.E is ( D 3 D) y 0 m3 m 0 m( m 3 1) 0 m 0. Engineering Mathematics II 15MAT21 3 14) Solve ( D D) y 2e x 4 cos x (Jan 2015) 3 Solution: Given ( D D) y 2e x 4cos x A.F. 1. m 2 1 0 m I1 m 0. 1 1 3x 2 1 x P. t 9 27 Using this in equation (ii). we get (D+5) (D+1)y +2(D+5)x=0. -3. the general solution of equation (iii) is 2 4 Y = C.. + P. x 2 ex cos 2 x 11 11 11 2D 1 1 3x 1 1 2D 1 e 1 D2 6D D 4 12 D 3 36 D 2 x2 ex cos 2 x 11 11 112 4D4 1 1 3x 2 1 1 2D 1 e x 2 12 x 72 ex cos 2 x 11 11 112 4( 22 ) 1 1 3 x 2 12 50 1 x e x x e ( 4sin 2 x cos 2 x) 11 11 121 17 Therefore. = (c1+ c2t) e-3t 1 1 1 P. Engineering Mathematics II 15MAT21 1 1 e3 x x 2 e3 x cos 2 x L( D 3) L( D 1) 1 1 e3 x 2 x2 ex cos 2 x ( D 3) 2 ( D 1) 2 2 1 1 e3 x 2 x2 ex 2 cos 2 x D 6 D 11 D 2D 3 1 1 1 e3 x 1 D2 6D 1 2 x ex 2 cos 2 x 11 11 2 2D 3 1 3x 1 1 2 1 e 1 D2 6D 2 D2 6 D . SJBIT Page 8 .F.. Therefore C. Using equation (i)..F. this becomes (D+5) (D+1)y +2(t+2y)=0. 2 ( 2t ) ( 2t ) ( D 3) 9 (1 D / 3) 2 2 2 D 2 2 2 2 1 (t ) 1 D . we get Department of Mathematics.+P. = (c1+ c2t)e-3t.=Acos 2 x B sin 2 x e x x e (4sin 2 x cos 2 x).... For this equation. the A. t t 9 3 9 3 9 3 Therefore..E.F... 11 11 121 17 16) Solve the simultaneous equation (D+5) x-2y=t and (D+1) y+2x=0 ( Jan2015) Solution: The given equations are (D+5) x-2y=t (D+1) y+2x=0 From equation (ii).I.. is (m+3)2 = 0 whose toots are -3.I.I . the general solution of the given equation is 1 3 x 2 12 50 1 x Y=C. +P.(D +1)y=-(D+1) c1 c2 e-3t . we get. 2 17) Solve ( D 2) y 8 e2 x sin 2 x (June 2014) 2 Solution: A. 4 1 1 0 c1 . 0 c1 c2 27 2 27 From these. Using this condition in expressions (iv) and (v).E is m 2 0 m 2.c2 c2t e 3t t 2 9 27 Expressions (iv) and (v) constitute the general solution of the given system. It is given that x=0 and y=0 when t = 0. SJBIT Page 9 .F y = c1 c2 x e 2 x 4 x 2e 2 x cos 2 x Department of Mathematics. c2 2 c1 2 27 27 27 9 Substituting these values of c1 and c2 into expressions (V) and (iV). we get 1 1 x (1 6t )e 3t (3t 1) 27 27 1 1 y (2 3t )e 3t (3t 2) 27 27 These constitute the required solution of the given system.F c1 c2 x e 2 x 8 e2 x sin 2 x e2 x sin 2 x e2 x sin 2 x P. we get 4 1 3 2 c1 .F. C. Engineering Mathematics II 15MAT21 2 4 2x = . 2. t 9 27 2 2 4 ( 3) c1 c2 e-3t c2 e 3t c1 c 2t e-3t t 9 9 27 2 2 ( 2) c1 c2 e-3t c2 e 3t t 9 27 3t 2 2 2c1 -c2 2ct e t 9 27 1 2 1 x c1 .I 8 2 8 2 2 f ( D) D 2 D 2 D 4D 4 xe 2 x sin 2 x x 2e2 x cos 2 x 8 8 2 D 2 4D 2 8 4 x 2 e 2 x cos 2 x The solution is y = C. ..... +P.E is m2 12m 37 0 m 6 i y e6t c1 cos t c2 sin t .. 2x 5 y 0 (June 2014) dt dt Sol : The given equations can be written as D 7 x y 0...(1) 2x D 5 y 0. D 5 D 7 2 y 0 or D 2 12 D 37 y 0 A.. SJBIT Page 10 ........I 2 x 2 2 D 2D 1 D 2D 1 D 2D 1 2 D 1 cos x 2 sin x x 2 x D 1 2D D 1 2 x sin x sin x x sin x ( D 1) sin x 2 D 1 2 D2 1 x sin x cos x sin x 1 x sin x cos x sin x 2 2 2 The solution is y = C..F 1 y = c1 c2 x e2 x x sin x cos x sin x 2 dx dy 19) Solve 7 x y 0.. Engineering Mathematics II 15MAT21 18) Solve: y 2y y x cos x (June 2014) Solution: A.. x 5y dt 2 dt 1 d 6t x e c1 cos t c2 sin t 5e 6t c1 cos t c2 sin t 2 dt Department of Mathematics....(3) dy 1 dy By considering 2x 5 y 0 we get.....E is m 2 2m 1 0 m 1..(2) Operating (2) by (D-7) and multiplying (1) by 2 we have 2 D 7 x 2y 0 2 D 7 x D 5 D 7 y 0 Adding we get .F c1 c2 x e x x cos x 2D 2 cos x P...... 1 C.F..... .................E is m2 4 0 x c1 cos 2t c2 sin 2t.(4) 2 (3) and (4) represents the general solution of given system of equations dx dy 20) Solve 2 y cos 2t ..(3) dx 1 dx By considering 2 y cos 2t we get.... Using this condition in expressions (3) and (4). we get sin 2t x cos 2t ....... 2 x sin 2t .. y cos 2t dt 2 dt 1 d y c1 cos 2t c2 sin 2t cos 2t 2 dt 1 2c1 sin 2t 2c2 cos 2t cos 2t 2 y c1 sin 2t c2 1/ 2 cos 2t....... y sin 2t 2 Department of Mathematics..(2) Operating (1) by D and multiplying (2) by 2 we have D 2 x 2 Dy 2sin 2t 4 x 2 Dy 2sin 2t Adding we get .... Engineering Mathematics II 15MAT21 1 6t e c1 sin t c2 cos t 6e6t c1 cos t c2 sin t 5e6t c1 cos t c2 sin t 2 1 x c1 c2 e6t cos t c2 c1 e6t sin t ....... from (3)1 c1 0 c1 1 from (4) 0 0 c2 1/ 2 c2 1/ 2 Substituting these values of c1 and c2 into expressions (3) and (4)..... SJBIT Page 11 .. we get..(4) (3) and (4) represents the general solution of given system of equations It is given that x=1 and y=0 when t = 0.... y 0 at t 0 (Dec2013) dt dt Sol: The given equations can be written as Dx 2 y cos 2t.. given that x 1.. D 2 4 x 0 A....(1) 2 x Dy sin 2t....... e / x 2 3x 3x A' .B(x) are to be found We have y1=e3x and y2=xe3x y2 x y1 ( x) A' . ( Dec 2013) Solution: Put logx=t or et = x d xy Dy.3 yc c1 c2 x e3 x y A Bx e3x be the complete solution of the given equation where A(x). B' e6 x e6 x 1 1 A' .E. x 2 y D( D 1) y where D dt D( D 1) y Dy y 2 cos 2 t D( D 1) D 1 2 cos 2 t D2 2D 1 y 2 cos 2 t Department of Mathematics. We have D 2 6 D 9 y x2 A.e3 x / x 2 e . SJBIT Page 12 . Dec 2013) e3 x Solution: . Engineering Mathematics II 15MAT21 e3 x 21) Using the method of variation of variation of parameters solve y 6y 9y . B' 2 x x Integrating we get 1 A log x k1 B k2 x Using the expression of A and B in in y A Bx e3x we have. x2 ( June 2014. 1 y log x k1 e3 x k2 xe3 x x y k1 k2 x e3 x e3 x log x 2 '' ' 2 22) Solve: x y xy y 2cos (log x) . m2 6m 9 0 m 3. B' W W xe3 x . Engineering Mathematics II 15MAT21 m 2 2m 1 0 m 1 yc (c1 c2t )et 2 cos 2 t yp D2 2D 1 1 cos 2t D2 2D 1 1 cos 2t 2 2 P1 P2 D 2D 1 D 2D 1 1 eox eox 1 P1 2 2 1 D 2 D 1 D 2 D 1 0 2(0) 1 1 cos 2t cos 2t cos 2t P2 2 D 2D 1 4 2D 1 2D 3 (2 D 3) cos 2t (2 D 3)(2 D 3) 2 D(cos 2t ) 3cos 2t (4 D 2 9) 4sin 2t 3cos 2t 25 4sin 2t 3cos 2t 25 Department of Mathematics. SJBIT Page 13 . Engineering Mathematics II 15MAT21 MODULE-2 DIFFERENTIAL EQUATIONS . y = 1 when t = 0 (Jan 2016) Sol: Given Dx 2 y sin t .......D D 1 D 2 2D 2 sin t sin t et 2 et 2 ( Dr 0) D 1 1 1 sin t et t cos t et t 2D 2 et t cos t GSis y et c1 cos t c2 sin t 2 x log x cos log x y x c1 cos(log x) c2 sin(log x) 2 Department of Mathematics. 2 x Dy cos t Solving the equations we get D 2 4y 3 sin t m2 4 0 m 2i yc c1 cos 2t c2 sin 2t 3 sin t yp sin t D2 4 y c1 cos 2t c2 sin 2t sin t dy Substituting in the given equation we get dt x c1 sin 2t c2 cos 2t cos t 2 2) Solve x y xy 2y x sin log x (Jan 2016) put x et t log x Sol: then we have D( D 1) D 2 y et sin t AE m 2 2m 2 0 m 1 i t yc e c1 cos t c2 sin t et sin t yp .... dt 2 x cos t 0 given that x = 0. SJBIT Page 14 .II dx dy 1) Solve the simultaneous equations dt 2 y sin t 0.... SJBIT Page 15 . Engineering Mathematics II 15MAT21 dy dx x y 3) Solve dx dy y x (Jan 2016.3 yc c1e 2t c2e3t a et yp 2 D 5D 6 et a 2 6 Department of Mathematics. June 2015) Sol: The given equation can be written as dy we note that p dx 1 x y p p y x x y p2 p 1 0 y x Solving for p we get x y p or p y x dy x now xdx ydy k dx y then x 2 y2 c 0 dy y dx dy Similarly k dx x x y then log x log y log c then xy c 0 G S is x 2 y 2 c xy c 0 2 x a y 4x ay 6y x (July 2015) 4) Solve put x et t log x Solution: put x a et t log x a then we have D( D 1) 4 D 6 y et a AE m2 5m 6 0 m 2. ..... SJBIT Page 16 ...(1) 1 p2 Differentiating both the sides dp dp 1 p2 p 2p dx 1 dp dy dy 2 dy 1 p 2 dy 1 p2 dp dp p2 1 1 dp dy dy 2 p 1 p 2 dy 1 p2 dp 1 1 p2 2 dy 1 p 2 1 p2 dp 2 2 dy 1 p2 2p dy dp....(2) t 1 p2 (1) and (2) arethe general solution Department of Mathematics...... Engineering Mathematics II 15MAT21 et a GS is y c1e2t c2e3t 2 6 2 3 x a a y c1 x a c2 x a 2 6 p 5) Solve p tan x 1 p2 (Jan 2016) Sol: The given equation can be written as p x tan 1 p .... put 1 p 2 t 2 2 1 p dy dt / t 2 1 1 y c y c.. Engineering Mathematics II 15MAT21 6) Find the general and singular solution of the equation y px p3 (Jan 2016) Solution: The given equation is Clairaut’s equation the general solution is obtained by replacing p by c y cx c 3 diff w.r.t c to get the singular solution x 0 x 3c 2 c i 3 x x x y ix i 3 3 3 7) Solve ( px y )( py x ) a 2 p by reducing toClairaut 's equati on. (June 2015) du Solution: Put u=x 2 du 2 xdx dx 2x dv Put v=y 2 dv 2 ydy dy 2y dy dv / 2 y x dv x dv p P where P dx du / 2 x y du y du substituting this in the given equation we get x x x Px y Py x a2 P y y y x2 P y 2 x Px x a2 P y y x( x 2 P y 2 )( P 1) Pa 2 x uP v ( P 1) Pa 2 Pa 2 v uP ......this is Clairaut ' s form P 1 TheGeneral Solution of this equation is ca 2 ca 2 v uc or y 2 cx 2 c 1 c 1 Department of Mathematics, SJBIT Page 17 Engineering Mathematics II 15MAT21 8) Solve (1 x) 2 y (1 x) y y 2sin log x (June 2015) Solution: Put 1 x e z z log(1 x) given equationbecomes D( D 1) y Dy y 2sin z D2 1 y 2sin z The AE is m2 1 0 m i yc c1 cos z c2 sin z c1 cos log(1 x) c2 sin log(1 x) 2sin z sin z yp 2 2z z cos z log(1 x) cos log(1 x) D 1 2D y c1 cos log(1 x) c2 sin log(1 x) log(1 x) cos log(1 x) 9) Solve y 2 px y 2 p3 by solving for x. (June 2015) y y 2 p3 Solution: x 2p differentiating w.r.t y we get dp dp p 1 y 2 .3 p 2 p 3 .2 y y y 2 p3 dx 1 dy dy dy 2 p2 1 1 dp dp dp or 2 p 3 y 2 p3 2 p4 y y y 2 p3 p 2p dy dy dy on cross multiplication dp 4 dp 2 p p 2 y 2 p3 2p y y dy dy dp or p 1 2 yp 3 y 1 2 yp 3 0 dy dp d p y 1 2 yp 3 0 or ( py ) 0 dy dy c on int egration p y Department of Mathematics, SJBIT Page 18 Engineering Mathematics II 15MAT21 substituting in given equation we get c c3 c c3 y 2. x y 2 . 3 y 2. x y y y y or y 2 2cx c3 which is the required solution d2y 2 dy 10) Solve x 2 4x 2 y ex (June 2015) dx dx Solution: Put x e z z log x given equation becomes z D( D 1) y 4 Dy 2 y 2sin z D 2 3D 2 y ee The AE is m 2 3m 2 0 m 1, 2 c1 c2 yc c1e z c2e 2 z x x2 z ee 1 1 z 1 z yp ee e z ee e z dz D 1 D 2 D 2 D 1 D 2 1 z ez 2z ez z 2z 2 z ez ex e e e e e e dz e e D 2 x2 c1 c2 ex y x x2 x2 dx dy 11) Solve 7x y 0, 2x 5 y 0 (June 2015) dt dt d Solution: Put D the given equations becomes dt Dx 7 x y 0 ; Dy 2 x 5 y 0 or ( D 7) x y 0; 2 x ( D 5) y 0 solving for y we get D 2 12 D 37 y 0 The AE is m 2 12m 37 0 m 6 i Department of Mathematics, SJBIT Page 19 .r.. (Jan 2015) Sol: The given equation is solvable for y only.t. dp dp 2p 4 x5 20 x 4 p 12 x 4 p 48 x3 y 0 dx dx dp p 2 4 x5 p (2 p 4 x5 ) 8 x3 ( xp ) 0 dx 2 x4 dp 2 p ( p 2 x5 ) ( p 2 x5 ) dx x dp 2p 0 dx x Integrating log p log x k p c2 x2 equation (1) becomes c 4 4c 2 x3 12 y Setting c 2 k the general solution becomes k 2 4kx3 12 y Differentiating w. obtain the singular solution also.. p ) 12 x 4 Differentiating (1) w.t k partially we get Department of Mathematics.. Engineering Mathematics II 15MAT21 y e6t c1 cos t c2 sin t dy substituting y in 2x 5 y 0 dt 1 dy 1 6t x 5y e c1 sin t c2 cos t 6e6t c1 cos t c2 sin t 5e6t c1 cos t c2 sin t 2 dt 2 e6t c1 c2 cos t c2 c1 sin t 2 2 5 4 12) Solve p 4 x p 12 x y 0 ...r. SJBIT Page 20 ... p 2 4 x5 p 12 x 4 y 0 .x..(1)_ 2 5 p 4x p y f ( x.. . taking the 13) Solve . by reducing into Clairaut's form.r. Engineering Mathematics II 15MAT21 2k 4 x 3 0 Using k 2 x3 in general solution we get x6 3 y 0 as the singular solution px y py x 2 p. dX Solution: Let X x2 2x dx dX Y x2 2y dy dy dy dY dX dY Now. . p.2 x or p P 2y y X p P Y Consider ( px y ) ( py x) 2p X X X P X Y P Y X 2 P Y Y Y ( PX Y )( P 1) 2P 2P Y PX P 1 Is in the Clairaut’s form and hence the associated general solution is 2c Y cX c 1 2c Thus the required general solution of the given equation is y 2 cx 2 c 1 3 2 14) Solve p 4 xyp 8 y 0 by solving for x. (Jan 2015) substitution X x2 . p and let P dx dY dX dx dx 1 x P . p 3 4 xyp 8 y 2 0 3 2 p 8y x f ( y. 4 px 16 y 0 dy dy p Department of Mathematics.t. y. SJBIT Page 21 . dp dp 1 3 p2 4 xy 4 yp. Y y2. (Jan 2015) Solution: The given equation is solvable for x only. p ) 4 yp Differentiating (1) w. SJBIT Page 22 . cy cy 4 xy cy 8 y 2 0 3 2 Dividingthroughout by y y y we have. Engineering Mathematics II 15MAT21 dp (3 p 2 4 xy ) 4 px 12 y dy 3 dp 2 p 8 y2 p3 8 y 2 3p 12 y dy p y dp 2 p 3 8 y 2 p3 4 y 2 dy p y 2 dp 3 2 p3 4 y 2 ( p 4y ) p dy y 2 dp 1 p dy y 2 log p log y log c U sin g P cy in (1) we have.. y x dx Department of Mathematics. p py x( x y) 0 y y2 4 x( x y) p 2 2 y 4x 4 xy y2 y (2 x y) p 2 2 2( y x) ie. dy x2 x y k dx 2 dy Also. c c 4x c 8 y 0 c (c 4 x ) 8 y Thus the general solution isc(c 4 x) 2 64 y 15) Solve p( p y ) x( x y ) (June 2014) 2 Sol: The given equation is. p x or p (y x) 2 We have. int egrating by parts. c we get Department of Mathematics. Q x.r.. e ex Hence ye x xe x dx c ie. (Dec 2013) Solution: The given equation can be written as xp 3 1 1 y 2 y px i s in theClairaut ' s form y px f ( p) p p2 whose general solution is y cx f (c) 1 Thus general solution is y cx 2 c Differnetiating partially w... ye x ( xe x e x ) c.e ( similar to the previous problem) dx P dx P 1...... y x..t y we get 2 1 y dp dp p y p p p 2 dy dy 1 y dp p 1 0 p p dy 1 dp Ignoring p which does not contain ..... SJBIT Page 23 .. is a linear d . Thus the general solution is given by (2 y x 2 c ) e x ( y x 1) c 0 16) Obtain the general solution and singular solution of the equation y 2 px p2 y . this gives p dy y dp dy dp 1 0 or 0 p dy y p Integrating we get yp c.(2) substituting for p from 2 in (1) y2 2cx c 2 3 17) Obtain the general solution and singular solution of the Clairaut’s equation xp yp 2 1 0 .(1) p Differentiating w. (June 2014) Solution: The given equation is solvable for x and it can be written as y 2x py.r. Engineering Mathematics II 15MAT21 dy ie.t... (Dec 2013) Solution: Dividing throughout by p2. the equation can be written as y2 2y cot x 1 adding cot 2 x to b. Engineering Mathematics II 15MAT21 1/3 2 2 0 x 3 c c x Thus general solution becomes 1/3 2/3 2 x y x 22/3 y 3x 2/3 x 2 or 4 y 3 27 x 2 18) Solve p 2 2 py cot x y 2 .s p2 p y2 2y cot x cot 2 x 1 cot 2 x p2 p 2 y or cot x cos ec 2 x p y cot x cos ecx p y cot x cos ecx dy / dx dy sin x dy sin x dx and y cos x 1 y cos x 1 Integrating these two equations we get y (cos x 1) c1 and y (cos x 1) c2 general solutionis y (cos x 1) c y (cos x 1) c 0 Department of Mathematics. SJBIT Page 24 . One end of the rod at x=0(Origin) and the other end of the rod at x=L. (Jan 2016) z Sol : g ( y ) yf '( x). July 2015) t x2 Sol: Consider a heat conducting homogeneous rod of length L placed along x-axis. Department of Mathematics. 2 u 2 u 2. Engineering Mathematics II 15MAT21 MODULE 3 PARTIAL DIFFERENTIAL EQUATION 1) Form the partial differential equation of Z = y f(x) + xg(y) where f and g are arbitrary functions. SJBIT Page 25 . x z xg '( y ) f ( x) y Substituting g'( y) and f '( x) 2 z z z xy x y [ xg ( y ) yf ( x)] x y x y 2 z z z xy x y z x y x y This is the required pde. (Jan 2016. Derive one dimensional heat equation as c . . the density of material.(1) x x x x x But the rate of increase of heat in the rod u s A x . From (1) & (2) Department of Mathematics. The rod is sufficiently thin so that the temperature is same at all points of any cross sectional area of the rod... The amount of heat crossing any section of the rod per second depends on the area A of the cross u section... Let u(x. t) be the temperature of the cross section at the point x at any time t.... Therefore q1 the quantity of heat flowing into the cross section at a distance x in unit time is u q1 kA per second x x Negative sign appears because heat flows in the direction of decreasing temperature (as x increases u decreases ) q2 the quantity of heat flowing out of the cross section at a distance x x (i... Engineering Mathematics II 15MAT21 Assume that the rod as constant density and uniform cross section A............ the rate of heat flow at cross section x x ) u q2 kA per second x x x The rate of change of heat content in segment of the rod between x and x+x must be equal to net heat flow into this segment of the rod is u u q1 q2 kA per second. the rate of change of temperature with respect to distance normal to the area.... Also assume that the rod is insulated laterally and therefore heat flows only in the x direction... SJBIT Page 26 .e.(2) t Where S is the specific heat.e.. the thermal conductivity k of the material of rod and the temperature gradient x i. . we get f f z (2 x) y 0 u v x f f z (2 y ) x 0 u v y f f Eliminating and from these equations... we get u v z 2x y x 0 z 2y x y z z or x x y y 0 y x z z or x y x2 y2 y x This is the required partial differential equation.... we have 2 2 u u k u s k 2 or t x t s x2 2 u u k or c2 .. Engineering Mathematics II 15MAT21 u u u s A x kA t x x x x x u u u x x x x x or s k t x Taking limit as x 0 .. Department of Mathematics..... From the function f(x2+ y2 ..... 3.t x and y.r. z-xy) = 0 form the partial differential equation .v) = 0 Differentiating this partially w......(3) where c2 t x2 s Is known as diffusivity constant. SJBIT Page 27 .... Equation (3) is the one dimensional heat equation which is second order homogenous and parabolic type.. (July 2015) Sol: Let u = x2+ y2 and v= z-xy so that the given relation is f(u.. i. Consider the motion of an element PQ of length s . the tensions T1 At P and T2 at Q are tangential to the curve. Let m be the mass per unit length of the string. The string is perfectly flexible and offers no resistance to bending Points on the string move only in the vertical direction. Since the string does not offer resistance to bending. We shall find the displacement y a function of the distance x and the time t. . The motion takes place entirely in the X Y plane .axis. Gravitational forces on the string are neglected. (July 2015) Sol: Consider a tightly stretched elastic string of length l stretched between two points O and A and displaced slightly from its equilibrium position OA. We shall obtain the equation of motion of string under the following assumptions.T1cos + T2cosβ=0 or T1cos= T2cosβ=T=constant….e.(1) Department of Mathematics. there is no motion in the horizontal direction.. some of the forces in the horizontal direction must be zero. Engineering Mathematics II 15MAT21 2 2 u u 2 c2 4) Derive one dimensional wave equation as t x2 .. Taking O as origin and OA as x axis and a perpendicular line through O as Y. SJBIT Page 28 . Since the is no motion in the horizontal direction. .. z The given PDF can be written as sin x sin y x y Department of Mathematics.. By Newton’s second law of motion . y T tan tan t2 m s 2 y T y y 2 t m s x x x x x ( s x to a first approximation and tan .. Engineering Mathematics II 15MAT21 Since gravitational force on the string is neglected . the only two forces acting on the string are the vertical components of tension . SJBIT Page 29 ..T1sin at P and T2sinβ at Q with up[ward direction takes as positive..(2) .... Mass of an element PQ is m s .........T1sin =m s 2 ...(3) where c 2 t m x2 t 2 x 2 m Which is the partial differential equation giving the transverse vibrations of the string ....... Equation (3) is the one dimensional wave equation which is second order homogenous and parabolic type. tan arethe slopes of the curve of the string at x and x x) y y 2 y T x x x x x 2 t m x Taking Limit as x 0 2 2 2 2 y T y y y T 2 or c2 . the equation of motion in the vertical direction is Resultant of forces = mass *acceleration 2 y T2sinβ ...... (Jan 2015) 2 Solution: Here we first find z by integration and apply the given conditions to determine the arbitrary functions occurring as constants of integration. 5) Solve zxy sin x sin y for which z y 2sin y when x 0 and z 0 when y is an odd multiple of . t 2 T sin T1 sin m s 2y gives 2 1 T2 cos T1 cos T t2 m s 2y or tan tan T t2 2 .. 2sin y when x 0.r.t y treating x as constant z cos x sin y dy f ( y ) dy g ( x) z cos x ( cos y ) F ( y ) g ( x). (2) becomes z cos x cos y cos y g ( x) U sin g the condition that z 0 if y (2n 1) in (3) we have 2 0 cos x cos(2n 1) cos x c(2n 1) g ( x) 2 2 But cos (2n 1) 0. The auxiliary equations are dx dy dz 2 2 2 x y z 2 xy 2 xz Taking the second and third terms we have. z each ratio in (1) is equal to Department of Mathematics.r. z sin y sin x dx f ( y) sin y cos x f ( y) y Integrating w. dy dz dy dz or 2 xy 2 xz y z Integrating we get. where F ( y ) f ( y ) dy.t x treating y as constant. log y = log z + log C1 Log (y/z) = logc1 Using multipliers x. y. and hence 0 0 0 g ( x) 2 Thus the solution of the PDE is given by z=cos x cosy + cosy 6) Solve: x 2 y2 z 2 P 2 xyq 2 xz (Jan 2015) Solution: The given equation is of the form Pp +Qq = R. SJBIT Page 30 . U sin g this in (1) y 2sin y ( sin y ).1 f ( y ) (cos 0 1) Hence F ( y ) f ( y ) dy sin y dy cos y With this. Engineering Mathematics II 15MAT21 Integrating w. Thus z cos x cos y F ( y ) g ( x) z Also by data. y / x 2 y2 z2) 0 x 7) Solve by the method of variables 3u x 2u y 0. SJBIT Page 31 . log y log ( x 2 y2 z 2 ) log c2 Thus a general solution of the PDE is given by ( y / z. giventhat u ( x. 0) 4e (Jan 2015) u u Solution: Given 3 2 0 x x Assume solution of (1) as U=XY where X=X(x). Engineering Mathematics II 15MAT21 xdx y dy z dz x dx y dy z dz x dx y dy z dz x3 xy 2 xz 2 2 xy 2 2 xz 2 x3 xy 2 xz 2 x x2 y2 z 2 Let us consider dy xdx y dy z dz 2 xy x( x 2 y 2 z 2 ) Integrating we get . Y Y ( y) u u 3 ( xy ) 2 ( xy ) 0 x x dX dY 3 dX 2 dY 3Y 2X 0 dx dy X dx Y dy 3 dX 3dX Let K kdx X dx X Kx 3log X kx c1 log X c1 3 kx c1 X e3 2 dY dY Kdy Let k Y dy Y 2 ky Kdy 2 c2 log Y c2 Y e 2 Substituting (2) & (3) in (1) x y K c1 c2 3 2 U e x Also u ( x1o) 4e Department of Mathematics. ..(1) x2 2 z s 2 f ( y 2 x) 2 g (2 y x).......... 4e Ae 4e Ae Comparing we get A 4 & K 3 x y 3 3 2 U 4e is required solution. Dec 2013) Solution: The given equation is of the form Pp +Qq = R..... The auxiliary equations are dx dy dz 2 2 2 x yz y zx z xy Equivalently we can write in the form...(3) y2 (1) 2 (2) 2r s 6 f ( y 2 x)......(2) x y 2 z t f ( y 2 x) 4 g (2 y x)......... dx dy dy dz dz dx x2 y2 z x y y2 z2 x y z z 2 x2 y z x Department of Mathematics.. 8) Form the partial differential equation by eliminating the arbitrary functions from z = f(y-2x)+g(2y-x) (June 2014) Solution: By data........ z = f(y-2x)+g(2y-x) z p 2 f ( y 2 x) g (2 y x) x z q f ( y 2 x) 2 g (2 y x) y 2 z r 4 f ( y 2 x) g (2 y x)...........(5) Nowdividing (4) by (5) we get 2r s 2 or 2r 5s 2t 0 2s t 2 2 2 z z z Thus 2 2 5 2 2 0 is the required PDE x x y y 9) Solve: x 2 yz P y2 zx q z2 xy (June 2014...e.... SJBIT Page 32 .(4) (2) 2 (3) 2 s t 3 f ( y 2 x).......... Engineering Mathematics II 15MAT21 2x kx k x 6 x 3 i. June2014) u u Solution: Given 4 3u x y Assume solution of (1) as u XY where X X ( x).. Y Y ( y) 4 ( XY ) ( XY ) 3 XY x y dX dY 4 dX 1 dY 4Y X 3 XY 3 dx dy X dx Y dy 4 dX 1 dY Let k. Engineering Mathematics II 15MAT21 dx dy dy dz dz dx i... y) 2e5 y dx y (Jan 2015. d x y d y z x y x y log log c1 c1 x y y z y z y z y z Similarly we get c2 z x Thus the general solution is x y y z .. 3 k X dx Y dy Separating var iables and int egrating we get kx log X c1 . giventhat u (0.... log Y 3 k y c2 4 kx c1 X e 4 and Y e3 k y c2 kx kx c1 c2 3 k y 3 k y Hence u XY e e 4 Ae 4 where A e c1 c2 put x 0 and u 2e5 y The general solution becomes 2e5 y Ae 3 k y A 2 and k 2 Particular solution is x 5y u 2e 2 Department of Mathematics.(1) x y y z z x Fromthe first and sec ond terms of (1) we have.... x y x y z y z x y z z x x y z dx dy dy dz dz dx or . 0 y z z x u u 11) Solve by the method of variables 4 3u.e.. SJBIT Page 33 . Engineering Mathematics II 15MAT21 2 z z 11) 12) Solve 2 z 0 giventhat when x 0. z e y and 1 (Dec 2013) x x Solution: Let us suppose that z is a function of x only..r.t x z f ( y ) sin x g ( y ) cos x 1 f ( y ) sin 0 g ( y ) cos 0 x g ( y) 1 equation (1) becomes z e y cos x sin x Department of Mathematics.E.. m2 1 0 m i The solution of theODE z c1 cos x c2 sin x z f ( y ) cos x g ( y ) sin x. the PDE assumes the form of ODE d 2z d 2 z 0 D 2 1 z 0 where D dx dx A.(1)by replacing c1 & c2 by functions of y Now put x 0 and z e y in (1) ey f ( y ) cos 0 g ( y ) sin 0 e y f ( y) z Again put x 0 and 1 x Differentiating (1) partially w. SJBIT Page 34 .. (July 2015) 0 x2 12 x Sol : Let I xy dy dx 0 x 2 To identify the region of integration R let us find the point of intersection of the curves y x 2 and y 2 x x 1. SJBIT Page 35 . Evaluate xy dy dx changing the order of integration. Evaluate x2 y2 z 2 dxdydz (Jan 2016) c b a c b a 2 2 z3 Solution : I x z y z dydx c b 3 a c b 2a 3 2ax 2 2ay 2 dydx c b 3 c b 2 3 2a 3 2ax y 2a y / 3 y dx c 3 b Department of Mathematics. x 2 and y 1. Engineering Mathematics II 15MAT21 MODULE-4 INTEGRAL CALCULUS 12 x 1. y 4 On changing the order we must have constant limits for y and variable limits for to cover thesame region 1 y 2 2 y I x y dx dy x y dx dy y 0x 0 y 1x 0 1 y 2 2 y x2 x2 y dy y dy y 0 2 x 0 y 1 2 x 0 1 2 2 y 1 dy y (2 y) 2 dy y 0 2 2y1 1 2 y3 1 4 y3 y4 2 y2 6 0 2 3 4 1 3 I 8 c b a 2. July2015 when y is an odd multiple of 2 z Sol : Sinx sin y.(3) again by data 0 cos x cos(2n 1) cos(2n 1) g ( x) 2 2 but cos(2n 1) 0..t y treating x as constant z .... for which 2sin y when x 0 and z 0 x y y Jan 2016.cos x( cos y ) F ( y ) g ( x) whereF ( y ) f ( y )dy z cos x cos y F ( y ) g ( x)..... x y Integrating w.(2) by given data (1) becomes -2siny -siny.(1) y integrating w.r. Engineering Mathematics II 15MAT21 c 2 3 4a 3b 4abx 4a b / 3 dx c 3 c 3 3 4a 3b 4ab x / 3 4a b / 3 x x 3 c 3 3 3 8abc 8ab c 8a bc 3 3 3 8abc 2 2 2 a b c 3 2 z z 3) Solve Sinx sin y.t x treating y as constant z sin y sin xdx f ( y ) y z sin y cos x f ( y )...1 f(y) f(y) -siny hence F(y) -sinydy cosy (2) z cos x cos y cos y g ( x). hence g ( x) 0 2 solution of given PDE is z cos y (cos x 1) Department of Mathematics....... SJBIT Page 36 .....r... where R is the region bounded by x. (Jan 2016) Sol: x2=4ay is a parabola symmetrical about the y-axis. SJBIT Page 37 . The point of intersection of this curve with x=2a is to be found. Evaluate R . Hence 4a2=4ay or y=a The point of intersection is (2a. (July 2015) 10 x z 1 z x z Sol : Let I x y z dy dx dz 10 x z 1 z x z y2 I xy zy dx dz 10 2 x z 1 z 4 xz 2 z 2 dx dz 10 1 z z (2 x 2 ) 2 z 2 (x) dz 0 1 1 4 z 3 dz 1 1 z4 1 0 xy dx dy 5. the ordinate x=2a and the parabola x2=4ay. Evaluate x y z dy dx dz .a ) Department of Mathematics.axis. Engineering Mathematics II 15MAT21 1 z x z 4. SJBIT Page 38 . Engineering Mathematics II 15MAT21 I xy dx dy R x2 2a 4a xy dy dx x 0 y 0 x2 2a 2 y 4a x dx x 0 2 0 2a 1 x5 dx 2x 0 16a 2 2a 1 x6 32a 2 6 0 4 a I 3 (x 2 y 2 ) 6. (Jan 2016) 0 0 Department of Mathematics. Evaluate e dx dy by changing into polar coordinates. 0) and (4a. Engineering Mathematics II 15MAT21 7) Find the area between the parabolas y2=4ax and x2 = 4ay (July 2015) Sol: We have y2=4ax ………………… (1) and x2 = 4ay…………………(2). Solving (1) and (2) we get the point of intersections (0. 4a Therefore the required area is 4a 4 ay 4a 4 ay dy dx dy x y2 0 y2 0 4a 4a Department of Mathematics. y=0 to A. to Q 4ay and then y varies from O. y=4a . The shaded portion in the figure is the required area divide the arc into horizontal strips of width y y2 x varies from p. SJBIT Page 39 .4a) . 1 / 2) but 1 1 1 2 (1/ 2. 0 4a 3 4a 3 2 0 3 4 a 1 3 4a 2 4a 3 12a 32 2 16 2 16 2 a a a 3 3 3 8) DefineGamma function and Beta function. (m. (n 0) is called Gamma function. Engineering Mathematics II 15MAT21 4a 3 4a y2 y 2 1 y3 4ay dy 4a . 0 1 (m.1/ 2) 2 sin cos d 2 1d 0 0 2 1/ 2 1/ 2 Department of Mathematics. n 0) is called Beta function. n) . 0 m n We have (m.1/ 2) 1/ 2 /2 nowconsider (m. SJBIT Page 40 . put m n 1/ 2 m n 1/ 2 1/ 2 (1/ 2. n) x m 1 (1 x) n 1 dx. Prove that 1/ 2 (Jan 2016) Sol : n e x x n 1dx. n) 2 sin 2 m 1 cos 2 n 1 d 0 /2 /2 0 0 (1/ 2. ....(1) 0 x2 ( n) 2 e x 2 n 1dx.. n) m n 16 2 10) Show that the area between the parabolas y 2 4ax.(5) 0 0 evaluating RHS by changing into polars put x r cos ..... (n) 4 e x 2 n 1 y 2 m 1dxdy...( 4) 0 x2 y 2 (m). Jan 2016) 3 Sol: Solving y 2 4ax and x 2 4ay we get 2 y :0 4a and x : y / 4a 4ay 4a 4 ay 4a 4 ay required area dy dx dy x y 2 /4 a 0 y 2 /4 a 0 4a 4ay y 2 / 4a dy 0 Department of Mathematics.... n) 2 sin 2 m 1 cos 2 n 1 d . (n) 4 e r cos r sin rdrd 0 0 /2 r2 2m 1 4 e r 2m 2n 1 cos 2 n 1 sin drd 0 0 /2 2 2 e r r2 m n 1 dr 2 sin 2 m 1 cos 2 n 1 d 0 0 (m n). y r sin 2 2 2 x y r ... SJBIT Page 41 ..... Engineering Mathematics II 15MAT21 m n 9) Prove that (m.. n) by using (1) and (4) m n Thus (m............... and x 2 4ay is a (July 2015.. r :0 and :0 /2 /2 r2 2n 1 2m 1 (5) (m)... dxdy rdrd .(2) 0 y2 ( m) 2 e y 2 m 1dx. (m.......(3) 0 2 ( m n) 2 e r r2 m n 1 dx. n) (Jan 2016) m n Solution :We have by the definition of Beta and the Gamma function /2 (m...... SJBIT Page 42 . required volume is a a2 x2 a2 x2 v dzdydx x ay a2 x2 z a2 x2 a2 x2 a a2 x2 z dydx a a 2 x 2 a 2 x 2 a a2 x2 2 a2 x 2 dydx a a2 x2 a2 x2 a a2 x 2 dy dx 2 a2 x2 a a a2 x2 2 a2 x2 y dx a a2 x2 a 2 a2 x2 2 a2 x 2 dx a Department of Mathematics. Engineering Mathematics II 15MAT21 4a y 3/2 1 y3 4a 3 4a 3 0 4 a 3/2 1 3 4a 4a 3 12a 32 2 16 2 a a 3 3 16 2 a 3 11) Find the volume common to the cylinders x2+y2=a2 and x2+z2=a2 (Jan 2016) Sol: In the given region z varies from a2 x 2 to a2 x 2 and y varies from a2 x2 to a2 x 2 . y=0 x varies from –a to a Therefore.for z=0. Engineering Mathematics II 15MAT21 a a 2 2 2 x3 4 a x dx 4 a x a 3 a 3 3 a a 4 a3 a3 3 3 2a 3 16a 3 4 2a 3 3 4 1 dx 12) Evaluate using beta and gamma function (June 2015) 0 1 x4 1 Solution: Put x 2 sin x sin dx cos d 2 sin /2 /2 1 cos 1 2 I d sin d 02 sin 1 sin 2 0 2 1 1 4 2 2 3 4 Department of Mathematics. SJBIT Page 43 . (Jan 2016) s 1 s 2 4s 5 Sol. Find L e sin 3t et t cos t . (Jan 2016) 3 Sol. L sin 3t 2 s 9 2t 3 L e sin 3t 2 s 4 s 12 s L cos t s2 1 d s L t cos t 1 2 ds s 1 s2 1 s (2 s ) 2 s2 1 1 s2 s2 1 s2 1 s2 1 2 t s 1 1 L e t cos t 2 s 1 1 2 s 2s 2 s 2s 2 2t 3 s 2 2s L e sin 3t e t t cos t s2 4s 12 s 2 2s 2 4s 5 2. Engineering Mathematics II 15MAT21 MODULE-5 LAPLACE TRANSFORMS 2t 1.L 1 2 s 1 s 2 4s 5 A B C 2 2 s 1 s 2 s 1 s 1 s 2 Department of Mathematics. SJBIT Page 44 . Find the inverse Laplace transform of 2 . A 3. 24 s 2 6s 9 L y (t) 3 s 3 24 L y (t) 5 s 3 24 y (t) L1 5 s 3 3t 1 t4 y (t) 24 e L1 5 24 e 3t s 4! 3t y (t) e t4 cos t .t 3e 2t 3. B 1. Engineering Mathematics II 15MAT21 2 4s 5 A s 1 s 2 B s 2 C s 1 on solving . C 3 4s 5 1 1 1 L1 2 3L 1 L1 2 3L 1 s 1 s 2 s 1 s 1 s 2 t 1 3e e tL1 3e 2t s2 t 3e e t . t 2 transform . t 2 in terms of unit step function and hence find its Laplace sin t . Solve y '' 6 y ' 9 y 12t 2e 3t by Laplace transform method with y(0) = 0 =y1(0). 0 t 4. Exp ress f (t ) 1 . (Jan 2016) Sol. y ''(t) 6 y '(t) 9 y (t) 12t 2e 3t L y ''(t) 6 L y '(t) 9 L y (t) 12 L e 3t t 2 12 2! s 2 L y (t) sy (0) y '(0) 6 sL y (t) y (0) 9 L y (t) 3 s 3 using given initial conditions we obtain. SJBIT Page 45 . (Jan 2016) Department of Mathematics. Solve y '' 6 y ' 9 y 12t 2e 3t by Laplace transform method with y(0) = 0 =y1(0).. y ''(t) 6 y '(t) 9 y (t) 12t 2e 3t L y ''(t) 6 L y '(t) 9 L y (t) 12 L e 3t t 2 12 2! s 2 L y (t) sy (0) y '(0) 6 sL y (t) y (0) 9 L y (t) 3 s 3 using given initial conditions we obtain.(1) Let F (t ) 1 cos t . SJBIT Page 46 .. Engineering Mathematics II 15MAT21 sol : f (t ) cos t (1 cos t )u (t ) (sin t 1)u (t 2 ) L{ f (t )} L {cos}t L{(1 cos t )u (t )} L{(sin t 1)u (t 2 )}. G (t 2 ) sin t 1 F (t ) 1 cos(t ) . G (t ) sin t 1 1 s 1 1 F ( s) 2 . (Jan 2016) Sol. G (t ) sin(t 2 ) 1 F (t ) 1 cos t ... 24 s 2 6s 9 L y (t) 3 s 3 24 L y (t) 5 s 3 24 y (t) L1 5 s 3 3t 1 t4 y (t) 24 e L1 5 24 e 3t s 4! 3t y (t) e t4 Department of Mathematics. G ( s) 2 s s 1 s 1 s s L[ F (t )u (t )] e F ( s ) 2 s L[G (t 2 )u (t 2 )] e G(s) s 1 s L[(1 cos t ) u (t )] e 2 s s 1 2 s 1 1 L[(sint 1)u (t 2 )] e 2 s 1 s equ (1) becomes s s 1 s 2 s 1 1 L{ f (t )} 2 e 2 e 2 s 1 s s 1 s 1 s 5. SJBIT Page 47 . Find L (Jan 2016) t Sol : Let f (t ) cosat cosbt s s f ( s) 2 2 s a s b2 2 f (t ) s s hence L ds t s s a2 2 s 2 b2 cosat cosbt 1 L log( s 2 a 2 ) log( s 2 b 2 ) t 2 s 1 log( s 2 a 2 ) 2 log( s 2 b 2 ) s 2 1 1 a 2 (s 2 a2 ) lim log s log 2 2 s 1 b 2 (s 2 b2 ) s 1 (s 2 a2 ) log1 log 2 (s 2 b2 ) 1 (s 2 b2 ) log 2 (s 2 a2 ) cosat cosbt (s 2 b2 ) L log . Find the Laplace transform of te sin 3t and (July 2015) t 3 4t 3 3 Sol :(i) L(sin 3t ) 2 L(te sin 3t ) 2 2 s 9 s 4 9 s 8s 25 4t d 3 3(2s 8) Hence L(te sin 3t ) 2 2 ds s 8s 25 s 2 8s 25 4t 6( s 4) thus L(te sin 3t ) 2 s 2 8s 25 (ii ) Let f (t ) e at e at 1 1 L{ f (t )} L(eat ) L(e at ) F ( s) s a s a Department of Mathematics. Engineering Mathematics II 15MAT21 cos at cos bt 6. t (s 2 a2 ) 4t eat e at 7. (1) Let F1 (t 2) 4t t 2 2 F1 (t) 4(t 2) t 2 4t 8 t 2 4t 4 t2 4 4 2 L{F1 (t )} s s3 4 2 L 4t t 2 u (t 2) e 2s ..... Engineering Mathematics II 15MAT21 eat e at 1 1 L ds log(s a) log(s a) s t s s a s a s a s a lim log log s s a s a a 1 s s a lim log log s a s a 1 s s a log s a 8.. Express f(t) in terms of unit step function and find its Laplace transform given that t2 .... 2 t 4 (July 2015) 8 . 3 e 2s SJBIT 3 4e 4s Page 48 s s s s2 s ..0 t 2 f (t ) t 4t .t 4 Sol : f (t ) f1 (t) f 2 (t) f1 (t) u (t a) f 3 (t) f 2 (t) u (t b) t2 4t t 2 u (t 2) 8 4t u (t 4) L{ f (t )} L{t 2 } L 4t t 2 u (t 2) L 8 4t u (t 4) ...(2) s s3 Let F2 (t 4) 8 4t F2 (t) 8 4 t 4 4t 8 4s 4 8 L 8 4t u (t 4) e s2 s 4 8 L{F2 (t )} 4e 4s 1 2 .(3) s2 s s2 s substituting (3) and (2)in (1) 2! 4 2 1 2 Department L{ f (t )} of Mathematics...... Find L 1 using convolution theorem. A periodic function f(t) with period 2 is defined by f (t ) find L{f(t)} 2 t . Engineering Mathematics II 15MAT21 1 9.1 t 2 (July 2015) T 1 st Sol :We haveT 2 and L f (t ) sT e f (t )dt 1 e 0 2 1 st L f (t ) sT e f (t )dt 1 e 0 1 2 1 st 2s te dt 2 t e st dt 1 e 0 1 Department of Mathematics. SJBIT Page 49 .0 t 1 10.sin 3udu 0 0 3 t t 1 t u 1 t eu (sin 3u 3cos 3u ) e e sin 3udu e 3 0 3 5 0 1 t t e e (sin 3t 3cos 3t ) 3 15 1 sin 3t 3cos 3t 3e t 15 t . (July 2015) s 1 s2 9 1 1 Sol : Let F ( s ) and G ( s ) 2 s 1 s 9 1 f (t ) L 1 F ( s ) L1 e t s 1 1 1 g (t ) L 1 G( s) L1 2 sin 3t s 9 3 1 L1 L 1 F ( s ) G( s ) f (t ) * g (t ) s 1 s2 9 t t t u 1 f (t u )g (u ) du e . (1) 2 2 t .. SJBIT Page 50 ... Engineering Mathematics II 15MAT21 st 1 2 1 e e st e st e st 2s t. Find L 1 2 log 2 1 (July 2015) 3s 4 s 8 s 5s 2 1 5s 2 1 Sol : L 1 2 log 2 1 L1 2 L 1 log 1 .. 3s 4s 8 2 2 2 4 8 2 20 consider 3s 4s 8 3 s s 3 s 3 3 3 9 2 16 5s 2 5 s 3 3 2 16 5 s 5s 2 1 1 3 3 L1 2 L 2 3s 4 s 8 3 2 20 s 3 9 2t 3 e s 16 1 1 5L 1 2 L 2 3 20 3 20 2 2 s s 3 3 2t 3 e 20 16 20 5cos t sin t .. ( 1) 2 1 e s s 0 s s 1 1 s 2s 2 2s 2e 1 e s 1 e 2 s 1 e s2 1 e s 1 e s s 1 e e as /2 e as /2 L f (t ) s2 1 e s s 2 e as /2 e as /2 2sinh(as / 2) 1 L f (t ) 2 tanh(as / 2) s ..(1) 3s 4s 8 s 3s 4s 8 s2 5s 2 L1 2 .(2) 3 3 20 3 Department of Mathematics.2 cosh(as / 2) s2 5s 2 1 11. y '(0) 2 dt 2 dt (July 2015) t Sol : y ''(t) 2 y '(t) y (t) te L y ''(t) 2 L y '(t) L y (t) L e tt 1 s 2 L y (t) sy (0) y '(0) 2 sL y (t) y (0) L y (t) 2 s 1 using given initial conditions we obtain....(3) t 2t 3 5s 2 1 e 20 16 20 2(1 cosh t ) (1) L1 2 log 2 1 5cos t sin t 3s 4s 8 s 3 3 20 3 t d2y dy 12. SJBIT Page 51 . 1 s 2 2s 1 L y (t) s 2 2 2 s 1 2 1 s 1 L y (t) s 2 s 1 s 1 L y (t) 2 4 s 1 s 1 s 1 1 1 y (t) 2 2 4 s 1 s 1 s 1 Department of Mathematics..... Solve using Laplace transform method 2 y te t with y (0) 1. Engineering Mathematics II 15MAT21 1 s2 f ( s) log log(1 s 2 ) 2 log s s2 1 s f '( s ) 2 2 s s 1 1 s L1 f '( s ) 2L 1 2 s s 1 2(1 cosh t ) f (t ) . ds 3 2s 2s 1 ( s 2 1) 2s 2 ( s 2 9) 2s 2 L t (sin 3 t cos3 t ) 3. 2 ( s 1) 2 2 ( s 2 9) 2 4 ( s 2 1) 2 ( s 2 9) 2 14) Express f(t) in terms of unit step function and hence find the Laplace transform t2 0 t 2 given that f (t ) 4t 2 t 4 (Jan 2015) 8 t 4 Solution: f (t ) t 2 (4t t 2 )u (t 2) (8 4t ) (t 4) L f (t ) L t2 L (4t t 2 )u (t 2) L (8 4t ) (t 4) 2 e 2 s L 4(t 2) (t 2) 2 e 4 s L 8 4 (t 4) s3 2 e 2 s L t 2 4 e 4 s L 4t 8 s3 2 4 2 4 8 3 e 2s 3 e 4s 2 s s s s s Department of Mathematics. Engineering Mathematics II 15MAT21 t 1 1 1 e L1 L1 L1 s s2 s4 t t3 y (t) e 1 t 6 13) Find L t (sin 3 t cos3 t ) (Jan 2015) 1 1 sin 3 t cos3 t (3sin t sin 3t ) (3cos t cos3t ) Solution: 4 4 1 3 3 1 3s s L(sin 3 t cos3 t ) 2 2 2 2 4 s 1 s 9 4 s 1 s 9 d U sin g the property : L tf (t ) f ( s) we have. SJBIT Page 52 . 4 ( s 1)2 2 ( s 9) 2 2 4 ( s 2 1)2 ( s 2 9) 2 Thus 3 3 3s 1 1 1 (1 s 2 ) 9 s2 L t (sin t cos t ) 3. 2 s 2 1 2 4 ds 2 s2 1 ds s2 1 3 2 d 2s 2 2s s 2 1 (4 s 2) 2 s 2 2 s .3 s 2 1 ds s 2 1 3 s2 1 6 4 s 3 8s 2 8s 2 4 s2 1 Putting s = 1 in this result. we get 1 e t t 3 sin tdt 0 8 This is the result as required. t 2 Solution: Here T = 2 . Therefore 2 1 st L f (t ) e f (t )dt 1 e2 s 0 2 1 st = te dt ( t )e st dt 1 e2 s 0 2 st st st st 1 e e e e 1 e2 s t t 1 s s2 0 s s2 Department of Mathematics. 0 t 16) Find Laplace transform of a periodic function f (t ) (Jan 2015) t. SJBIT Page 53 . t.2 2 s. Engineering Mathematics II 15MAT21 3 t 15) Find the value of t e sin tdt using Laplace transforms (Jan 2015) 0 Solution: We have 3 d3 d3 1 e st t 3 sin tdt L(t 3 sin t ) 1 L sin t 0 ds 3 3 2 ds s 1 2 d2 2s d s 2 1 . e. by u sin g definition a st a 1 e 1 s(a ) as e e s a s s as 1 e L (t a ) e s But L (t a ) L lim (t a) lim L (t a) 0 0 s as 1 e as e lim e ( By applying L ' Hospital ' s rule) 0 s t. a t 2a s 2 (June 2014) 2a 1 st Solution: Here T = 2 a. showthat L f (t ) tanh 2a t .. SJBIT Page 54 . where f (t 2a) f (t ). 0 t a E as 18) If f (t ) . Engineering Mathematics II 15MAT21 s 1 e e s 1 e s e 2 s e s 1 e2 s s s2 s2 s s2 s2 s 2 s 1 2 s s 1 2e e e e 1 e2 s s s2 as 17) Prove that L (t a ) e (June 2014) Solution: We shall first find the Laplace transform of (t a) st L (t a ) e (t a )dt 0 a a st st st e (t a)dt e (t a )dt e (t a )dt 0 a a a st 1 i. L (t a) e dt . Therefore L f (t ) e f (t )dt 1 e 2 as 0 a 2a 1 st = te dt (2a t )e st dt 1 e 2as 0 a Department of Mathematics. 3 1 4 tan (4 / s ) s s 4 4s Hence L 1 f (s) L1 4 s 4 4s tf (t ) L1 4 s 4 Department of Mathematics. SJBIT Page 55 . 0 t 1 19) Express f (t ) t.& Dr by e as /2 ) s 2 1 e as 2 s e as /2 e as /2 2sinh(as / 2) 1 L f (t ) 2 tanh(as / 2) s . t 2 Laplacetransform (June 2014) 2 Solution: f (t ) 1 (t 1)u (t 1) (t t )u (t 2) L f (t ) L 1 L (t 1)u (t 1) L (t 2 t )u (t 2) 1 e s L t e 2 s L t 2 3t 2 s 1 1 2 3 2 e s 2 e 2s 3 s s s s2 s 1 2 20) Find the inverse Laplace transform of tan (2 / s ) (June 2014) Solution: Let f ( s) tan 1 (2 / s 2 ) 1 4 4s f ( s) 4 . Engineering Mathematics II 15MAT21 a 2a st st st st 1 e e e e 1 e 2 as t 2a t 1 s s2 0 s s2 a as 1 ae e as 1 e 2 as ae as e as 1 e 2 as s s2 s2 s2 s s2 1 as 2 as (1 e as ) 2 2 2 as 1 2e e s 1 e s 2 1 e as 1 e as (1 e as ) (eas /2 e as /2 ) ( multiplied Nr. 1 t 2 int erms of unit step function and hence find its t2.2 cosh(as / 2) s2 1. H . Engineering Mathematics II 15MAT21 Now s 4 4 ( s 2 2) 2 4s 2 ( s 2 2 2s) ( s 2 2 2s) Also 4s ( s 2 2 2s ) ( s 2 2 2s ) 4s ( s 2 2 2s) ( s 2 2 2s) Hence s4 4 ( s 2 2 2s) ( s 2 2 2s) 1 1 2 2 s 2 2s s 2 2s 4s 1 1 L1 4 L1 2 L1 2 s 4 s 2s 2 s 2s 2 U sin g (1) in L.S we have. 1 1 tf (t ) L 1 L1 ( s 1) 2 1 ( s 1) 2 1 1 1 tf (t ) et L 1 2 e tL 1 2 s 1 s 1 tf (t ) et sin t e t sin t sin t (et e t) tf (t ) sont. SJBIT Page 56 . ( s 1)( s 2 4) 1 s F(s) = c1 ( s) 2 then s-1 (s 4) f (t ) L 1 F ( s ) e t and g (t ) L 1 G ( s ) cos 2t S L1 L 1 F (s) G(s) f (t )* g (t ) ( s 1)( s 2 4) t t f (t u ) g (u )du et u cos 2u du 0 0 t et e u cos 2u du 0 Department of Mathematics.2sinh t 2sin t sin h t f (t ) t S 21) Find L 1 using convolution theorem (June 2014) ( s 1)(s 2 4) Solution: S L1 using convolution theorem. 1 L y (t ) s 2 4s 4 s 1 1 y (t ) L 1 ( s 1)( s 2) 2 1 A B C Let 2 ( s 1)( s 2) 2 s 1 s 2 s 2 Multiplying with ( s 1) ( s 2) 2 we obtain 1 A ( s 2) 2 B ( s 1) ( s 2) C ( s 1) Putting s 1 we get A 1 Putting s 2 we get C 1 Putting s 0 we have 1 1(4) B(2) 1(1) B 1 1 1 1 1 Hence ( s 1)( s 2) 2 s 1 s 2 s 22 Department of Mathematics. 1 L y (t ) 4 L y (t ) 4 L y (t ) L e 1 s 2 L y (t ) sy (0) y (0) 4 sL y (t ) y (0)) 4 L y (t ) s 1 using the given initial conditions we obtain. SJBIT Page 57 . y (0) 0. Engineering Mathematics II 15MAT21 u t t e e ( 1) cos 2u 2sin 2u 12 22 0 t e e t (2sin 2t cos 2t ) 1 5 1 (2sin 2t cos 2t e t ) 5 22) Solve the following initial value problem by using Laplace transforms: d2y dy 2 4 4 y e t . y (0) 0 dt dt (Dec 2013) t Solution: The given equation is y (t ) 4 y (t ) 4 y (t ) e Taking Laplace transform on both sides we have. C 2. g (t ) L 1 g (s) cos bt Now applying convolution theorem we have.cos(bt bu )du s2 a2 s2 a2 u 0 Department of Mathematics. B 1 5s 3 1 3 ( s 1) s 1 s 2 2s 5 s 1 s 1 2 4 5s 3 1 3 s 1 Thus L 1 L1 L1 s 1 s 2 2s 5 s 1 s 1 2 4 1 s et e tL 1 3 s et e t 3L 1 2 L1 2 s2 4 s 4 s 4 5s 3 3 L1 et e t . t s2 L1 cos au. sin 2t cos 2t s 1 s 2 2s 5 2 1 s2 24) U sin g convolution theorem evaluateL s2 a2 s2 a2 (Dec 2013) Solution : f (t ) L 1 f (s) cos at . SJBIT Page 58 . Engineering Mathematics II 15MAT21 1 1 1 1 L1 L1 L1 L1 2 ( s 1)( s 2) 2 s 1 s 2 s 2 t 2t 2t 1 y (t ) e e e L1 s2 t 2t Thus y (t ) e e e 2t t e t (1 t ) e 2t 1 5s 3 23) Find L L s 1 s 2 2s 5 (Dec 2013) 5s 3 A Bs C Solution: = 2 s 1 s 2 2s 5 s 1 s 2s 5 5s 3 A s2 2s 5 ( Bs C )( s 1) put s 1 and s 0 we get A 1. Engineering Mathematics II 15MAT21 t t 1 f (t ) g (t ) cos(au bt bu ) cos(au bt bu )du 2u 0 u 0 t 1 sin(au bt bu ) sin(au bt bu ) 2 a b a b u 0 1 1 1 sin at sin bt sin at sin bt 2 a b a b 1 2a 2b sin at. b. 2 a b a b2 2 1 a b s2 Thus L f (t ) g (t ) a. SJBIT Page 59 . f ( s). 2 2 sin bt.g ( s) a2 b2 s2 a2 s2 b2 s2 a2 s2 a2 s2 a sin at b sin bt Thus L 1 s 2 a 2 s 2 b 2 a 2 b2 ''' 25) Solve y 2 y '' y ' 2 y 0 given y(0) y' (0) 0 and y'' (0) 6 by u sin g Laplace transform method (Dec 2013) ''' '' ' Solution: Taking laplace on both sides L [ y (t )] L[2 y (t )] L[ y (t )] 2 L[ y (t )] l (0) s 3 l[ y (t ) s 2 y (0) sy ' (0) y '' (0) 2 s 2 Ly (t ) sy (0) y' (0) sL[ y(t ) y (0 2 L[ y (t )] 0 U sin g initial conditions we have 6 Ly (t ) s 2 s 1 s 1 6 y (t ) L1 s 2 s 1 s 1 6 A B C s 2 s 1 s 1 s 2 s 1 s 1 Department of Mathematics. we get 2 s3 y s 2 y (0) sy '(0) y "(0) 3 s 2 y sy(0) y '(0) 3 sy y(0) y 2. y "(0) 2 Solution : Taking the laplace transform of the given equation. C 3 6 2 1 3 s 2 s 1 s 1 s 2 s 1 s 1 6 1 1 1 L1 2L 1 L1 3L 1 s 2 s 1 s 1 s 2 s 1 s 1 2t y (t ) 2e et 3e t 7) Solve the initial value problem (D3 3D 2 3D 1) y 2t 2 et . B 1. y (0) 1. y '(0) 0. 3 s 1 4 U sin g given conditions we get s 3 3s 2 3s 1 y (s 2 3s 1) 3 s 1 3 4 s 1 y (s 2 3s 1) 3 s 1 2 (s 2 3s 1) 4 s 1 ( s 1) 1 4 y 3 6 3 6 s 1 s 1 s 1 s 1 taking laplace transform we get 1 2 1 5 y et 1 t t t 2 30 Department of Mathematics. SJBIT Page 60 . Engineering Mathematics II 15MAT21 solving we get A 2.
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