09M1 XYCHAPTER 4 Difference Equations INTRODUCTION In this chapter, we shall discuss basic concepts of difference equation, homogeneous and non-homogeneous linear difference equation with constant coefficient. 4.1. DIFFERENCE EQUATIONS An equation which contain independent variable, dependent variable and the successive differences of the dependent variable is called the difference equation. Examples (1) yn+2 – 6yn+1 + 9yn= 0 (2) ( E 2 + 6 E + 9) yn = 2 n i.e., yn + 2 + 6 yn +1 + yn = 2 n (3) (∆2 + 3∆ + 2)yn = 1 or ∆2yn + 3∆yn + 2yn = 1 ( yn + 2 − 2 yn +1 + yn ) + 3( yn +1 − yn ) + 2 yn = 1 yn + 2 + yn +1 = 1 . or or 4.2. ORDER OF DIFFERENCE EQUATION The order of the difference equation is the difference between the largest and smallest arguments occurring in the difference equation divided by the unit of argument. Thus, the order of the difference equation = Largest argument − Smallest argument Unit of argument The order of the difference equation yn + 2 − 7 yn = 5 is ( n + 2) − n = 2. 1 119 Solution. 4. Example 2. SOLUTION OF DIFFERENCE EQUATION Any function which satisfies the given difference equation is called the solution of the difference equation. The order and degree of yn + 3 + 3 yn + 2 + 3 yn +1 + yn = 0 are respectively. We have ∆yn = yn+1 – yn ∆2yn = yn+2 – 2yn+1 + yn ∆3yn = yn + 3 − 3 yn + 2 + 3 yn +1 − yn Putting in given difference equation ( yn +3 − 3 yn + 2 + 3 yn +1 − yn ) + ( yn + 2 − 2 yn +1 + yn ) + ( yn +1 − yn ) + yn = 0 or or or yn+3 – 2yn+2 + 2yn+1 = 0 E3yn – 2E2yn + 2Eyn = 0 (E3 – 2E2 + 2E)yn = 0. Example 1. Form the difference equation from the equation y = Ax2 – Bx. 1 n+3−n = 3 and 2 1 2 Example 2. A solution in which the number of arbitrary constants is equal to the order of the difference equation is called general solution of the difference equation. FORMATION OF DIFFERENCE EQUATION Example 1.09M1 XY 120 ADVANCED MATHEMATICS 4.4. 4. A solution which is obtained from the general solution by assigning particular values is called particular solution. Write the difference equation ∆3yn + ∆2yn + ∆yn + yn = 0 in the subscript notation. We have ∆y = ∆(Ax2 – Bx) = A∆x2 – B∆x = A{(x + 1)2 – x2} – B {(x + 1) – x} ∆y = A(2x + 1) – B and ∆2y = ∆ {A(2x + 1) – B} = A ∆ (2x + 1) – B∆ = A{2(x + 1) + 1 – (2x + 1)} . DEGREE OF DIFFERENCE EQUATION The highest degree of yn’s in the difference equation is called the degree of the difference equation. Solution...3.(i) . The order and degree of yn+2 + 4yn+1 + 4yn = 2n are n+2−n = 2 and 1 respectively.5. occur to the first degree only and are not multiplied together is called the linear difference equation..6. A linear difference equations of order k is a0 yk+n + a1 yk + n −1 + a 2 yk + n − 2 + .(4. Form by yn = A2n + B(– 2)n a difference equation by eliminating the constants A and B.. 4.. Solution.. we get ⇒ yn – A2n – B(– 2)n ⇒ yn+1 – 2A2n + 2B(– 2)n ⇒ yn+2 – 4A2n – 4B(– 2)n yn y n +1 yn + 2 −1 −1 −2 2 =0 −4 −4 or yn+2 – 4yn = 0..1) is called the linear homogeneous difference equation otherwise it is called non-homogeneous difference equation. Ans.1) If R(n) = 0 then the equation (4.. Example 3. Ans.. + a k yn = R( n ) . LINEAR DIFFERENCE EQUATION A difference equation in which yn... .09M1 XY DIFFERENCE EQUATIONS 121 ∆2y = 2A 1 2 ∆y 2 1 Put in (i). We have yn = A2n + B(– 2)n yn+1 = 2A2n – 2B(– 2)n yn+2 = 4A2n + 4B(– 2)n Eliminate A and B. we get . yn+1.. yn+2 .... we get B = ∆2y (2x + 1) – ∆y 2 1 B = (2x + 1)∆2y – ∆y 2 Putting these values in given equation.(ii) By (ii) A= or y= 2y = [– x2 – x] ∆2y + 2x ∆y or or or LM 1 ∆ yOP x − LM 1 (2 x + 1) ∆ y − ∆yOP x N2 Q N2 Q L1 1 O = M x − (2 x + 1) x P ∆ y + x∆y 2 2 N Q 2 2 2 2 2 (x2 + x)∆2y – x ∆y + 2y = 0 (x2 + x) ( yn + 2 − 2 yn +1 + yn ) − 2 x ( yn +1 − yn ) + 2 yn = 0 (x2 + x)yn+2 – (2x2 + 4x)yn+1 + (x2 + 3x + 2)yn = 0.. 2) ( a0 E k + a1 E k −1 + .e. mk are real and distinct roots of auxiliary equation (4. an are all constants. When the roots are complex number We know that if complex roots occur then they must be conjugate complex number i. + an m n = 0 ⇒ ( a0 m k + a1 m k −1 + . then we have a0 m n + k + a1m n + k −1 + ........+ ck nk–1] (m1)n. m1 = m2 = m3.......09M1 XY 122 ADVANCED MATHEMATICS 4. If three roots be equal i.. + an I ) yn = 0 or φ(E)yn = 0 where a0..(4.4) Equation (4.3) which show that mn is the solution of (4.4) is known as auxiliary equation.... + ck(mk)n.+ ck(mk)n If k roots be equal i. A2 = i(c1 – c2) Case IV.. then the solution is yn = c1(m1)n + c2(m2)n .. then the solution is yn = [c1 + c2n + c3n2 +.e... a1. Case I. Case II.3) if m satisfies ... m1 = m2 = m3 . m2 ... When some of the roots are equal If two roots be equal i. m2........ is known as homogeneous linear difference equation with constant coefficient Let yn = mn be the solution of the difference equation (4.. + ck(mk)n... Here some cases arise. When roots are all real and distinct If m1... + an = 0 .... ... if (α + iβ) is the root then (α – iβ) is also a root where α and β are real.7.e... + an )m n = 0 a0 m k + a1 m k −1 + .4)..(4.(4..... When some of the complex roots are equal Let the root (α ± i β) becomes twice then the solution... m1 = m2.e...... .mk.. then the solution is yn = (c1 + c2n) (m1)n + c3(m3)n + . then the solution is yn = (c1 + c2n + c3n2) (m1)n + c4(m4)n +.2). This equation has k roots which we take m1.... Case III. HOMOGENEOUS LINEAR DIFFERENCE EQUATION WITH CONSTANT COEFFICIENT An equation of the form ....... and A1 = c1 + c2. = mk .. Then the solution is yn = c1(α + iβ)n + c2(α – iβ)n which can be written as yn = γ n (A1 cos nθ + A2 sin nθ) where γ= θ= α 2 + β2 tan–1 (β/α). left hand side 1 R(n).. + a n I ) yn = R( n ) .09M1 XY DIFFERENCE EQUATIONS 123 yn = γ n [(A1 + A2n) cos nθ + (A3 + A4n) sin nθ] where γ= α 2 + β2 and θ = tan–1 (β/α).e.I. ( n − ( k − 1)) n − k n a k a = k! ( E − a) 1 an = nan–1 E−a 1 an = n(n – 1)a n–2. ( E − a) 2 1 1 e ian − e −ian sin an = φ( E ) φ( E ) 2i Case II.I. = . When R(n) = sin an or cos an ∴ P... = Particular case: When k = 1 P. an are all constants is known as non-homogeneous linear difference equation with constant coefficient..I. P. complementary function and particular integral Complementary function is the general solution of the homogeneous equation i.... = 1 R(n).... a1.5) or φ(E)yn = R(n) where a0.) = Case I. NON-HOMOGENEOUS LINEAR DIFFERENCE EQUATION WITH CONSTANT COEFFICIENT The equation of the form k −1 (a0E k + a1 E + . k is positive integer.I.. .. = 1 n( n − 1)( n − 2) . The general solution of (4.I.5) and particular integral = φ( E ) Rules for Obtaining the Particular Integral The particular integral (P. φ( E ) 1 1 an = an provided φ(a) ≠ 0 φ( E ) φ( a) If φ (a) = 0. a2. 4. When R(n) = an ∴ P.. of (4..I..5) consists of two parts..8. = When k = 2 P. then for the equation (E – a)k yn = an ..(4. where F(n) being a polynomial in n. ∴ P. Case III. Solution. we expand [φ(1 + ∆)]–1 in ascending power of ∆ by the Binomial theorem as far as the term in ∆k. Example 3. When R(n) = an F(n). then express nk in the factorial notation and distribute it each term of the expansion. When R(n) = nk ∴ P. Solve yn+3 + yn+2 – 8yn+1 – 12yn = 0.I. Example 2. Case IV. 3 Hence. Solution. Solve the difference equation yn+3 – 2yn+2 – 5yn+1 + 6yn = 0. Ans. The given difference equation can be written as (E3 – 2E2 – 5E + 6)yn = 0 The auxiliary equation is m3 – 2m2 – 5m + 6 = 0 The roots are m = 1. First. Solve the difference equation yn+2 – 2yn+1 – 8yn = 0. 4 Hence. the solution is yn = c1(1)n + c2(– 2)n + c33n. Ans. Solution. = 1 1 a n F( n ) = a n F( n) .I.I. = Provided φ (eia) ≠ 0 and φ (e–ia) ≠ 0. the solution is yn = c1 (– 2)n + c24n. The given equation can be written as (E2 – 2E – 8) yn = 0 The auxiliary equation is m2 – 2m – 8 = 0 The roots are m = – 2. = 1 1 1 (e ia ) n − (e −ia ) n φ( E ) 2 i φ( E ) ia ) e ian − 1 1 1 1 cos an = e ian + e − ian φ( E ) 2 φ(e ia ) φ(e −ia ) R S T 1 e − ian φ(e − ia ) U V W U V W U V W [3 E = 1 + ∆] 1 1 nk = nk φ( E ) φ(1 + ∆) = [φ (1 + ∆)]–1 nk.09M1 XY 124 ADVANCED MATHEMATICS R S T 1 R 1 = S 2 i T φ( e = Similarly P. The given difference equation can be written as (E3 + E2 – 8E – 12)yn = 0 The auxiliary equation is m3 + m2 – 8m – 12 = 0 . – 2. φ( E ) φ( aE ) SOLVED EXAMPLES Example 1. Solve the difference equation 9yn+2 – 6yn+1 + yn = 0. Solve yn+2 + 16yn = 0. 2 Q N 2 n n 2 1 2 1 2 1 2 Example 6. 3 Hence. – 2. The given difference equation can be written as (9E2 – 6E + 1)yn = 0 The auxiliary equation is The roots are m = 1 1 .09M1 XY DIFFERENCE EQUATIONS 125 The roots are m = – 2. Solution. the solution is yn = c1(4i)n + c2(– 4i)n yn = 4n {c1(i)n + c2(– i)n} = 4n 1 m2 + 16 = 0 = 4n = 4n = 4n Rc F cos π + i sin π I + c F cos π − i sin π I U S H 2 V H 2 2K 2K W T Rc F cos nπ + i sin nπ I + c F cos nπ − i sin nπ I U S H 2 V H 2 2K 2 KW T R(c + c ) cos nπ + i (c − c ) sin nπ U S V 2 2 W T LM A cos nπ + B sin nπ OP . Solution. Example 5. The given difference equation can be written as (E2 + 16)yn = 0 The auxiliary equation is The roots are m = ± 4i Hence. 3 3 9m2 – 6m + 1 = 0 Hence. the solution is yn = (c1 + c2n) 1 3 FI HK n . the solution is yn = (c1 + c2n)(– 2)n + c33n. H 2K . Ans. Example 4. Solve yn+2 – 4yn+1 + 13yn = 0. Solution. The given difference equation can be written as (E2 – 4E + 13)yn = 0 The auxiliary equation is m2 – 4m + 13 = 0 The roots are Let m= 4 ± 16 − 4 × 13 4 ± 6i = = 2 ± 3i 2 2 2 + 3i = r (cos θ + i sin θ) ⇒ r = (13)1/2 and θ = tan–1 F 3I H 2K Therefore solution is yn = (13)n/2 {c1 cos nθ + c2 sin nθ} where θ = tan–1 F 3I . 3n – 4 . yn = cos 3 3 Example 8.. 3 ⇒ B = 1 2 2 2 Put A = 1 and B = 1 in (i).(i) But y0 = 1 and y1 = ∴ and 1+ 3 2 1 = A cos 0 + B sin 0 ⇒ A=1 π π 1+ 3 = A cos + B sin 3 3 2 1+ 3 = 1. 7n. Solve yn+2 – yn+1 + yn = 0 given that y0 = 1. y1 = The auxiliary equation is m2 – m + 1 = 0 The roots are m= 1 ± 1 − 4 1 ± 3i = 2 2 Hence the solution is yn = c1 = c1 = c1 F1+ i 3I + c F1− i 3I GH 2 JK GH 2 JK F cos π + i sin π I + c F cos π − i sin π I H 3 H 3 3K 3K F cos nπ + i sin nπ I + c F cos nπ − i sin nπ I H 3 H 3 3K 3K n n 2 n n 2 2 nπ nπ + i (c1 – c2) sin 3 3 nπ nπ yn = A cos + B sin 3 3 = (c1 + c2) cos . 7n The auxiliary equation is m2 – 2m + 5 = 0 The roots are Let m= 2± 4−4×5 = 1 ± 2i 2 ⋅1 1 + 2i = r (cos θ + i sin θ) ⇒ r = 5 and θ = tan–1 2 Therefore complementary solution is = (5)n/2 (c1 cos nθ + c2 sin nθ) where θ = tan–1 2 . Solution. 1 + B . 2 Solution. The given difference equation can be written as (E2 – E + 1) yn = 0 Example 7. 3n – 4 . The given difference equation can be written as (E2 – 2E + 5)yn = 2 . Solve the difference equation yn+2 – 2yn+1 + 5yn = 2 ..09M1 XY 126 ADVANCED MATHEMATICS 1+ 3 . we get nπ nπ + sin . F. The auxiliary equation of given difference equation is m2 – 3m – 4 = 0 The roots are m = – 1. . 1 n 1 n 3 − 7 . 2. Therefore complementary solution is c1 (– 1)n + c2 4n. = (5)n/2 {c1 cos nθ + c2 sin nθ} + Solution. 2n E 2 − 3E + 2 [3 F(2) = 0] 1 1 1 6 .2n. yn = c1 + c22n + 3n2n. 2n = 6 2n ( E − 2) ( E − 1)( E − 2) ( 2 − 1)( E − 2) = 6.I. The given difference equation can be written as (E2 – 3E + 2)yn = 6 . Hence solution is yn = C. 2n = 6 .I.2n The auxiliary equation is The roots are m = 1. + P. Therefore the complementary solution is c1(1)n + c2(2)n. The particular solution is given by P.F. 4 yn = C. = c1(– 1)n + c24n – y(n + 2) – 3y(n + 1) + 2y(n) = 6 . Solve the difference equation (E2 – 3E – 4)yn = 3n.09M1 XY DIFFERENCE EQUATIONS 127 The particular solution is given by P. 3 n – 4 . = = m2 – 3m + 2 = 0 1 6 . Solve the difference equation or yn+2 – 3yn+1 + 2yn = 6 . + 4. Solution.F. 4 10 Example 9.I. n 2n–1 = 3n2n.I. = Hence solution is 1 1 1 3n = 3n = − 3n 9−9−4 4 E 2 − 3E − 4 1 n 3.2n Ans. + P. Ans. = 1 2 .I. 7n E − 2E + 5 2 =2 =2 1 1 3n − 4 2 7n E − 2E + 5 E − 2E + 5 2 Hence solution is 1 n 1 n 1 1 n 3 −4 7 = 3n − 7 8 40 4 10 yn = C. The particular solution is given by P. Ans.I. Example 10. + P. I.. = 1 6 .2n .) .1 4(1 + ∆ ) − 6(1 + ∆ ) + 2 2 2 1 1 1 = 6. 2n Example 11.. Therefore the complementary solution is c1(ia)n + c2(– ia)n = an {c1(i)n + c2(– i)n) = an c1 cos = an = an The particular solution is given by R F π + i sin π I + c F cos π − i sin π I U S H 2 V H 2 2K 2K W T Rc F cos nπ + i sin nπ I + c F cos nπ − i sin nπ I U V S H 2 H 2 2K 2 KW T R(c + c ) cos nπ + i (c − c ) sin nπ U = a RA cos nπ + B sin nπ U .1 2 4(1 + 2∆ + ∆ ) − 6(1 + ∆ ) + 2 4∆ + 2∆ 1 1 = 3 .I.. 2n n.I. 2n . 1 = 3 . 2n 1 1 ..1 2 4E − 6E + 2 ( 2 E ) − 3( 2 E) + 2 1 . 2n E − 3E + 2 2 =6 1 2n . V V S 2 S 2 W 2 2 W T T n n 2 1 2 1 2 1 2 n e ian + e −ian 1 1 cos an = 2 P. Solve yn+2 + a2yn = cos an. 1 E − 3E + 2 2 2 [3 F (2) = 0] = 6 .09M1 XY 128 ADVANCED MATHEMATICS Aliter : P. 2n .1 = 3 . = 3 .1 = 6 . = 3 .. 2n (1 + 2∆)–1 ..) ∆ ∆ P. Solution. = 2 2 E + a2 (E + a2 ) R S T 1R = S 2 Te = 1 1 1 (e ia ) n + 2 ( e − ia ) n 2 2 2 E +a (E + a2 ) 2 ia 1 1 e ian + −2 ia e − ian 2 +a + a2 e U V W U V W . 2n = 6 .1 ∆ ∆(1 + 2 ∆ ) 1 1 (1 – 2∆ + 4∆2 .. 2n (1 – 0 + 0. The given difference equation can be written as (E2 + a2)yn = cos an The auxiliary equation is m2 + a2 = 0 The roots are m = ± ia. 2n = 6 . F.I. + P. The given difference equation can be written as (E2 – 2 cos α E + 1)yn = cos α n .I. yn = an A cos R S T nπ nπ cos a( n − 2) + a 2 cos an + B sin + 2 2 1 + 2 a 2 cos 2a + a 4 U V W Aliter: P. 1 + 2 a 2 cos 2 a + a 4 Example 12. Solve yn+2 – 2 cos α yn+1 + yn = cos αn.09M1 XY DIFFERENCE EQUATIONS 129 R S T 1 Re = S 2T = = Hence solution 1 e ian (e −2 ia + a 2 ) + e − ian ( e 2 ia + a 2 ) 2 (e 2 ia + a 2 ) ( e −2 ia + a 2 ) ia ( n − 2 ) U V W + e −ia( n− 2) + a 2 ( e ian + e − ian ) 1 + a 2 (e 2ia + e −2ia ) + a 4 U V W 1 2 cos a ( n − 2) + a 2 2 cos an cos a( n − 2) + a 2 cos an = 2 1 + 2 a 2 cos 2a + a 4 1 + 2 a 2 cos 2a + a 4 yn = C. = 1 cos an E + a2 2 = Real part of = Real part of = Real part of 1 (cos an + i sin an) E + a2 2 1 eian E + a2 2 1 (eia)n E + a2 2 1 = Real part of e 2 ia a 2 eian + = Real part of = Real part of = Real part of = Real part of = (e 2 ia e −2 ia + a 2 e ian + a 2 ) (e −2 ia + a 2 ) e ia ( n − 2 ) + a 2 e ian 1 + a 2 (e 2 ia + e −2 ia ) + a 4 cos a (n − 2) + i sin a ( n − 2) + a 2 (cos an + i sin an) 1 + 2 a 2 cos 2a + a 4 cos a( n − 2) + a 2 cos an + i (sin a ( n − 2) + a 2 sin an) 1 + 2 a 2 cos 2 a + a 4 cos a (n − 2) + a 2 cos an . Solution. I.09M1 XY 130 ADVANCED MATHEMATICS The auxiliary equation is m2 – 2 cos α m + 1 = 0 2 cos α ± 4 cos 2 α − 4 = (cos α ± i sin α) 2. = = 1 cos nα E − 2 cos αE + 1 2 e inα + e −inα 1 (e iα + e − iα ) 2 E2 − 2 E +1 2 E 2 − (e iα 1 e inα + e − inα 2 + e − iα ) E + 1 = = R U S V T W U 1R 1 1 (e ) + (e ) V = S 2 T ( E − e )( e − e ) (e − e )( E − e ) W U 1R 1 1 (e ) + (e ) V = S 2 T ( E − e ) 2i sin α ( − 2i sin α ) ( E − e ) W R 1 ( e ) − 1 (e ) U 1 = S V 4i sin α T E − e E−e W = 1 1 1 ( e iα ) n + (e −iα ) n 2 ( E − e iα )( E − e −iα ) ( E − e iα )( E − e − iα ) iα iα − iα iα n − iα iα −iα − iα n iα iα n − iα − iα n iα iα n − iα − iα n 1 1 {(e iα ) n + (e −iα ) n } iα 2 ( E − e )( E − e −iα ) = = Hence solution n 1 (e i ( n −1)α − e − i( n −1)α ) n ( e iα ) n −1 − n ( e − iα ) n −1 = 4i sin α 4i sin α n sin ( n − 1)α n 2i sin ( n − 1) α = 2 sin α 4i sin α n sin ( n − 1)α .F. 2 sin α n s yn = C. yn = A cos nα + B sin nα + . Ans. + P.1 Therefore complementary solution is The roots are m= c1 (cos α + i sin α)n + c2 (cos α – i sin α)n = c1 (cos nα + i sin nα) + c2 (cos nα – i sin nα) = (c1 + c2) cos nα + i(c1 – c2) sin nα = A cos nα + B sin nα The particular solution is given by P.I. The given difference equation can be written as (8E2 – 6E + 1)yn = 5 sin F nπ I H 2K The auxiliary equation is 8m2 – 6m + 1 = 0 The roots are m= 1 1 . = 2 c1 F 1I H 4K n + c2 F 1I H 2K n 1 nπ 5 sin 8E − 6 E + 1 2 JJ K FF I F I I 5 1 5 1 = G H e K − GH e JK JJ = 2i 8E − 6E + 1 (i − (− i) ) 2i 8 E − 6 E + 1 G G J H K 5 L = M 1 i − 8E −16 E + 1 ( − i) OPQ 2i N 8 E − 6 E + 1 5 L = M 1 i − 8(− i) −16(− i) + 1 (− i) OPQ 2i N 8i − 6i + 1 5 L = M− 1 i + 7 −1 6i (− i) OPQ 2i N 7 + 6i LM − (7 − 6i) F cos π + i sin π I + (7 + 6i) F cos π − i sin π I OP H 2 H 2 5 2K 2K P M = ( 7 + 6i)( 7 − 6i ) 2i M PP MN Q L − (7 − 6i) F cos nπ + i sin nπ I + (7 + 6i) F cos nπ − i sin nπ I OP H 2 H 2 5 M 2K 2K = PP MM 2i 49 + 36 PQ MN e 1 = 5 2 8E − 6 E + 1 iπ 2 n − iπ 2 n n n 2 2 n n 2 2 n n 2 2 n n n n F GG GH F I H K F nπ I H2K −e 2i −i i F nπ I I H2KJ . Solve 8yn+2 – 6yn+1 + yn = 5 sin F nπ I . H 2K Solution.09M1 XY DIFFERENCE EQUATIONS 131 Example 13. 4 2 Therefore complementary solution is The particular solution is given by P.I. 6 cos − 7 sin 17 2 2 Example 14.F.. (3n(2) + 3n(1) + 2) IJ K = . Solve the difference equation + c2 + yn+2 – 6yn+1 + 8yn = 3n2 + 2. + P. 3 9 FG IJ (3n + 2) H K 1F (4 ∆ − ∆ ) 4∆ − ∆ = G1 + + 3H 3 9 1 4 ∆ − ∆2 1− = 3 3 2 2 FG1 − 4∆ − ∆ IJ (3n H 3 K 2 1 2 + 2) 2 2 + . Solution...I. The given difference equation can be written as (E2 – 6E + 8)yn = 3n2 + 2 The auxiliary equation is m2 – 6m + 8 = 0 The roots are m = 2. Ans. = c12n + c24n + n2 + .. 4 Therefore the complementary solution is c12n + c24n The particular solution is given by P. = = = F 1I H 4K n F 1I H 2K n R S T U V W 1 1 (3n2 + 2) = (3n 2 + 2) 2 (1 + ∆ ) − 6(1 + ∆ ) + 8 E − 6E + 8 2 1 (3n2 + 2) 1 + 2∆ + ∆ − 6 − 6∆ + 8 2 1 1 (3n2 + 2) = ∆ − 4∆ + 3 3 2 −1 1 4(6 n + 3) − 6 16 × 6 (3n ( 2 ) + 3n (1) + 2 ) + + 3 3 9 8 44 1 48 = 3n( n − 1) + 3n + 2 + 8n + 2 + = n2 + n + 3 9 3 3 8n 44 + Hence solution yn = C.I.09M1 XY 132 ADVANCED MATHEMATICS = 1 nπ nπ nπ nπ nπ − 7 cos − 7i sin + 6i cos − 6 sin + 7 cos 34i 2 2 2 2 2 − 7i sin nπ nπ nπ + 6i cos + 6 sin 2 2 2 LM N =– Hence solution is yn = c1 1 nπ nπ i nπ nπ 6 cos − 7 sin = − 14i sin + 12i cos 17 2 2 34 2 2 R S T U V W R S T U V W OP Q 1 nπ nπ .. 09M1 XY DIFFERENCE EQUATIONS 133 Example 15.. Solution. – 2 Therefore the complementary solution is c1(2)n + c2(– 2)n The particular solution is given by 1 1 ( n 2 + n − 1) = (n 2 + n − 1) E2 − 4 (1 + ∆ ) 2 − 4 1 (n 2 + n − 1) = 2 ∆ + 2∆ − 3 FG1 − 2∆ + ∆ IJ (n + n − 1) −3 H 3 K −1 F + = GH1 − 2∆ 3 ∆ IJK (n + n − 1) 3 F 2∆ + ∆ IJ + . =– yn = c12n + c2(– 2)n – n 2 7n 17 ... Solve the difference equation yn+2 – 4yn = n2 + n – 1. The given difference equation can be written as (E2 – 4)yn = n2 + n – 1 The auxiliary equation is m2 – 4 = 0 The roots are m = 2.. − − 3 9 27 .I.F.IJ (n 1F 2∆ + ∆ +G = − G1 + 3H 3 H 3 K K = 2 2 2 −1 2 2 2 2 1 (2 ) + 2n (1) − 1) FG H 1F 4 8 n( n − 1) + 2 n − 1 + n + I =− 3H 3 9K = Hence solution [3 n2 + n – 1 = n(2) + 2n(1) – 1] 2(2n (1) + 2) + 2 4 . + P. 2 − 1 (2) + n + 2n (1) − 1 + 3 3 9 n 2 7n 17 − − 3 9 27 yn = C.. 21. yn + 2 − 4 yn = 5 . 2 with y0 = 1. (∆ – 2)2 (∆ – 5)yn = 0 6. 2yn+2 – 5yn+1 + 2yn = 0 4. yn = c1 F 1I H 2K n + c22n and yn = – 2 1 3 2 F I H K n + 2 n (2 ) 3 14. yn = c1 + c23n + 15. yn = cos 1 n 4 3 12. (∆2 – 5∆ + 4)yn = 0 7. yn+ 2 − 4 yn+1 + 3yn = 4n . 4 yn + 2 + 25 yn = 0 9. f(n) = c1(7)n + c2(– 8)n 5. y1 = 1 Solve the following non-homogeneous difference equations 13. yn = c1 F 1I H 2K n + c2 (2)n 3. yn + 2 + yn = 5 . yn + 2 − 7 yn +1 + 12 yn = cos n 20. 3n. F 5 I LMc cos nπ + c sin nπ OP H 2K N 2 2 Q F πI F πI y = 2 c cos H − K n + c sin H − K n 3 3 n 1 n n 1 2 8. 16. Solve 2 yn + 2 − 5 yn +1 + 2 yn = 0 with y0 = 0. yn = (c1 + c2n) 2n and yn = (n + 2)2n–1 13. yn + 2 − 2 yn +1 + 5 yn = 0 11.09M1 XY 134 ADVANCED MATHEMATICS EXERCISE 4. yn +2 + yn = 4 cos 2n 22. yn = (c1 + c2n + c3n2 + c4n3)(1)n 10. (E2 – 2E – 8)yn = 0 3. Solve the difference equation yn + 2 − 4 yn +1 + 4 yn = 0 and find the particular solution satisfying the initial conditions y0 = 1 and y1 = 3 12. yn +3 + yn + 2 − yn +1 − yn = 0 8. yn = c12n + c2(– 2)n + 3n 16. yn = c1( + 2)n + c2 ( − 4)n 2. f(n) = c1 + c22n + 1 n 4 6 πn + 2n 2 . yn+2 + yn+1 + yn = n2 + n + 1 ANSWERS 1. 15.1 Solve the following homogeneous difference equations 1. yn = 5n (c1 cos nθ + c2 sin nθ) where θ = tan–1 2 11. f(n + 2) – 3f(n + 1) + 2f(n) = 4n 18. yn + 4 − 4 yn +3 + 6 yn + 2 − 4 yn +1 + yn = 0 10. yn = (c1 + c2n) 3n + c36n 6. yn = (c1 + c2n)(– 1)n + c3 2 9. yn = c12n + c25n 7. yn + 2 + 2 yn +1 + 4 yn = 0 2. y1 = 0 yn + 2 − 4 yn +1 + 4 yn = 2n yn + 2 − 16 yn = cos (n/2) yn+ 2 + a2 yn = sin an n 14. 17. f(n + 2) + f(n + 1) – 56 f(n) = 0 5. yn = 4. 19. yn = c1 cos F 2 nπ I + c H3K 2 sin F 2 nπ I + 1 F n H 3 K 3H 2 −n+ 1 3 I K . yn = (c1 + c2n)2n + n(n – 1) 2n–3 cos 18. an c1 cos (257 − 32 cos 1) F n − 1I − 16 cos n H2 K 2 nπ nπ 2[cos (2 n − 4 ) + cos 2 n ] + c2 sin + 2 2 1 + cos 4 F H nπ nπ sin a (n − 2 ) + a 2 sin an + c2 sin + 2 2 1 + 2 a 2 cos 2 a + a 4 I K 22. yn = c13n + c24n + 19.09M1 XY DIFFERENCE EQUATIONS 135 1 (18 cos n – 77 sin n) 170 17. yn = c1 cos 21. yn = c14n + c2 (– 4)n + 20.