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X6n+1 = T X6n-4 = X6n-3 = X6n-2 = Ibn = n-1 f3i+2p+f3i+3h) II f3i+3p+f3i+4h, i=0 h II i=0 n-1 k II i=0 T i=0 n-1 a II i=0 f3iq+f3i+1k f3i+19+f3i+2k, f3i-2p+f3i-1h) f3i-1p+f3ih F) (19 n-1 ™ II i=0 n-1 T6n-1 = p (f3i+1P+ f3i+2h\ (5 ( f3i-19+f3ik) f3i9+f3i+1k, f3ip + f3i+1h (J3i+1P + J3i+2h) (13i-29 + fai-1 k f3i-19+f3ik f3i-1p+f3ih f3iq+f3i+1k f3ip+f3i+1h, f3i+19+f3i+2k, f3i-19-f3ik f3i9-f3i+1k) f3i+2p+f3i+3h f3ip f3i+1h - (Sa+ap + Farah) (Sa+19+f34+2k). f3i+1p+f3i+2h,

X6n+1 = T X6n-4 = X6n-3 = X6n-2 = Ibn = n-1 f3i+2p+f3i+3h) II f3i+3p+f3i+4h, i=0 h II i=0 n-1 k II i=0 T i=0 n-1 a II i=0 f3iq+f3i+1k f3i+19+f3i+2k, f3i-2p+f3i-1h) f3i-1p+f3ih F) (19 n-1 ™ II i=0 n-1 T6n-1 = p (f3i+1P+ f3i+2h\ (5 ( f3i-19+f3ik) f3i9+f3i+1k, f3ip + f3i+1h (J3i+1P + J3i+2h) (13i-29 + fai-1 k f3i-19+f3ik f3i-1p+f3ih f3iq+f3i+1k f3ip+f3i+1h, f3i+19+f3i+2k, f3i-19-f3ik f3i9-f3i+1k) f3i+2p+f3i+3h f3ip f3i+1h - (Sa+ap + Farah) (Sa+19+f34+2k). f3i+1p+f3i+2h,

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Bxn-10n-2 Xn+1 = axn-2+ п 3D 0, 1, ..., (1) YXn-1 + dxn-4
Transcribed Image Text:Bxn-10n-2 Xn+1 = axn-2+ п 3D 0, 1, ..., (1) YXn-1 + dxn-4
The solution of the following special case of Eq.(1) has been obtained Xn-1Xn-2 Xn+1 = Xn-2 (10) Xn-1+ Xn-4 where the initial conditionsx –4, x –3, x–2, x_1, xo are arbitrary non zero real numbers. Theorem 6. Let {xn}-4 be a solution of Eq.(10). Then for n = 0, 1, 2, ... П J3i+3p + fa+ah ) (g + fai+1k п-1 fzi+2P + f3i+3h X6n+1 fai+19 + f3i+2k , i=0 п-1 f3i-2p + f3i–1h hII f3i-1p + f3ih f3i-19 + f3ik` f3iq + f3i+1k ) X6n-4 i=0 n-1 f3ip+ f3i+1h (fai-29 + f3i-1k kII X6n-3 fsi+1P+ fsi+2h ) ( fsi-19+ f;k i=0 п-1 f3i-1p + f3ih` П f3ip + f3i+1h, f3iq + f3i+1k X6n-2 f3i+19+ f3i+2k ) i=0 п-1 f3i-14 – faik` fi9 – fai+ik) f3i+1P+ f3i+2h pII X6n-1 = f3i+2P + f3i+3h, i=0 п-1 f3:P + f3i+1h II i+1P+ f3i+2h, (fai+19 + f3i+2k` f3i+29 + f3i+3k, X6n i=0
Transcribed Image Text:The solution of the following special case of Eq.(1) has been obtained Xn-1Xn-2 Xn+1 = Xn-2 (10) Xn-1+ Xn-4 where the initial conditionsx –4, x –3, x–2, x_1, xo are arbitrary non zero real numbers. Theorem 6. Let {xn}-4 be a solution of Eq.(10). Then for n = 0, 1, 2, ... П J3i+3p + fa+ah ) (g + fai+1k п-1 fzi+2P + f3i+3h X6n+1 fai+19 + f3i+2k , i=0 п-1 f3i-2p + f3i–1h hII f3i-1p + f3ih f3i-19 + f3ik` f3iq + f3i+1k ) X6n-4 i=0 n-1 f3ip+ f3i+1h (fai-29 + f3i-1k kII X6n-3 fsi+1P+ fsi+2h ) ( fsi-19+ f;k i=0 п-1 f3i-1p + f3ih` П f3ip + f3i+1h, f3iq + f3i+1k X6n-2 f3i+19+ f3i+2k ) i=0 п-1 f3i-14 – faik` fi9 – fai+ik) f3i+1P+ f3i+2h pII X6n-1 = f3i+2P + f3i+3h, i=0 п-1 f3:P + f3i+1h II i+1P+ f3i+2h, (fai+19 + f3i+2k` f3i+29 + f3i+3k, X6n i=0
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