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i=0 - (2²p+A) II (Jurisp + farrak) (foutag + Saurak) 7h) foi+8p+f6i+7h f6i+49 foi+3k P+h foi+7p+ foi+6h, foi+39 foi+2k, X6n-5 = r Now, it follows from Eq. (8) that Czan- X6n-4X6n-7+ X6n-6X6n-7 X6n-6 + X6n-9

i=0 - (2²p+A) II (Jurisp + farrak) (foutag + Saurak) 7h) foi+8p+f6i+7h f6i+49 foi+3k P+h foi+7p+ foi+6h, foi+39 foi+2k, X6n-5 = r Now, it follows from Eq. (8) that Czan- X6n-4X6n-7+ X6n-6X6n-7 X6n-6 + X6n-9

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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п-1 feip + f6i-1h foi-1p+ fei-2h ) \ foi+19 + foik foi+29 + föi+1k` hII X6n-4 %3D i=0 n-1 foi+4P + fei+3h` foi+3P + foi+2h ) ( foi-19 + fei-2k, feig + fei-1k kII X6n-3 %3D i=0 n-1 П foi+2P + foi+1h\ foi+1P + feih (foit49 + foi+3k` föi+39 + f6i+2k, X6n-2 i=0 n-1 ( foi+6P+ fsi+5h\( foi+29 + föi+1k \ foi+5P + föi+4h ) ( foi+aP + foi+3" ) ( fatus4 + foi+ak , X6n-1 foi+19 + feik i=0 n-1 ( föi+6q + fsi+5k` П föi+3P + foi+2h i=0 п-1 ( foi+8P + f6i+7h II fei+7P + fei+oh ) (foi+39 + foi+2k, 2р + h foi+49 + fei+3k' ). X6n+1 p+h i=0 where x-4 = h, x_3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1 = {1,0, 1, 1, 2, 3, 5, 8, ...}. Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n - 2. That is; n-2 föi+4p+ föi+3h` feiq + fei-1k kII föi+3P+ f6i+2h X6n-9 foi-19 + fei-2k, i=0 ( foi+2P + foi+1h (! П п-2 föi+49 + fsi+3k` föi+39 + fei+2k, X6n-8 fei+1p + feih i=0 foi+oP + foi+s" ) fo+19 + foik PII( fei+5P+ f6i+ah п-2 foi+29 + f6i+1k X6n-7 i=0 п-2 П ( fei+4P + foi+3h\ (fei+69 + foi+sk` foi+3P + f6i+2h )foi+59 + foi+ak , X6n-6 i=0 п-2 ( föi+8P+ foi+7h' П foi+7P + f6i+6h (föi+49 + f6i+3k` (Foi+39 + foi+2k, 2р + h X6n-5 = r p+h i=0 Now, it follows from Eq.(8) that X6n-6X6n-7 X6n-4 = x6n-7+ X6n-6 + x6n-9 11 || ||
Transcribed Image Text:п-1 feip + f6i-1h foi-1p+ fei-2h ) \ foi+19 + foik foi+29 + föi+1k` hII X6n-4 %3D i=0 n-1 foi+4P + fei+3h` foi+3P + foi+2h ) ( foi-19 + fei-2k, feig + fei-1k kII X6n-3 %3D i=0 n-1 П foi+2P + foi+1h\ foi+1P + feih (foit49 + foi+3k` föi+39 + f6i+2k, X6n-2 i=0 n-1 ( foi+6P+ fsi+5h\( foi+29 + föi+1k \ foi+5P + föi+4h ) ( foi+aP + foi+3" ) ( fatus4 + foi+ak , X6n-1 foi+19 + feik i=0 n-1 ( föi+6q + fsi+5k` П föi+3P + foi+2h i=0 п-1 ( foi+8P + f6i+7h II fei+7P + fei+oh ) (foi+39 + foi+2k, 2р + h foi+49 + fei+3k' ). X6n+1 p+h i=0 where x-4 = h, x_3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1 = {1,0, 1, 1, 2, 3, 5, 8, ...}. Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n - 2. That is; n-2 föi+4p+ föi+3h` feiq + fei-1k kII föi+3P+ f6i+2h X6n-9 foi-19 + fei-2k, i=0 ( foi+2P + foi+1h (! П п-2 föi+49 + fsi+3k` föi+39 + fei+2k, X6n-8 fei+1p + feih i=0 foi+oP + foi+s" ) fo+19 + foik PII( fei+5P+ f6i+ah п-2 foi+29 + f6i+1k X6n-7 i=0 п-2 П ( fei+4P + foi+3h\ (fei+69 + foi+sk` foi+3P + f6i+2h )foi+59 + foi+ak , X6n-6 i=0 п-2 ( föi+8P+ foi+7h' П foi+7P + f6i+6h (föi+49 + f6i+3k` (Foi+39 + foi+2k, 2р + h X6n-5 = r p+h i=0 Now, it follows from Eq.(8) that X6n-6X6n-7 X6n-4 = x6n-7+ X6n-6 + x6n-9 11 || ||
Brn-1n-2 YIn-1 + 8xn-4' = 0, 1, .., In+1 = aIn-2+ (1) The following special case of Eq.(1) has been studied Xn-1In-2 In+1 = In-2 + (8) In-1 + In-4' where the initial conditions r-4, x-3, x-2, x-1,and ro are arbitrary non zero real numbers. Theorem 4. Let {In}-4 be a solution of Eq.(8). Then for n = 0,1, 2, .. п-1 feip + fei-ih ( fei+29 + foi+1k hII fo X6n-4 i-1p+ foi-2h fei+19 + feik i=0 п-1 fei+4p+ fei+3h feig + foi-ik foi+3P + fei+2h) ( Fei-19 + fei-2k) n-1 ( fei+2P + Joi+" a39 + foi+2k , X6n-3 %3D i=0 fei+49+ fei+3k X6n-2 fei+1P + feih i=0 n-1 foi+24+ fei+1k foi+19 + feik foi+6p+ fei+5h PII foi+5p + fei+ah X6n-1 %3D i=0 n-1 П (fei+4p+ foi+3h\ ( foi+69 + föi+5k foi+3p + fei+2h) X6n %3D foi+59 + foi+ak, i=0 п-1 ( föi+8P+ fei+7h П foi+7P + foi+sh foi+49 + f6i+3k fei+39 + f6i+2k) 2р + h T6n+1 %3D p+h i=0 where x-4 = h, x-3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1 Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 2. That is; 100 {1,0, 1, 1, 2, 3, 5, 8, ...}). п-2 foi+4p + fei+3h fei+3p + fei+2h) ( foi-19 + fei-2k ) foig + föi-ik X6n-9 %3D i=0 п-2 fei+2P + fei+1h( fei+49 + foi+3k I. fei+1p + feih ) X6n-8 %3D fei+39 + fei+2k, i=0 п-2 П ( foi+6P + fei+sh\ ( foi+29 + foi+ik fsi+sP+ fei+ah, X6n-7 %3D fei+14 + foik i=0 п-2 foi+4P + foi+3h föi+3P + fei+2h ) \foi+59 + foi+ak ) foi+69 + foi+sk` qII X6n-6 %3D i=0 n-2 föi+8P + foi+7h foi+7P + fei+sh föi+49 + fei+3k ) Foi+:39 + foi+2k, 2p + h T6n-5 p+h i=0 Now, it follows from Eq.(8) that X6n-6X6n-7 X6n-4 = x6n-7 + X6n-6 + X6n-9 11
Transcribed Image Text:Brn-1n-2 YIn-1 + 8xn-4' = 0, 1, .., In+1 = aIn-2+ (1) The following special case of Eq.(1) has been studied Xn-1In-2 In+1 = In-2 + (8) In-1 + In-4' where the initial conditions r-4, x-3, x-2, x-1,and ro are arbitrary non zero real numbers. Theorem 4. Let {In}-4 be a solution of Eq.(8). Then for n = 0,1, 2, .. п-1 feip + fei-ih ( fei+29 + foi+1k hII fo X6n-4 i-1p+ foi-2h fei+19 + feik i=0 п-1 fei+4p+ fei+3h feig + foi-ik foi+3P + fei+2h) ( Fei-19 + fei-2k) n-1 ( fei+2P + Joi+" a39 + foi+2k , X6n-3 %3D i=0 fei+49+ fei+3k X6n-2 fei+1P + feih i=0 n-1 foi+24+ fei+1k foi+19 + feik foi+6p+ fei+5h PII foi+5p + fei+ah X6n-1 %3D i=0 n-1 П (fei+4p+ foi+3h\ ( foi+69 + föi+5k foi+3p + fei+2h) X6n %3D foi+59 + foi+ak, i=0 п-1 ( föi+8P+ fei+7h П foi+7P + foi+sh foi+49 + f6i+3k fei+39 + f6i+2k) 2р + h T6n+1 %3D p+h i=0 where x-4 = h, x-3 = k, x-2 = r, x-1 = p, xo = q, {fm}m=-1 Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n – 2. That is; 100 {1,0, 1, 1, 2, 3, 5, 8, ...}). п-2 foi+4p + fei+3h fei+3p + fei+2h) ( foi-19 + fei-2k ) foig + föi-ik X6n-9 %3D i=0 п-2 fei+2P + fei+1h( fei+49 + foi+3k I. fei+1p + feih ) X6n-8 %3D fei+39 + fei+2k, i=0 п-2 П ( foi+6P + fei+sh\ ( foi+29 + foi+ik fsi+sP+ fei+ah, X6n-7 %3D fei+14 + foik i=0 п-2 foi+4P + foi+3h föi+3P + fei+2h ) \foi+59 + foi+ak ) foi+69 + foi+sk` qII X6n-6 %3D i=0 n-2 föi+8P + foi+7h foi+7P + fei+sh föi+49 + fei+3k ) Foi+:39 + foi+2k, 2p + h T6n-5 p+h i=0 Now, it follows from Eq.(8) that X6n-6X6n-7 X6n-4 = x6n-7 + X6n-6 + X6n-9 11
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