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X6n-4 = 1 The following special case of Eq. (1) has been studied X6n-3 = X6n-1 = (8) where the initial conditions x-4, X-3, X-2, X-1,and zo are arbitrary non zero real numbers. Theorem 4. Let {n}-4 be a solution of Eq. (8). Then for n = 0, 1, 2, ... I6n = In+1 = αIn-2 + X6n-2 = T X6n+1 = X6n-9 = X6n-7 = n-1 hII X6n-6 i=0 n-1 k⇓I i=0 X6n-5 = n-1 PII i=0 n-1 II i=0 n-1 9 II i=0 X6n-8 = T n-2 k II i=0 n-2 ™II i=0 Bxn-1n-2 Yn-1 +6xn-4' = 9 n-2 PII i=0 n-2 In+1 = In-2 + i=0 In-1n-2 In-1 + In-4 foip+f6i-1h foi-1p+ foi-2h, where x_4 = h, x_3= k, x_2= r, x_1=p, xo = q, {fm}=-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; h) (fi (f6i+4P+f6i+3h) f6i+3P + fei+2h, f6i+2p+f6i+1h) foi+1P + foih n-1 foi+8p+f6i+7h` 7 (²P + h) II (faisap + feir7h ) ( Sai+49 + Sei+3k) T f6i+39 +f6i+2k) i=0 n = 0, 1, ..., (f6i+49 + f6i+3k) foi+39 + foi+2k, (f6i+29+ foi+1k\ f6i+5p+f6i+4h) f6i+19+f6ik f6i+4P+f6i+3h) (f6i+69+f6i+5k foi+3p+f6i+2h/ f6i+59 + f6i+4k) f6i+29+f6i+1k) foi+19+f6ik (fai+2P + fei+¹h) ( n-2 r (2²p+h) ( T i=0 foiq+f6i-1k fei-19+f6i-2k f6i+4P+f6i+3h f6iq+f6i-1k foi+3p+f6i+2h, f6i-19 + foi-2k) Now, it follows from Eq. (8) that f6i+6p+f6i+5h) (f6i+29+f6i+1k) f6i+5P+f6i+4h, f6i+19+ foik f6i+4P+f6i+3h) (f6i+69+f6i+5k foi+3P+f6i+2h) f6i+59 + f6i+4k) X6n-4X6n-7 + (f6i+49 +f6i+3k \f6i+39 +f6i+2k, 7 f6i+8p+f6i+7h Joi+7p + feitch) (f6i+49 + foi+3h) f6i+39 + f6i+2k) X6n-66n-7 X6n-6 + X6n-9 (1) 7

X6n-4 = 1 The following special case of Eq. (1) has been studied X6n-3 = X6n-1 = (8) where the initial conditions x-4, X-3, X-2, X-1,and zo are arbitrary non zero real numbers. Theorem 4. Let {n}-4 be a solution of Eq. (8). Then for n = 0, 1, 2, ... I6n = In+1 = αIn-2 + X6n-2 = T X6n+1 = X6n-9 = X6n-7 = n-1 hII X6n-6 i=0 n-1 k⇓I i=0 X6n-5 = n-1 PII i=0 n-1 II i=0 n-1 9 II i=0 X6n-8 = T n-2 k II i=0 n-2 ™II i=0 Bxn-1n-2 Yn-1 +6xn-4' = 9 n-2 PII i=0 n-2 In+1 = In-2 + i=0 In-1n-2 In-1 + In-4 foip+f6i-1h foi-1p+ foi-2h, where x_4 = h, x_3= k, x_2= r, x_1=p, xo = q, {fm}=-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; h) (fi (f6i+4P+f6i+3h) f6i+3P + fei+2h, f6i+2p+f6i+1h) foi+1P + foih n-1 foi+8p+f6i+7h` 7 (²P + h) II (faisap + feir7h ) ( Sai+49 + Sei+3k) T f6i+39 +f6i+2k) i=0 n = 0, 1, ..., (f6i+49 + f6i+3k) foi+39 + foi+2k, (f6i+29+ foi+1k\ f6i+5p+f6i+4h) f6i+19+f6ik f6i+4P+f6i+3h) (f6i+69+f6i+5k foi+3p+f6i+2h/ f6i+59 + f6i+4k) f6i+29+f6i+1k) foi+19+f6ik (fai+2P + fei+¹h) ( n-2 r (2²p+h) ( T i=0 foiq+f6i-1k fei-19+f6i-2k f6i+4P+f6i+3h f6iq+f6i-1k foi+3p+f6i+2h, f6i-19 + foi-2k) Now, it follows from Eq. (8) that f6i+6p+f6i+5h) (f6i+29+f6i+1k) f6i+5P+f6i+4h, f6i+19+ foik f6i+4P+f6i+3h) (f6i+69+f6i+5k foi+3P+f6i+2h) f6i+59 + f6i+4k) X6n-4X6n-7 + (f6i+49 +f6i+3k \f6i+39 +f6i+2k, 7 f6i+8p+f6i+7h Joi+7p + feitch) (f6i+49 + foi+3h) f6i+39 + f6i+2k) X6n-66n-7 X6n-6 + X6n-9 (1) 7

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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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
Joi+7P + fotneh ) (Gi+19 + foi+3k` föi+39 + föi+2k , п-2 2p + h П foi+sP + fei+7h` X6n-2 p+h i=0 п-2 2p+h pth ) 11 (Foi+7p+f6i+6h f6i+8p+f6i+7h h r f6i+49+f6i+3k f6i+39+f6i+2k i=0 fênq+fên-1k h h + fon+19+fönk p п-2 foi+8P + foi+7h` П foi+7P+ föi+6h , foi+49 + f6i+3k` föi+39 + föi+2k , 2р + h X6n-2 p+ h i=0 п-2 2p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h föi+49+f6i+3k f6i+39+f6i+2k h Jr p+h i=0 + h|1+ fönq+fên-1k| fön+19+f6nk Soi+7P + feiseh ) (Bi+49 + foi+3k' fsi+39 + f6i+2k , п-2 2p + h ´fei+sP + fei+7h` X6n-2 p+h i=0 п-2 2р+h p+h II f6i+8p+f6i+7h f6i+7p+f6i+6h f6i+4q+f6i+3k` f6i+39+f6i+2k i=0 + fên+19+fénk+fênq+fén–1k fön+19+fönk (föi+8P+ f6i+7h` r (2p + 4) Ti fai sp + faiszh) (feireg + foisak \ foi+7P + f6i+6h п-2 П foi+49 + föi+3k` föi+39 + foi+2k, p+h i=0 п-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h f6i+49+f6i+3k f6i+3q+f6i+2k i=0 + fén+29+f6n+1k fon+19+ fönk (foi+49 + föi+3k fei+sP+ foi+7! (g+ foi+2k , п-2 ( 2p +h П föi+7P+ f6i+6h, fön+19 + fonk 1+ p+h fon+24 + fön+1k ] i=0 п-2 (2p + h foi+sP + fei+7h` П fei+7P + f6i+6h, (fei+49 + fei+3k\ [ fon+29 + fon+1k + fon+19 + fonk foi+39 + foi+2k ) | p+h fen+29 + fen+1k i=0 Joi+7p + fei+6h) fea ad t o)fon+39 + fön+2k] [ fön+29 + fön+1k п-2 ( 2p +h' foi+sp+ fei+7h` (föi+49 + f6i+3k` П = r p+h i=0 16 Therefore п-1 foi+49 + föi+3k` fsi+39 + f6i+2k, foi+2P + fói+1h X6n-2 = r i=0 Also, other formulas can be proved similarly. Hence, the proof is completed.
Transcribed Image Text:Joi+7P + fotneh ) (Gi+19 + foi+3k` föi+39 + föi+2k , п-2 2p + h П foi+sP + fei+7h` X6n-2 p+h i=0 п-2 2p+h pth ) 11 (Foi+7p+f6i+6h f6i+8p+f6i+7h h r f6i+49+f6i+3k f6i+39+f6i+2k i=0 fênq+fên-1k h h + fon+19+fönk p п-2 foi+8P + foi+7h` П foi+7P+ föi+6h , foi+49 + f6i+3k` föi+39 + föi+2k , 2р + h X6n-2 p+ h i=0 п-2 2p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h föi+49+f6i+3k f6i+39+f6i+2k h Jr p+h i=0 + h|1+ fönq+fên-1k| fön+19+f6nk Soi+7P + feiseh ) (Bi+49 + foi+3k' fsi+39 + f6i+2k , п-2 2p + h ´fei+sP + fei+7h` X6n-2 p+h i=0 п-2 2р+h p+h II f6i+8p+f6i+7h f6i+7p+f6i+6h f6i+4q+f6i+3k` f6i+39+f6i+2k i=0 + fên+19+fénk+fênq+fén–1k fön+19+fönk (föi+8P+ f6i+7h` r (2p + 4) Ti fai sp + faiszh) (feireg + foisak \ foi+7P + f6i+6h п-2 П foi+49 + föi+3k` föi+39 + foi+2k, p+h i=0 п-2 2p+h p+h П f6i+8p+f6i+7h f6i+7p+f6i+6h f6i+49+f6i+3k f6i+3q+f6i+2k i=0 + fén+29+f6n+1k fon+19+ fönk (foi+49 + föi+3k fei+sP+ foi+7! (g+ foi+2k , п-2 ( 2p +h П föi+7P+ f6i+6h, fön+19 + fonk 1+ p+h fon+24 + fön+1k ] i=0 п-2 (2p + h foi+sP + fei+7h` П fei+7P + f6i+6h, (fei+49 + fei+3k\ [ fon+29 + fon+1k + fon+19 + fonk foi+39 + foi+2k ) | p+h fen+29 + fen+1k i=0 Joi+7p + fei+6h) fea ad t o)fon+39 + fön+2k] [ fön+29 + fön+1k п-2 ( 2p +h' foi+sp+ fei+7h` (föi+49 + f6i+3k` П = r p+h i=0 16 Therefore п-1 foi+49 + föi+3k` fsi+39 + f6i+2k, foi+2P + fói+1h X6n-2 = r i=0 Also, other formulas can be proved similarly. Hence, the proof is completed.
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