AEM_3e_Chapter_05

AEM_3e_Chapter_05 - 5 Series Solutions of Linear...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
5 5 Series Solutions of Linear Differential Equations EXERCISES 5.1 Solutions About Ordinary Points 3. By the ratio test, lim n →∞ ¯ ¯ ¯ ¯ ( x 5) n +1 / 10 n +1 ( x 5) n / 10 n ¯ ¯ ¯ ¯ = lim n →∞ 1 10 | x 5 | = 1 10 | x 5 | . The series is absolutely convergent for 1 10 | x 5 | < 1, | x 5 | < 10, or on ( 5 , 15). The radius of convergence is R = 10. At x = 5, the series n =1 ( 1) n ( 10) n / 10 n = n =1 1 diverges by the n th term test. At x = 15, the series n =1 ( 1) n 10 n / 10 n = n =1 ( 1) n diverges by the n th term test. Thus, the series converges on ( 5 , 15). 6. e x cos x = µ 1 x + x 2 2 x 3 6 + x 4 24 −··· ¶µ 1 x 2 2 + x 4 24 =1 x + x 3 3 x 4 6 + ··· 9. Let k = n + 2 so that n = k 2 and X n =1 nc n x n +2 = X k =3 ( k 2) c k 2 x k . 12. X n =2 n ( n 1) c n x n +2 X n =2 n ( n 1) c n x n 2 +3 X n =1 nc n x n =2 · 2 · 1 c 2 x 0 · 3 · 2 c 3 x 1 · 1 · c 1 x 1 + X n =2 n ( n 1) c n x n | {z } k = n +2 X n =4 n ( n 1) c n x n 2 | {z } k = n 2 +3 X n =2 nc n x n | {z } k = n =4 c 2 +(3 c 1 +12 c 3 ) x + X k =2 k ( k 1) c k x k X k =2 ( k + 2)( k +1) c k +2 x k X k =2 kc k x k c 2 c 1 c 3 ) x + X k =2 ±( k ( k 1)+3 k ) c k +2( k + 2)( k c k +2 ² x k c 2 c 1 c 3 ) x + X k =2 ± k ( k +2) c k k + 1)( k c k +2 ² x k 15. The singular points of ( x 2 25) y 0 xy 0 + y = 0 are 5 and 5. The distance from 0 to either of these points is 5. The distance from 1 to the closest of these points is 4. 75
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
5.1 Solutions About Ordinary Points 18. Substituting y = n =0 c n x n into the diferential equation we have y 0 + x 2 y = X n =2 n ( n 1) c n x n 2 | {z } k = n 2 + X n =0 c n x n +2 | {z } k = n +2 = X k =0 ( k + 2)( k +1) c k +2 x k + X k =2 c k 2 x k =2 c 2 +6 c 3 x + X k =2 [( k + 2)( k c k +2 + c k 2 ] x k =0 . Thus c 2 = c 3 ( k + 2)( k c k +2 + c k 2 and c k +2 = 1 ( k + 2)( k c k 2 ,k , 3 , 4 ,... . Choosing c 0 = 1 and c 1 = 0 we Fnd c 4 = 1 12 c 5 = c 6 = c 7 c 8 = 1 672 and so on. ±or c 0 = 0 and c 1 = 1 we obtain c 4 c 5 = 1 20 c 6 = c 7 = c 8 c 9 = 1 1440 and so on. Thus, two solutions are y 1 =1 1 12 x 4 + 1 672 x 8 −··· and y 2 = x 1 20 x 5 + 1 1440 x 9 . 21. Substituting y = n =0 c n x n into the diferential equation we have y 0 + x 2 y 0 + xy = X n =2 n ( n 1) c n x n 2 | {z } k = n 2 + X n =1 nc n x n +1 | {z } k = n +1 + X n =0 c n x n +1 | {z } k = n +1 = X k =0 ( k + 2)( k c k +2 x k + X k =2 ( k 1) c k 1 x k + X k =1 c k 1 x k c 2 +(6 c 3 + c 0 ) x + X k =2 [( k + 2)( k c k +2 + kc k 1 ] x k . Thus c 2 6 c 3 + c 0 ( k + 2)( k c k +2 + k 1 76
Background image of page 2
5.1 Solutions About Ordinary Points and c 2 =0 c 3 = 1 6 c 0 c k +2 = k ( k + 2)( k +1) c k 1 ,k =2 , 3 , 4 ,... . Choosing c 0 = 1 and c 1 = 0 we fnd c 3 = 1 6 c 4 = c 5 c 6 = 1 45 and so on. For c 0 = 0 and c 1 = 1 we obtain c 3 c 4 = 1 6 c 5 = c 6 c 7 = 5 252 and so on. Thus, two solutions are y 1 =1 1 6 x 3 + 1 45 x 6 −··· and y 2 = x 1 6 x 4 + 5 252 x 7 . 24. Substituting y = n =0 c n x n into the di±erential equation we have ( x +2) y 0 + xy 0 y = X n =2 n ( n 1) c n x n 1 | {z } k = n 1 + X n =2 2 n ( n 1) c n x n 2 | {z } k = n 2 + X n =1 nc n x n | {z } k = n X n =0 c n x n | {z } k = n = X k =1 ( k kc k +1 x k + X k =0 2( k + 2)( k c k +2 x k + X k =1 k x k
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.

Page1 / 20

AEM_3e_Chapter_05 - 5 Series Solutions of Linear...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online