{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

AEM_3e_Chapter_05

# AEM_3e_Chapter_05 - 5 Series Solutions of Linear...

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

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 = 1 x + x 2 2 x 3 6 + x 4 24 − · · · 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 n =1 nc n x n +2 = k =3 ( k 2) c k 2 x k . 12. n =2 n ( n 1) c n x n + 2 n =2 n ( n 1) c n x n 2 + 3 n =1 nc n x n = 2 · 2 · 1 c 2 x 0 + 2 · 3 · 2 c 3 x 1 + 3 · 1 · c 1 x 1 + n =2 n ( n 1) c n x n k = n +2 n =4 n ( n 1) c n x n 2 k = n 2 +3 n =2 nc n x n k = n = 4 c 2 + (3 c 1 + 12 c 3 ) x + k =2 k ( k 1) c k x k + 2 k =2 ( k + 2)( k + 1) c k +2 x k + 3 k =2 kc k x k = 4 c 2 + (3 c 1 + 12 c 3 ) x + k =2 ( k ( k 1) + 3 k ) c k + 2( k + 2)( k + 1) c k +2 x k = 4 c 2 + (3 c 1 + 12 c 3 ) x + k =2 k ( k + 2) c k + 2( k + 1)( k + 2) c k +2 x k 15. The singular points of ( x 2 25) y + 2 xy + 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

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

View Full Document
5.1 Solutions About Ordinary Points 18. Substituting y = n =0 c n x n into the differential equation we have y + x 2 y = n =2 n ( n 1) c n x n 2 k = n 2 + n =0 c n x n +2 k = n +2 = k =0 ( k + 2)( k + 1) c k +2 x k + k =2 c k 2 x k = 2 c 2 + 6 c 3 x + k =2 [( k + 2)( k + 1) c k +2 + c k 2 ] x k = 0 . Thus c 2 = c 3 = 0 ( k + 2)( k + 1) c k +2 + c k 2 = 0 and c k +2 = 1 ( k + 2)( k + 1) c k 2 , k = 2 , 3 , 4 , . . . . Choosing c 0 = 1 and c 1 = 0 we find c 4 = 1 12 c 5 = c 6 = c 7 = 0 c 8 = 1 672 and so on. For c 0 = 0 and c 1 = 1 we obtain c 4 = 0 c 5 = 1 20 c 6 = c 7 = c 8 = 0 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 differential equation we have y + x 2 y + xy = n =2 n ( n 1) c n x n 2 k = n 2 + n =1 nc n x n +1 k = n +1 + n =0 c n x n +1 k = n +1 = k =0 ( k + 2)( k + 1) c k +2 x k + k =2 ( k 1) c k 1 x k + k =1 c k 1 x k = 2 c 2 + (6 c 3 + c 0 ) x + k =2 [( k + 2)( k + 1) c k +2 + kc k 1 ] x k = 0 . Thus c 2 = 0 6 c 3 + c 0 = 0 ( k + 2)( k + 1) c k +2 + kc k 1 = 0 76
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 find c 3 = 1 6 c 4 = c 5 = 0 c 6 = 1 45 and so on. For c 0 = 0 and c 1 = 1 we obtain c 3 = 0 c 4 = 1 6 c 5 = c 6 = 0 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 − · · · .

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### 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
Ask a homework question - tutors are online