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AEM_3e_Chapter05

AEM_3e_Chapter05 - 5 1 lim n Series Solutions of Linear...

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5 5 Series Solutions of Linear Differential Equations EXERCISES 5.1 Solutions About Ordinary Points 1. lim n →∞ 2 n +1 x n +1 / ( n + 1) 2 n x n /n = lim n →∞ 2 n n + 1 | x | = 2 | x | The series is absolutely convergent for 2 | x | < 1 or | x | < 1 2 . The radius of convergence is R = 1 2 . At x = 1 2 , the series n =1 ( 1) n /n converges by the alternating series test. At x = 1 2 , the series n =1 1 /n is the harmonic series which diverges. Thus, the given series converges on [ 1 2 , 1 2 ). 2. lim n →∞ 100 n +1 ( x + 7) n +1 / ( n + 1)! 100 n ( x + 7) n /n ! = lim n →∞ 100 n + 1 | x + 7 | = 0 The radius of convergence is R = . The series is absolutely convergent on ( −∞ , ). 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). 4. lim n →∞ ( n + 1)!( x 1) n +1 n !( x 1) n = lim n →∞ ( n + 1) | x 1 | = , x = 1 0 , x = 1 The radius of convergence is R = 0 and the series converges only for x = 1. 5. sin x cos x = x x 3 6 + x 5 120 x 7 5040 + · · · 1 x 2 2 + x 4 24 x 6 720 + · · · = x 2 x 3 3 + 2 x 5 15 4 x 7 315 + · · · 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 + · · · 7. 1 cos x = 1 1 x 2 2 + x 4 4! x 6 6! + · · · = 1 + x 2 2 + 5 x 4 4! + 61 x 6 6! + · · · Since cos( π/ 2) = cos( π/ 2) = 0, the series converges on ( π/ 2 , π/ 2). 8. 1 x 2 + x = 1 2 3 4 x + 3 8 x 2 3 16 x 3 + · · · Since the function is undefined at x = 2, the series converges on ( 2 , 2). 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 . 252
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5.1 Solutions About Ordinary Points 10. Let k = n 3 so that n = k + 3 and n =3 (2 n 1) c n x n 3 = k =0 (2 k + 5) c k +3 x k . 11. n =1 2 nc n x n 1 + n =0 6 c n x n +1 = 2 · 1 · c 1 x 0 + n =2 2 nc n x n 1 k = n 1 + n =0 6 c n x n +1 k = n +1 = 2 c 1 + k =1 2( k + 1) c k +1 x k + k =1 6 c k 1 x k = 2 c 1 + k =1 [2( k + 1) c k +1 + 6 c k 1 ] 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 13. y = n =1 ( 1) n +1 x n 1 , y = n =2 ( 1) n +1 ( n 1) x n 2 ( x + 1) y + y = ( x + 1) n =2 ( 1) n +1 ( n 1) x n 2 + n =1 ( 1) n +1 x n 1 = n =2 ( 1) n +1 ( n 1) x n 1 + n =2 ( 1) n +1 ( n 1) x n 2 + n =1 ( 1) n +1 x n 1 = x 0 + x 0 + n =2 ( 1) n +1 ( n 1) x n 1 k = n 1 + n =3 ( 1) n +1 ( n 1) x n 2 k = n 2 + n =2 ( 1) n +1 x n
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