AEM_3e_Chapter05

AEM_3e_Chapter05 - Series Solutions of Linear Differential...

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Unformatted text preview: Series Solutions of Linear Differential Equations 5 EXERCISES 5.1 Solutions About Ordinary Points 1. lim n→∞ 2n+1 xn+1 /(n + 1) 2n = lim |x| = 2|x| n xn /n n→∞ n + 1 2 The series is absolutely convergent for 2|x| < 1 or |x| < 1 . The radius of convergence is R = 1 . At x = − 1 , the 2 2 2 ∞ ∞ series n=1 (−1)n /n converges by the alternating series test. At x = 1 , the series n=1 1/n is the harmonic 2 series which diverges. Thus, the given series converges on [− 1 , 1 ). 2 2 2. lim n→∞ 100n+1 (x + 7)n+1 /(n + 1)! 100 = lim |x + 7| = 0 n→∞ n + 1 100n (x + 7)n /n! The radius of convergence is R = ∞. The series is absolutely convergent on (−∞, ∞). 3. By the ratio test, lim n→∞ (x − 5)n+1 /10n+1 1 1 = lim |x − 5| = |x − 5|. n→∞ 10 (x − 5)n /10n 10 The series is absolutely convergent for R = 10. At x = −5, the series series ∞ n n n n=1 (−1) 10 /10 = 1 10 |x n ∞ n=1 (−1) ∞ n n=1 (−1) − 5| < 1, |x − 5| < 10, or on (−5, 15). The radius of convergence is ∞ n=1 (−10)n /10n = 1 diverges by the nth term test. At x = 15, the diverges by the nth term test. Thus, the series converges on (−5, 15). ∞, x = 1 (n + 1)!(x − 1)n+1 = lim (n + 1)|x − 1| = n n→∞ n→∞ n!(x − 1) 0, x = 1 The radius of convergence is R = 0 and the series converges only for x = 1. 4. lim 5. sin x cos x = 6. e−x cos x = x− x3 x5 x7 + − + ··· 6 120 5040 1− x2 x4 x6 + − + ··· 2 24 720 x2 x3 x4 − + − ··· 2 6 24 1− x2 x4 + − ··· 2 24 1−x+ =x− =1−x+ 1 1 5x4 61x6 x2 = + + + ··· =1+ 2 4 6 cos x 2 4! 6! 1 − x + x − x + ··· 2 4! 6! Since cos(π/2) = cos(−π/2) = 0, the series converges on (−π/2, π/2). 1−x 1 3 3 3 8. = − x + x2 − x3 + · · · 2+x 2 4 8 16 Since the function is undefined at x = −2, the series converges on (−2, 2). 7. 9. Let k = n + 2 so that n = k − 2 and ∞ ∞ (k − 2)ck−2 xk . ncn xn+2 = n=1 k=3 252 2x3 2x5 4x7 + − + ··· 3 15 315 x3 x4 − + ··· 3 6 5.1 Solutions About Ordinary Points 10. Let k = n − 3 so that n = k + 3 and ∞ ∞ (2n − 1)cn xn−3 = n=3 ∞ k=0 ∞ n=1 ∞ ∞ 6cn xn+1 = 2 · 1 · c1 x0 + 2ncn xn−1 + 11. (2k + 5)ck+3 xk . n=0 2ncn xn−1 + n=2 6cn xn+1 n=0 k=n−1 k=n+1 ∞ ∞ 2(k + 1)ck+1 xk + = 2c1 + 6ck−1 xk k=1 k=1 ∞ [2(k + 1)ck+1 + 6ck−1 ]xk = 2c1 + k=1 ∞ ∞ ∞ n(n − 1)cn xn + 2 12. n(n − 1)cn xn−2 + 3 n=2 n=2 ncn xn n=1 ∞ = 2 · 2 · 1c2 x + 2 · 3 · 2c3 x + 3 · 1 · c1 x + 0 1 ∞ n(n − 1)cn x +2 1 n=2 ∞ n(n − 1)cn x n n−2 n=2 k=n ∞ k=n−2 ∞ k(k − 1)ck xk + 2 = 4c2 + (3c1 + 12c3 )x + k=2 ∞ ncn xn +3 n=4 k=n ∞ (k + 2)(k + 1)ck+2 xk + 3 k=2 kck xk k=2 k(k − 1) + 3k ck + 2(k + 2)(k + 1)ck+2 xk = 4c2 + (3c1 + 12c3 )x + k=2 ∞ k(k + 2)ck + 2(k + 1)(k + 2)ck+2 xk = 4c2 + (3c1 + 12c3 )x + k=2 ∞ ∞ (−1)n+1 xn−1 , 13. y = (−1)n+1 (n − 1)xn−2 y = n=1 n=2 ∞ ∞ (−1)n+1 (n − 1)xn−2 + (x + 1)y + y = (x + 1) n=2 (−1)n+1 xn−1 n=1 ∞ ∞ (−1)n+1 (n − 1)xn−1 + = n=2 ∞ (−1)n+1 (n − 1)xn−2 + n=2 n=1 ∞ = −x0 + x0 + ∞ ∞ (−1)n+1 (n − 1)xn−1 + n=2 (−1)n+1 (n − 1)xn−2 + n=3 ∞ k=n−2 ∞ (−1)k+2 kxk + k=1 ∞ ∞ (−1)k+3 (k + 1)xk + k=1 (−1)k+2 xk k=1 (−1)k+2 k − (−1)k+2 k − (−1)k+2 + (−1)k+2 xk = 0 = k=1 ∞ 14. y = (−1)n 2n 2n−1 x , 22n (n!)2 n=1 ∞ y = (−1)n 2n(2n − 1) 2n−2 x 22n (n!)2 n=1 253 (−1)n+1 xn−1 n=2 k=n−1 = (−1)n+1 xn−1 k=n−1 5.1 Solutions About Ordinary Points ∞ xy + y + xy = ∞ ∞ (−1)n 2n(2n − 1) 2n−1 (−1)n 2n 2n−1 (−1)n 2n+1 x + x + x 22n (n!)2 22n (n!)2 22n (n!)2 n=1 n=1 n=0 k=n ∞ k=n = k=1 ∞ = k=1 ∞ k k−1 (−1)k (2k)2 (−1)k x2k−1 − 2k−2 22k (k!)2 2 [(k − 1)!]2 (−1)k = k=n+1 (−1) 2k(2k − 1) (−1) 2k (−1) x2k−1 + 2k + 2k−2 2k (k!)2 2 2 2 (k!) 2 [(k − 1)!]2 k k=1 (2k)2 − 22 k 2 2k−1 x =0 22k (k!)2 15. The singular points of (x2 − 25)y + 2xy + 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. 16. The singular points of (x2 − 2x + 10)y + xy − 4y = 0 are 1 + 3i and 1 − 3i. The distance from 0 to either of √ these points is 10 . The distance from 1 to either of these points is 3. 17. Substituting y = ∞ n=0 cn xn into the differential equation we have ∞ y − xy = ∞ n(n − 1)cn xn−2 − n=2 ∞ ∞ (k + 2)(k + 1)ck+2 xk − cn xn+1 = n=0 k=n−2 k=0 ck−1 xk k=1 k=n+1 ∞ [(k + 2)(k + 1)ck+2 − ck−1 ]xk = 0. = 2c2 + k=1 Thus c2 = 0 (k + 2)(k + 1)ck+2 − ck−1 = 0 and ck+2 = 1 ck−1 , (k + 2)(k + 1) k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 6 c4 = c5 = 0 1 c6 = 180 c3 = and so on. For c0 = 0 and c1 = 1 we obtain c3 = 0 1 c4 = 12 c5 = c6 = 0 1 c7 = 504 and so on. Thus, two solutions are 1 1 6 y1 = 1 + x3 + x + ··· 6 180 and 254 y2 = x + 1 4 1 7 x + x + ··· . 12 504 5.1 18. Substituting y = ∞ n=0 Solutions About Ordinary Points cn xn into the differential equation we have ∞ ∞ n(n − 1)cn x 2 n−2 y +x y = n=2 ∞ n+2 + cn x = n=0 k=n−2 ∞ k ck−2 xk (k + 2)(k + 1)ck+2 x + k=0 k=2 k=n+2 ∞ [(k + 2)(k + 1)ck+2 + ck−2 ]xk = 0. = 2c2 + 6c3 x + k=2 Thus c2 = c3 = 0 (k + 2)(k + 1)ck+2 + ck−2 = 0 and ck+2 = − 1 ck−2 , (k + 2)(k + 1) k = 2, 3, 4, . . . . Choosing c0 = 1 and c1 = 0 we find 1 12 c5 = c6 = c7 = 0 1 c8 = 672 c4 = − and so on. For c0 = 0 and c1 = 1 we obtain c4 = 0 1 20 c6 = c7 = c8 = 0 1 c9 = 1440 c5 = − and so on. Thus, two solutions are y1 = 1 − 19. Substituting y = ∞ n=0 1 4 1 8 x + x − ··· 12 672 and y2 = x − 1 5 1 9 x + x − ··· . 20 1440 cn xn into the differential equation we have ∞ ∞ y − 2xy + y = ∞ n(n − 1)cn xn−2 − 2 n=2 ncn xn + n=1 k=n−2 n=0 k=n ∞ k=n ∞ (k + 2)(k + 1)ck+2 xk − 2 = cn xn k=0 ∞ kck xk + k=1 ck xk k=0 ∞ [(k + 2)(k + 1)ck+2 − (2k − 1)ck ]xk = 0. = 2c2 + c0 + k=1 Thus 2c2 + c0 = 0 (k + 2)(k + 1)ck+2 − (2k − 1)ck = 0 255 5.1 Solutions About Ordinary Points and 1 c2 = − c0 2 2k − 1 ck+2 = ck , (k + 2)(k + 1) k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 2 c3 = c5 = c7 = · · · = 0 1 c4 = − 8 7 c6 = − 240 c2 = − and so on. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 1 c3 = 6 1 c5 = 24 1 c7 = 112 and so on. Thus, two solutions are 1 1 7 6 y1 = 1 − x2 − x4 − x − ··· 2 8 240 20. Substituting y = ∞ n=0 1 1 1 7 y2 = x + x3 + x5 + x + ··· . 6 24 112 and cn xn into the differential equation we have ∞ ∞ y − xy + 2y = ∞ n(n − 1)cn xn−2 − n=2 ncn xn + 2 n=1 k=n−2 n=0 k=n ∞ k=n ∞ (k + 2)(k + 1)ck+2 xk − = cn xn k=0 ∞ kck xk + 2 k=1 ck xk k=0 ∞ [(k + 2)(k + 1)ck+2 − (k − 2)ck ]xk = 0. = 2c2 + 2c0 + k=1 Thus 2c2 + 2c0 = 0 (k + 2)(k + 1)ck+2 − (k − 2)ck = 0 and c2 = −c0 ck+2 = k−2 ck , (k + 2)(k + 1) 256 k = 1, 2, 3, . . . . 5.1 Solutions About Ordinary Points Choosing c0 = 1 and c1 = 0 we find c2 = −1 c3 = c5 = c7 = · · · = 0 c4 = 0 c6 = c8 = c10 = · · · = 0. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 1 c3 = − 6 1 c5 = − 120 and so on. Thus, two solutions are y1 = 1 − x2 21. Substituting y = ∞ n=0 and 1 1 5 y2 = x − x3 − x − ··· . 6 120 cn xn into the differential equation we have ∞ ∞ ∞ n(n − 1)cn xn−2 + y + x2 y + xy = n=2 ncn xn+1 + n=1 k=n−2 n=0 k=n+1 ∞ k=n+1 ∞ k=0 ∞ (k − 1)ck−1 xk + (k + 2)(k + 1)ck+2 xk + = cn xn+1 k=2 ck−1 xk k=1 ∞ [(k + 2)(k + 1)ck+2 + kck−1 ]xk = 0. = 2c2 + (6c3 + c0 )x + k=2 Thus c2 = 0 6c3 + c0 = 0 (k + 2)(k + 1)ck+2 + kck−1 = 0 and c2 = 0 1 c3 = − c0 6 ck+2 = − k ck−1 , (k + 2)(k + 1) Choosing c0 = 1 and c1 = 0 we find 1 6 c4 = c5 = 0 1 c6 = 45 c3 = − 257 k = 2, 3, 4, . . . . 5.1 Solutions About Ordinary Points and so on. For c0 = 0 and c1 = 1 we obtain c3 = 0 1 6 c5 = c6 = 0 5 c7 = 252 c4 = − and so on. Thus, two solutions are 1 1 y1 = 1 − x3 + x6 − · · · 6 45 22. Substituting y = ∞ n=0 1 5 7 x − ··· . y2 = x − x4 + 6 252 and cn xn into the differential equation we have ∞ ∞ ∞ n(n − 1)cn xn−2 + 2 y + 2xy + 2y = n=2 ncn xn + 2 n=1 k=n−2 n=0 k=n ∞ k=n ∞ (k + 2)(k + 1)ck+2 xk + 2 = cn xn k=0 ∞ kck xk + 2 k=1 ck xk k=0 ∞ [(k + 2)(k + 1)ck+2 + 2(k + 1)ck ]xk = 0. = 2c2 + 2c0 + k=1 Thus 2c2 + 2c0 = 0 (k + 2)(k + 1)ck+2 + 2(k + 1)ck = 0 and c2 = −c0 ck+2 = − 2 ck , k+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find c2 = −1 c3 = c5 = c7 = · · · = 0 c4 = 1 2 c6 = − 1 6 and so on. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 2 c3 = − 3 4 c5 = 15 8 c7 = − 105 258 5.1 and so on. Thus, two solutions are 1 1 y1 = 1 − x2 + x4 − x6 + · · · 2 6 23. Substituting y = ∞ n=0 Solutions About Ordinary Points 2 4 8 7 x + ··· . and y2 = x − x3 + x5 − 3 15 105 cn xn into the differential equation we have ∞ ∞ (x − 1)y + y = ∞ n(n − 1)cn xn−1 − n=2 n(n − 1)cn xn−2 + n=2 ncn xn−1 n=1 k=n−1 k=n−2 ∞ k=n−1 ∞ (k + 1)kck+1 xk − = k=1 ∞ (k + 2)(k + 1)ck+2 xk + k=0 (k + 1)ck+1 xk k=0 ∞ = −2c2 + c1 + [(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 ]xk = 0. k=1 Thus −2c2 + c1 = 0 (k + 1)2 ck+1 − (k + 2)(k + 1)ck+2 = 0 and 1 c1 2 k+1 = ck+1 , k+2 c2 = ck+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find c2 = c3 = c4 = · · · = 0. For c0 = 0 and c1 = 1 we obtain 1 1 1 c2 = , c3 = , c4 = , 2 3 4 and so on. Thus, two solutions are 1 1 1 y1 = 1 and y2 = x + x2 + x3 + x4 + · · · . 2 3 4 24. Substituting y = ∞ n=0 cn xn into the differential equation we have ∞ ∞ (x + 2)y + xy − y = ∞ n(n − 1)cn xn−1 + n=2 n=2 k=n−1 k=1 cn xn n=0 k=n ∞ (k + 1)kck+1 xk + = ncn xn − n=1 k=n−2 ∞ ∞ 2n(n − 1)cn xn−2 + k=n ∞ k=0 ∞ kck xk − 2(k + 2)(k + 1)ck+2 xk + k=1 ck xk k=0 ∞ = 4c2 − c0 + (k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck xk = 0. k=1 Thus 4c2 − c0 = 0 (k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck = 0, and 1 c0 4 (k + 1)kck+1 + (k − 1)ck , =− 2(k + 2)(k + 1) k = 1, 2, 3, . . . c2 = ck+2 259 k = 1, 2, 3, . . . . 5.1 Solutions About Ordinary Points Choosing c0 = 1 and c1 = 0 we find c1 = 0, c2 = 1 , 4 c3 = − 1 , 24 c4 = 0, c5 = 1 480 and so on. For c0 = 0 and c1 = 1 we obtain c2 = 0 c3 = 0 c4 = c5 = c6 = · · · = 0. Thus, two solutions are 1 1 1 5 y1 = c0 1 + x2 − x3 + x + ··· 4 24 480 25. Substituting y = ∞ n=0 and y2 = c1 x. cn xn into the differential equation we have ∞ ∞ y − (x + 1)y − y = ∞ n(n − 1)cn xn−2 − n=2 n=1 k=n−2 ncn xn−1 − n=1 k=n ∞ k=0 cn xn n=0 k=n−1 ∞ (k + 2)(k + 1)ck+2 xk − = ∞ ncn xn − k=n ∞ kck xk − k=1 ∞ (k + 1)ck+1 xk − k=0 ck xk k=0 ∞ = 2c2 − c1 − c0 + [(k + 2)(k + 1)ck+2 − (k + 1)ck+1 − (k + 1)ck ]xk = 0. k=1 Thus 2c2 − c1 − c0 = 0 (k + 2)(k + 1)ck+2 − (k + 1)(ck+1 + ck ) = 0 and c1 + c0 2 ck+1 + ck = , k+2 c2 = ck+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find c2 = 1 , 2 c3 = 1 , 6 c4 = 1 , 6 1 , 2 c3 = 1 , 2 c4 = 1 , 4 and so on. For c0 = 0 and c1 = 1 we obtain c2 = and so on. Thus, two solutions are 1 1 1 y1 = 1 + x2 + x3 + x4 + · · · 2 6 6 and 260 1 1 1 y2 = x + x2 + x3 + x4 + · · · . 2 2 4 5.1 26. Substituting y = ∞ n=0 Solutions About Ordinary Points cn xn into the differential equation we have ∞ x2 + 1 y − 6y = ∞ ∞ n(n − 1)cn xn + n=2 n(n − 1)cn xn−2 − 6 n=2 n=0 k=n k=n−2 ∞ k=n ∞ ∞ k(k − 1)ck xk + = cn xn k=2 (k + 2)(k + 1)ck+2 xk − 6 k=0 ck xk k=0 ∞ = 2c2 − 6c0 + (6c3 − 6c1 )x + k 2 − k − 6 ck + (k + 2)(k + 1)ck+2 xk = 0. k=2 Thus 2c2 − 6c0 = 0 6c3 − 6c1 = 0 (k − 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0 and c2 = 3c0 c3 = c1 ck+2 = − k−3 ck , k+1 k = 2, 3, 4, . . . . Choosing c0 = 1 and c1 = 0 we find c2 = 3 c3 = c5 = c7 = · · · = 0 c4 = 1 c6 = − 1 5 and so on. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 c3 = 1 c5 = c7 = c9 = · · · = 0. Thus, two solutions are 1 y1 = 1 + 3x2 + x4 − x6 + · · · 5 27. Substituting y = ∞ n=0 and y2 = x + x3 . cn xn into the differential equation we have ∞ x2 + 2 y + 3xy − y = ∞ n(n − 1)cn xn + 2 n=2 n=2 k=n ∞ k=2 ∞ ncn xn − n=1 k=n−2 ∞ k(k − 1)ck xk + 2 = ∞ n(n − 1)cn xn−2 + 3 k=1 k=n ∞ kck xk − (k + 2)(k + 1)ck+2 xk + 3 k=0 n=0 k=n ∞ cn xn ck xk k=0 ∞ = (4c2 − c0 ) + (12c3 + 2c1 )x + 2(k + 2)(k + 1)ck+2 + k 2 + 2k − 1 ck xk = 0. k=2 261 5.1 Solutions About Ordinary Points Thus 4c2 − c0 = 0 12c3 + 2c1 = 0 2(k + 2)(k + 1)ck+2 + k 2 + 2k − 1 ck = 0 and 1 c0 4 1 c3 = − c1 6 k 2 + 2k − 1 ck+2 = − ck , 2(k + 2)(k + 1) c2 = k = 2, 3, 4, . . . . Choosing c0 = 1 and c1 = 0 we find 1 4 c3 = c5 = c7 = · · · = 0 7 c4 = − 96 c2 = and so on. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 1 c3 = − 6 7 c5 = 120 and so on. Thus, two solutions are 1 7 y1 = 1 + x2 − x4 + · · · 4 96 28. Substituting y = ∞ n=0 1 7 5 y2 = x − x3 + x − ··· . 6 120 and cn xn into the differential equation we have ∞ x2 − 1 y + xy − y = ∞ ∞ n(n − 1)cn xn − n=2 n=2 k=n k=2 n=1 cn xn n=0 k=n ∞ k=n ∞ k(k − 1)ck xk − = ncn xn − k=n−2 ∞ ∞ n(n − 1)cn xn−2 + k=0 ∞ kck xk − (k + 2)(k + 1)ck+2 xk + k=1 ck xk k=0 ∞ = (−2c2 − c0 ) − 6c3 x + −(k + 2)(k + 1)ck+2 + k 2 − 1 ck xk = 0. k=2 Thus −2c2 − c0 = 0 −6c3 = 0 −(k + 2)(k + 1)ck+2 + (k − 1)(k + 1)ck = 0 262 5.1 and Solutions About Ordinary Points 1 c2 = − c0 2 c3 = 0 ck+2 = k−1 ck , k+2 k = 2, 3, 4, . . . . Choosing c0 = 1 and c1 = 0 we find 1 2 c3 = c5 = c7 = · · · = 0 1 c4 = − 8 c2 = − and so on. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 c3 = c5 = c7 = · · · = 0. Thus, two solutions are 1 1 y1 = 1 − x2 − x4 − · · · 2 8 29. Substituting y = ∞ n=0 and y2 = x. cn xn into the differential equation we have ∞ ∞ (x − 1)y − xy + y = n(n − 1)cn xn−1 − n=2 ∞ n=2 k=n−1 ∞ k=1 ncn xn + n=1 k=n−2 ∞ (k + 2)(k + 1)ck+2 xk − k=0 cn xn n=0 k=n ∞ (k + 1)kck+1 xk − = ∞ n(n − 1)cn xn−2 − k=n ∞ kck xk + k=1 ck xk k=0 ∞ = −2c2 + c0 + [−(k + 2)(k + 1)ck+2 + (k + 1)kck+1 − (k − 1)ck ]xk = 0. k=1 Thus −2c2 + c0 = 0 −(k + 2)(k + 1)ck+2 + (k − 1)kck+1 − (k − 1)ck = 0 and 1 c0 2 kck+1 (k − 1)ck = − , k+2 (k + 2)(k + 1) c2 = ck+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 1 , c3 = , c4 = 0, 2 6 and so on. For c0 = 0 and c1 = 1 we obtain c2 = c3 = c4 = · · · = 0. Thus, c2 = 1 1 y = C1 1 + x2 + x3 + · · · + C2 x 2 6 and 1 y = C1 x + x2 + · · · + C2 . 2 263 5.1 Solutions About Ordinary Points The initial conditions imply C1 = −2 and C2 = 6, so 1 1 y = −2 1 + x2 + x3 + · · · + 6x = 8x − 2ex . 2 6 ∞ n=0 30. Substituting y = cn xn into the differential equation we have (x+1)y − (2 − x)y + y ∞ ∞ ∞ n(n − 1)cn xn−1 + = n=2 n=2 k=n−1 n=1 k=1 ncn xn + n=1 k=n−1 ∞ k=0 cn xn n=0 k=n ∞ (k + 2)(k + 1)ck+2 xk − 2 (k + 1)kck+1 xk + = ∞ ncn xn−1 + k=n−2 ∞ ∞ n(n − 1)cn xn−2 − 2 k=n ∞ (k + 1)ck+1 xk + k=0 ∞ kck xk + k=1 ∞ = 2c2 − 2c1 + c0 + [(k + 2)(k + 1)ck+2 − (k + 1)ck+1 + (k + 1)ck ]xk = 0. k=1 Thus 2c2 − 2c1 + c0 = 0 (k + 2)(k + 1)ck+2 − (k + 1)ck+1 + (k + 1)ck = 0 and 1 c2 = c1 − c0 2 1 1 ck+1 − ck , ck+2 = k+2 k+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 c2 = − , 2 1 c3 = − , 6 c4 = 1 , 12 and so on. For c0 = 0 and c1 = 1 we obtain c2 = 1, c3 = 0, 1 c4 = − , 4 and so on. Thus, 1 1 1 1 y = C1 1 − x2 − x3 + x4 + · · · + C2 x + x2 − x4 + · · · 2 6 12 4 and 1 1 y = C1 −x − x2 + x3 + · · · + C2 1 + 2x − x3 + · · · . 2 3 The initial conditions imply C1 = 2 and C2 = −1, so 1 1 1 1 y = 2 1 − x2 − x3 + x4 + · · · − x + x2 − x4 + · · · 2 6 12 4 1 5 = 2 − x − 2x2 − x3 + x4 + · · · . 3 12 264 ck xk k=0 5.1 31. Substituting y = ∞ n=0 Solutions About Ordinary Points cn xn into the differential equation we have ∞ ∞ y − 2xy + 8y = ∞ n(n − 1)cn xn−2 − 2 n=2 ncn xn + 8 n=1 k=n−2 n=0 k=n ∞ k=n ∞ (k + 2)(k + 1)ck+2 xk − 2 = cn xn k=0 ∞ kck xk + 8 k=1 ck xk k=0 ∞ [(k + 2)(k + 1)ck+2 + (8 − 2k)ck ]xk = 0. = 2c2 + 8c0 + k=1 Thus 2c2 + 8c0 = 0 (k + 2)(k + 1)ck+2 + (8 − 2k)ck = 0 and c2 = −4c0 ck+2 = 2(k − 4) ck , (k + 2)(k + 1) k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find c2 = −4 c3 = c5 = c7 = · · · = 0 4 c4 = 3 c6 = c8 = c10 = · · · = 0. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 c3 = −1 1 c5 = 10 and so on. Thus, 4 y = C1 1 − 4x2 + x4 3 + C2 x − x3 + 1 5 x + ··· 10 and y = C1 −8x + 16 3 x 3 1 + C2 1 − 3x2 + x4 + · · · . 2 The initial conditions imply C1 = 3 and C2 = 0, so 4 y = 3 1 − 4x2 + x4 3 265 = 3 − 12x2 + 4x4 . 5.1 Solutions About Ordinary Points 32. Substituting y = ∞ n=0 cn xn into the differential equation we have ∞ ∞ ∞ n(n − 1)cn xn + (x2 + 1)y + 2xy = n=2 n(n − 1)cn xn−2 + n=2 n=1 k=n k=n−2 ∞ k=n ∞ k(k − 1)ck xk + = 2ncn xn k=2 ∞ (k + 2)(k + 1)ck+2 xk + k=0 ∞ 2kck xk k=1 k(k + 1)ck + (k + 2)(k + 1)ck+2 xk = 0. = 2c2 + (6c3 + 2c1 )x + k=2 Thus 2c2 = 0 6c3 + 2c1 = 0 k(k + 1)ck + (k + 2)(k + 1)ck+2 = 0 and c2 = 0 1 c3 = − c1 3 k ck+2 = − ck , k+2 k = 2, 3, 4, . . . . Choosing c0 = 1 and c1 = 0 we find c3 = c4 = c5 = · · · = 0. For c0 = 0 and c1 = 1 we obtain 1 3 c4 = c6 = c8 = · · · = 0 1 c5 = − 5 1 c7 = 7 c3 = − and so on. Thus 1 1 1 y = C0 + C1 x − x3 + x5 − x7 + · · · 3 5 7 and y = c1 1 − x2 + x4 − x6 + · · · . The initial conditions imply c0 = 0 and c1 = 1, so 1 1 1 y = x − x3 + x5 − x7 + · · · . 3 5 7 33. Substituting y = ∞ n=0 cn xn into the differential equation we have ∞ y + (sin x)y = 1 1 5 n(n − 1)cn xn−2 + x − x3 + x − ··· 6 120 n=2 c0 + c1 x + c2 x2 + · · · 1 = 2c2 + 6c3 x + 12c4 x2 + 20c5 x3 + · · · + c0 x + c1 x2 + c2 − c0 x3 + · · · 6 1 = 2c2 + (6c3 + c0 )x + (12c4 + c1 )x2 + 20c5 + c2 − c0 x3 + · · · = 0. 6 266 5.1 Solutions About Ordinary Points Thus 2c2 = 0 6c3 + c0 = 0 12c4 + c1 = 0 1 20c5 + c2 − c0 = 0 6 c2 = 0 and 1 c3 = − c0 6 1 c4 = − c1 12 1 1 c0 . c5 = − c2 + 20 120 Choosing c0 = 1 and c1 = 0 we find 1 c3 = − , 6 c2 = 0, c4 = 0, c5 = 1 120 and so on. For c0 = 0 and c1 = 1 we obtain c2 = 0, c4 = − c3 = 0, 1 , 12 c5 = 0 and so on. Thus, two solutions are 1 1 5 y1 = 1 − x3 + x + ··· 6 120 34. Substituting y = ∞ n=0 and y2 = x − 1 4 x + ··· . 12 cn xn into the differential equation we have ∞ y + ex y − y = n(n − 1)cn xn−2 n=2 1 1 + 1 + x + x2 + x3 + · · · 2 6 ∞ c1 + 2c2 x + 3c3 x2 + 4c4 x3 + · · · − cn xn n=0 = 2c2 + 6c3 x + 12c4 x + 20c5 x + · · · 2 3 1 + c1 + (2c2 + c1 )x + 3c3 + 2c2 + c1 x2 + · · · − [c0 + c1 x + c2 x2 + · · ·] 2 1 = (2c2 + c1 − c0 ) + (6c3 + 2c2 )x + 12c4 + 3c3 + c2 + c1 x2 + · · · = 0. 2 Thus 2c2 + c1 − c0 = 0 6c3 + 2c2 = 0 1 12c4 + 3c3 + c2 + c1 = 0 2 267 5.1 Solutions About Ordinary Points and 1 1 c0 − c1 2 2 1 c3 = − c2 3 1 1 1 c4 = − c3 + c2 − c1 . 4 12 24 c2 = Choosing c0 = 1 and c1 = 0 we find c2 = 1 , 2 1 c3 = − , 6 c4 = 0 and so on. For c0 = 0 and c1 = 1 we obtain 1 c2 = − , 2 c3 = 1 , 6 c4 = − 1 24 and so on. Thus, two solutions are 1 1 y1 = 1 + x2 − x3 + · · · 2 6 1 1 1 y2 = x − x2 + x3 − x4 + · · · . 2 6 24 and 35. The singular points of (cos x)y + y + 5y = 0 are odd integer multiples of π/2. The distance from 0 to either ±π/2 is π/2. The singular point closest to 1 is π/2. The distance from 1 to the closest singular point is then π/2 − 1. 36. Substituting y = n=0 cn xn into the first differential equation leads to ∞ y − xy = ∞ n(n − 1)cn xn−2 − n=2 ∞ n=0 k=n−2 ∞ (k + 2)(k + 1)ck+2 xk − cn xn+1 = k=0 k=n+1 ∞ [(k + 2)(k + 1)ck+2 − ck−1 ]xk = 1. = 2c2 + k=1 Thus 2c2 = 1 (k + 2)(k + 1)ck+2 − ck−1 = 0 and c2 = ck+2 = 1 2 ck−1 , (k + 2)(k + 1) k = 1, 2, 3, . . . . Let c0 and c1 be arbitrary and iterate to find 1 2 1 c3 = c0 6 1 c4 = c1 12 1 1 c5 = c2 = 20 40 c2 = 268 ck−1 xk k=1 5.1 Solutions About Ordinary Points and so on. The solution is 1 1 1 1 y = c0 + c1 x + x2 + c0 x3 + c1 x4 + c5 + · · · 2 6 12 40 1 1 1 1 = c0 1 + x3 + · · · + c1 x + x4 + · · · + x2 + x5 + · · · . 6 12 2 40 Substituting y = ∞ n=0 cn xn into the second differential equation leads to ∞ ∞ y − 4xy − 4y = ∞ n(n − 1)cn xn−2 − n=2 4ncn xn − n=1 k=n−2 n=0 k=n ∞ k=n ∞ (k + 2)(k + 1)ck+2 xk − = 4cn xn k=0 ∞ 4kck xk − k=1 4ck xk k=0 ∞ = 2c2 − 4c0 + (k + 2)(k + 1)ck+2 − 4(k + 1)ck xk k=1 ∞ = ex = 1 + k=1 1 k x . k! Thus 2c2 − 4c0 = 1 (k + 2)(k + 1)ck+2 − 4(k + 1)ck = 1 k! and 1 + 2c0 2 1 4 = + ck , (k + 2)! k + 2 c2 = ck+2 k = 1, 2, 3, . . . . Let c0 and c1 be arbitrary and iterate to find 1 + 2c0 2 4 4 1 1 + c1 = + c1 c3 = 3! 3 3! 3 c2 = c4 = 1 1 13 4 1 + c2 = + + 2c0 = + 2c0 4! 4 4! 2 4! c5 = 4 4 16 1 1 17 16 + c3 = + + c1 = + c1 5! 5 5! 5 · 3! 15 5! 15 c6 = 1 1 261 4 4 4 · 13 8 + c4 = + + c0 = + c0 6! 6 6! 6 · 4! 6 6! 3 c7 = 4 4 · 17 64 64 1 1 409 + c5 = + + c1 = + c1 7! 7 7! 7 · 5! 105 7! 105 and so on. The solution is 269 5.1 Solutions About Ordinary Points y = c0 + c1 x + + 1 + 2c0 x2 + 2 261 4 + c0 x6 + 6! 3 1 4 + c1 x3 + 3! 3 13 + 2c0 x4 + 4! 17 16 + c1 x5 5! 15 64 409 + c1 x7 + · · · 7! 105 4 4 16 64 7 = c0 1 + 2x2 + 2x4 + x6 + · · · + c1 x + x3 + x5 + x + ··· 3 3 15 105 1 1 13 17 261 6 409 7 + x2 + x3 + x4 + x5 + x + x + ··· . 2 3! 4! 5! 6! 7! 37. We identify P (x) = 0 and Q(x) = sin x/x. The Taylor series representation for sin x/x is 1 − x2 /3! + x4 /5! − · · · , for |x| < ∞. Thus, Q(x) is analytic at x = 0 and x = 0 is an ordinary point of the differential equation. 38. If x > 0 and y > 0, then y = −xy < 0 and the graph of a solution curve is concave down. Thus, whatever portion of a solution curve lies in the first quadrant is concave down. When x > 0 and y < 0, y = −xy > 0, so whatever portion of a solution curve lies in the fourth quadrant is concave up. 39. (a) Substituting y = ∞ n n=0 cn x into the differential equation we have ∞ ∞ ∞ n(n − 1)cn xn−2 + y + xy + y = n=2 ncn xn + n=1 k=n−2 n=0 k=n ∞ cn xn k=n ∞ (k + 2)(k + 1)ck+2 xk + = k=0 ∞ kck xk + k=1 ck xk k=0 ∞ (k + 2)(k + 1)ck+2 + (k + 1)ck xk = 0. = (2c2 + c0 ) + k=1 Thus 2c2 + c0 = 0 (k + 2)(k + 1)ck+2 + (k + 1)ck = 0 and 1 c2 = − c0 2 1 ck+2 = − ck , k+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 2 c3 = c5 = c7 = · · · = 0 1 1 1 c4 = − − = 2 4 2 2 ·2 1 1 1 c6 = − =− 3 6 22 · 2 2 · 3! c2 = − and so on. For c0 = 0 and c1 = 1 we obtain 270 5.1 Solutions About Ordinary Points c2 = c4 = c6 = · · · = 0 1 2 c3 = − = − 3 3! 1 1 1 4·2 c5 = − − = = 5 3 5·3 5! 6·4·2 1 4·2 c7 = − =− 7 5! 7! and so on. Thus, two solutions are ∞ y1 = k=0 ∞ (−1)k 2k x 2k · k! and y2 = k=0 (−1)k 2k k! 2k+1 x . (2k + 1)! (b) For y1 , S3 = S2 and S5 = S4 , so we plot S2 , S4 , S6 , S8 , and S10 . y y 4 2 -4 -2 -2 y 4 N=2 x -4 -2 -2 -4 2 x 2 4 -4 -2 -2 N=4 -4 2 4 x 4 2 N=6 y 4 4 2 2 4 y 2 -4 -2 -2 -4 2 4 x N=8 -4 -2 -2 -4 2 4 x 2 4 x -4 y N=10 y For y2 , S3 = S4 and S5 = S6 , so we plot S2 , S4 , S6 , S8 , and S10 . y y 4 4 2 -4 -2 -2 2 x 2 4 N=2 y 4 N=4 -4 -2 -2 -4 2 x 2 4 -4 -2 -2 -4 (c) 2 2 4 x N=6 -4 -2 -2 -4 y1 2 4 2 x -4 -2 -2 -4 N=10 -4 4 2 4 N=8 y2 4 -4 4 2 -2 2 4 x -4 -2 2 -2 x -2 -4 4 -4 The graphs of y1 and y2 obtained from a numerical solver are shown. We see that the partial sum representations indicate the even and odd natures of the solution, but don’t really give a very accurate representation of the true solution. Increasing N to about 20 gives a much more accurate representation on [−4, 4]. ∞ k −x (d) From ex = k=0 x /k! we see that e Section 3.2 we have 2 /2 = ∞ 2 k k=0 (−x /2) /k! 271 = ∞ k 2k k k=0 (−1) x /2 k! . From (5) of 5.1 Solutions About Ordinary Points e− y2 = y1 x dx 2 y1 ∞ = k=0 (−1)k 2k x 2k k! ∞ = k=0 2 e−x /2 dx = e−x /2 (e−x2 /2 )2 2 dx = e−x 2 ∞ k=0 1 x2k dx = k k! 2 ∞ (−1)k 2k x 2k k! /2 k=0 ∞ k=0 2 ∞ 1 2k k! k=0 2 ex /2 dx x2k dx 1 x2k+1 (2k + 1)2k k! 1 1 1 = 1 − x2 + 2 x4 − 3 x6 + · · · 2 2 ·2 2 · 3! =x− (−1)k 2k x 2k k! 2 e−x /2 dx = e−x /2 e−x2 x+ 2 3 4·2 5 6·4·2 7 x + x − x + ··· = 3! 5! 7! ∞ k=0 1 3 1 1 x + x5 + x7 + · · · 3·2 5 · 22 · 2 7 · 23 · 3! (−1)k 2k k! 2k+1 . x (2k + 1)! 40. (a) We have y + (cos x)y = 2c2 + 6c3 x + 12c4 x2 + 20c5 x3 + 30c6 x4 + 42c7 x5 + · · · x2 x4 x6 + − + · · · (c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + c5 x5 + · · · ) 2! 4! 6! 1 1 = (2c2 + c0 ) + (6c3 + c1 )x + 12c4 + c2 − c0 x2 + 20c5 + c3 − c1 x3 2 2 1 1 1 1 + 30c6 + c4 + c0 − c2 x4 + 42c7 + c5 + c1 − c3 x5 + · · · . 24 2 24 2 + 1− Then 30c6 + c4 + 1 1 c0 − c2 = 0 24 2 and 42c7 + c5 + 1 1 c1 − c3 = 0, 24 2 which gives c6 = −c0 /80 and c7 = −19c1 /5040. Thus 1 1 1 y1 (x) = 1 − x2 + x4 − x6 + · · · 2 12 80 and 1 1 19 7 y2 (x) = x − x3 + x5 − x + ··· . 6 30 5040 (b) From part (a) the general solution of the differential equation is y = c1 y1 + c2 y2 . Then y(0) = c1 + c2 · 0 = c1 and y (0) = c1 · 0 + c2 = c2 , so the solution of the initial-value problem is 1 1 1 1 1 19 7 y = y1 + y2 = 1 + x − x2 − x3 + x4 + x5 − x6 − x + ··· . 2 6 12 30 80 5040 272 5.2 (c) y Solutions About Singular Points y y 4 4 4 2 2 2 -6 -4 -2 2 4 6 x -6 -4 -2 2 4 6 x 2 -6 -4 -2 -2 -2 -4 x 2 4 6 x -4 y y y 4 4 4 2 2 2 -6 -4 -2 2 4 6 x -6 -4 -2 2 4 6 x -6 -4 -2 -2 -2 -2 -4 (d) 6 -2 -4 4 -4 -4 y 6 4 2 -6 -4 -2 -2 2 4 6 x -4 -6 EXERCISES 5.2 Solutions About Singular Points 1. Irregular singular point: x = 0 2. Regular singular points: x = 0, −3 3. Irregular singular point: x = 3; regular singular point: x = −3 4. Irregular singular point: x = 1; regular singular point: x = 0 5. Regular singular points: x = 0, ±2i 6. Irregular singular point: x = 5; regular singular point: x = 0 273 5.2 Solutions About Singular Points 7. Regular singular points: x = −3, 2 8. Regular singular points: x = 0, ±i 9. Irregular singular point: x = 0; regular singular points: x = 2, ±5 10. Irregular singular point: x = −1; regular singular points: x = 0, 3 11. Writing the differential equation in the form y + 5 x y + y=0 x−1 x+1 we see that x0 = 1 and x0 = −1 are regular singular points. For x0 = 1 the differential equation can be put in the form x(x − 1)2 (x − 1)2 y + 5(x − 1)y + y = 0. x+1 In this case p(x) = 5 and q(x) = x(x − 1)2 /(x + 1). For x0 = −1 the differential equation can be put in the form (x + 1)2 y + 5(x + 1) x+1 y + x(x + 1)y = 0. x−1 In this case p(x) = (x + 1)/(x − 1) and q(x) = x(x + 1). 12. Writing the differential equation in the form y + x+3 y + 7xy = 0 x we see that x0 = 0 is a regular singular point. Multiplying by x2 , the differential equation can be put in the form x2 y + x(x + 3)y + 7x3 y = 0. We identify p(x) = x + 3 and q(x) = 7x3 . 13. We identify P (x) = 5/3x + 1 and Q(x) = −1/3x2 , so that p(x) = xP (x) = Then a0 = 5 , b0 = − 1 , and the indicial equation is 3 3 5 3 + x and q(x) = x2 Q(x) = − 1 . 3 5 2 1 1 1 1 r(r − 1) + r − = r2 + r − = (3r2 + 2r − 1) = (3r − 1)(r + 1) = 0. 3 3 3 3 3 3 and −1. Since these do not differ by an integer we expect to find two series solutions using the method of Frobenius. The indicial roots are 1 3 14. We identify P (x) = 1/x and Q(x) = 10/x, so that p(x) = xP (x) = 1 and q(x) = x2 Q(x) = 10x. Then a0 = 1, b0 = 0, and the indicial equation is r(r − 1) + r = r2 = 0. The indicial roots are 0 and 0. Since these are equal, we expect the method of Frobenius to yield a single series solution. 15. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain ∞ 2xy − y + 2y = 2r2 − 3r c0 xr−1 + [2(k + r − 1)(k + r)ck − (k + r)ck + 2ck−1 ]xk+r−1 = 0, k=1 which implies 2r2 − 3r = r(2r − 3) = 0 and (k + r)(2k + 2r − 3)ck + 2ck−1 = 0. 274 5.2 Solutions About Singular Points The indicial roots are r = 0 and r = 3/2. For r = 0 the recurrence relation is ck = − 2ck−1 , k(2k − 3) k = 1, 2, 3, . . . , and c2 = −2c0 , c1 = 2c0 , c3 = 4 c0 , 9 and so on. For r = 3/2 the recurrence relation is ck = − 2ck−1 , (2k + 3)k k = 1, 2, 3, . . . , and 2 c1 = − c0 , 5 and so on. The general solution on (0, ∞) is c2 = 2 c0 , 35 c3 = − 4 c0 , 945 2 4 4 3 2 y = C1 1 + 2x − 2x2 + x3 + · · · + C2 x3/2 1 − x + x2 − x + ··· . 9 5 35 945 16. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain 2xy + 5y + xy = 2r2 + 3r c0 xr−1 + 2r2 + 7r + 5 c1 xr ∞ [2(k + r)(k + r − 1)ck + 5(k + r)ck + ck−2 ]xk+r−1 + k=2 = 0, which implies 2r2 + 3r = r(2r + 3) = 0, 2r2 + 7r + 5 c1 = 0, and (k + r)(2k + 2r + 3)ck + ck−2 = 0. The indicial roots are r = −3/2 and r = 0, so c1 = 0 . For r = −3/2 the recurrence relation is ck = − ck−2 , (2k − 3)k k = 2, 3, 4, . . . , and 1 c2 = − c0 , 2 and so on. For r = 0 the recurrence relation is ck = − c3 = 0, ck−2 , k(2k + 3) c4 = 1 c0 , 40 k = 2, 3, 4, . . . , and 1 c0 , 14 and so on. The general solution on (0, ∞) is c2 = − c3 = 0, c4 = 1 c0 , 616 1 1 4 1 1 y = C1 x−3/2 1 − x2 + x4 + · · · + C2 1 − x2 + x + ··· . 2 40 14 616 275 5.2 Solutions About Singular Points 17. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain 1 4xy + y + y = 2 7 4r2 − r c0 xr−1 + 2 ∞ 1 4(k + r)(k + r − 1)ck + (k + r)ck + ck−1 xk+r−1 2 k=1 = 0, which implies 7 7 4r2 − r = r 4r − 2 2 =0 and 1 (k + r)(8k + 8r − 7)ck + ck−1 = 0. 2 The indicial roots are r = 0 and r = 7/8. For r = 0 the recurrence relation is ck = − 2ck−1 , k(8k − 7) and c1 = −2c0 , c2 = k = 1, 2, 3, . . . , 2 c0 , 9 c3 = − 4 c0 , 459 and so on. For r = 7/8 the recurrence relation is ck = − and c1 = − 2ck−1 , (8k + 7)k 2 c0 , 15 c2 = k = 1, 2, 3, . . . , 2 c0 , 345 c3 = − 4 c0 , 32,085 and so on. The general solution on (0, ∞) is 2 2 4 3 4 2 2 y = C1 1 − 2x + x2 − x + · · · + C2 x7/8 1 − x + x − x3 + · · · . 9 459 15 345 32,085 18. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain 2x2 y − xy + x2 + 1 y = 2r2 − 3r + 1 c0 xr + 2r2 + r c1 xr+1 ∞ [2(k + r)(k + r − 1)ck − (k + r)ck + ck + ck−2 ]xk+r + k=2 = 0, which implies 2r2 − 3r + 1 = (2r − 1)(r − 1) = 0, 2r2 + r c1 = 0, and [(k + r)(2k + 2r − 3) + 1]ck + ck−2 = 0. The indicial roots are r = 1/2 and r = 1, so c1 = 0. For r = 1/2 the recurrence relation is ck = − and ck−2 , k(2k − 1) 1 c2 = − c0 , 6 k = 2, 3, 4, . . . , c3 = 0, 276 c4 = 1 c0 , 168 5.2 Solutions About Singular Points and so on. For r = 1 the recurrence relation is ck = − ck−2 , k(2k + 1) k = 2, 3, 4, . . . , and 1 c0 , 10 and so on. The general solution on (0, ∞) is c2 = − c3 = 0, c4 = 1 c0 , 360 1 1 4 1 1 4 y = C1 x1/2 1 − x2 + x + · · · + C2 x 1 − x2 + x + ··· . 6 168 10 360 19. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain 3xy + (2 − x)y − y = 3r2 − r c0 xr−1 ∞ [3(k + r − 1)(k + r)ck + 2(k + r)ck − (k + r)ck−1 ]xk+r−1 + k=1 = 0, which implies 3r2 − r = r(3r − 1) = 0 and (k + r)(3k + 3r − 1)ck − (k + r)ck−1 = 0. The indicial roots are r = 0 and r = 1/3. For r = 0 the recurrence relation is ck−1 ck = , k = 1, 2, 3, . . . , 3k − 1 and 1 1 1 c2 = c3 = c0 , c0 , c0 , 2 10 80 and so on. For r = 1/3 the recurrence relation is ck−1 ck = , k = 1, 2, 3, . . . , 3k and 1 1 1 c1 = c0 , c2 = c3 = c0 , c0 , 3 18 162 and so on. The general solution on (0, ∞) is c1 = 1 1 1 1 3 1 1 y = C1 1 + x + x2 + x3 + · · · + C2 x1/3 1 + x + x2 + x + ··· . 2 10 80 3 18 162 20. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain x2 y − x − 2 y= 9 r2 − r + 2 9 ∞ 2 (k + r)(k + r − 1)ck + ck − ck−1 xk+r 9 c0 xr + k=1 = 0, which implies r2 − r + 2 = 9 r− 2 3 r− 1 3 =0 and (k + r)(k + r − 1) + 277 2 ck − ck−1 = 0. 9 5.2 Solutions About Singular Points The indicial roots are r = 2/3 and r = 1/3. For r = 2/3 the recurrence relation is ck = 3ck−1 , 3k 2 + k 3 c0 , 4 c2 = k = 1, 2, 3, . . . , and c1 = 9 c0 , 56 c3 = 9 c0 , 560 and so on. For r = 1/3 the recurrence relation is ck = 3ck−1 , 3k 2 − k 3 c0 , 2 c2 = k = 1, 2, 3, . . . , and c1 = 9 c0 , 20 c3 = 9 c0 , 160 and so on. The general solution on (0, ∞) is 3 3 9 3 9 3 9 9 y = C1 x2/3 1 + x + x2 + x + · · · + C2 x1/3 1 + x + x2 + x + ··· . 4 56 560 2 20 160 21. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain ∞ 2xy − (3 + 2x)y + y = 2r2 − 5r c0 xr−1 + [2(k + r)(k + r − 1)ck k=1 − 3(k + r)ck − 2(k + r − 1)ck−1 + ck−1 ]xk+r−1 = 0, which implies 2r2 − 5r = r(2r − 5) = 0 and (k + r)(2k + 2r − 5)ck − (2k + 2r − 3)ck−1 = 0. The indicial roots are r = 0 and r = 5/2. For r = 0 the recurrence relation is ck = (2k − 3)ck−1 , k(2k − 5) k = 1, 2, 3, . . . , and c1 = 1 c0 , 3 1 c2 = − c0 , 6 1 c3 = − c0 , 6 and so on. For r = 5/2 the recurrence relation is ck = 2(k + 1)ck−1 , k(2k + 5) k = 1, 2, 3, . . . , and c1 = 4 c0 , 7 c2 = 4 c0 , 21 c3 = 32 c0 , 693 and so on. The general solution on (0, ∞) is 1 32 3 1 4 1 4 y = C1 1 + x − x2 − x3 + · · · + C2 x5/2 1 + x + x2 + x + ··· . 3 6 6 7 21 693 278 5.2 22. Substituting y = ∞ n=0 Solutions About Singular Points cn xn+r into the differential equation and collecting terms, we obtain x2 y + xy + x2 − 4 9 y= r2 − 4 9 5 9 c0 xr + r2 + 2r + ∞ c1 xr+1 4 (k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r 9 + k=2 = 0, which implies r2 − 4 = 9 r+ 2 3 r2 + 2r + r− 5 9 2 3 = 0, c1 = 0, and (k + r)2 − 4 ck + ck−2 = 0. 9 The indicial roots are r = −2/3 and r = 2/3, so c1 = 0. For r = −2/3 the recurrence relation is ck = − 9ck−2 , 3k(3k − 4) k = 2, 3, 4, . . . , and 3 c2 = − c0 , 4 and so on. For r = 2/3 the recurrence relation is ck = − and c2 = − c3 = 0, 9ck−2 , 3k(3k + 4) 3 c0 , 20 c3 = 0, c4 = 9 c0 , 128 k = 2, 3, 4, . . . , c4 = 9 c0 , 1,280 and so on. The general solution on (0, ∞) is 3 3 9 4 9 y = C1 x−2/3 1 − x2 + x + · · · + C2 x2/3 1 − x2 + x4 + · · · . 4 128 20 1,280 23. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain ∞ 9x2 y + 9x2 y + 2y = 9r2 − 9r + 2 c0 xr + [9(k + r)(k + r − 1)ck + 2ck + 9(k + r − 1)ck−1 ]xk+r = 0, k=1 which implies 9r2 − 9r + 2 = (3r − 1)(3r − 2) = 0 and [9(k + r)(k + r − 1) + 2]ck + 9(k + r − 1)ck−1 = 0. The indicial roots are r = 1/3 and r = 2/3. For r = 1/3 the recurrence relation is ck = − and (3k − 2)ck−1 , k(3k − 1) 1 c1 = − c0 , 2 c2 = 1 c0 , 5 279 k = 1, 2, 3, . . . , c3 = − 7 c0 , 120 5.2 Solutions About Singular Points and so on. For r = 2/3 the recurrence relation is ck = − (3k − 1)ck−1 , k(3k + 1) and 1 c1 = − c0 , 2 and so on. The general solution on (0, ∞) is c2 = k = 1, 2, 3, . . . , 5 c0 , 28 c3 = − 1 c0 , 21 1 1 7 3 1 1 5 y = C1 x1/3 1 − x + x2 − x + · · · + C2 x2/3 1 − x + x2 − x3 + · · · . 2 5 120 2 28 21 24. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain ∞ 2x2 y + 3xy + (2x − 1)y = 2r2 + r − 1 c0 xr + [2(k + r)(k + r − 1)ck + 3(k + r)ck − ck + 2ck−1 ]xk+r = 0, k=1 which implies 2r2 + r − 1 = (2r − 1)(r + 1) = 0 and [(k + r)(2k + 2r + 1) − 1]ck + 2ck−1 = 0. The indicial roots are r = −1 and r = 1/2. For r = −1 the recurrence relation is ck = − 2ck−1 , k(2k − 3) k = 1, 2, 3, . . . , and c2 = −2c0 , c1 = 2c0 , c3 = 4 c0 , 9 and so on. For r = 1/2 the recurrence relation is ck = − 2ck−1 , k(2k + 3) and 2 c1 = − c0 , 5 and so on. The general solution on (0, ∞) is c2 = k = 1, 2, 3, . . . , 2 c0 , 35 c3 = − 4 c0 , 945 2 2 4 4 3 y = C1 x−1 1 + 2x − 2x2 + x3 + · · · + C2 x1/2 1 − x + x2 − x + ··· . 9 5 35 945 25. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain ∞ xy + 2y − xy = r + r c0 x 2 r−1 2 [(k + r)(k + r − 1)ck + 2(k + r)ck − ck−2 ]xk+r−1 = 0, r + r + 3r + 2 c1 x + k=2 which implies r2 + r = r(r + 1) = 0, r2 + 3r + 2 c1 = 0, and (k + r)(k + r + 1)ck − ck−2 = 0. 280 5.2 Solutions About Singular Points The indicial roots are r1 = 0 and r2 = −1, so c1 = 0. For r1 = 0 the recurrence relation is ck−2 ck = , k = 2, 3, 4, . . . , k(k + 1) and 1 c0 3! c3 = c5 = c7 = · · · = 0 1 c4 = c0 5! 1 c2n = c0 . (2n + 1)! c2 = For r2 = −1 the recurrence relation is ck = and ck−2 , k(k − 1) k = 2, 3, 4, . . . , 1 c0 2! c3 = c5 = c7 = · · · = 0 1 c4 = c0 4! 1 c0 . c2n = (2n)! c2 = The general solution on (0, ∞) is ∞ ∞ y = C1 1 1 x2n + C2 x−1 x2n (2n + 1)! (2n)! n=0 n=0 ∞ ∞ = = 26. Substituting y = ∞ n=0 1 1 1 C1 x2n+1 + C2 x2n x (2n + 1)! (2n)! n=0 n=0 1 [C1 sinh x + C2 cosh x]. x cn xn+r into the differential equation and collecting terms, we obtain x2 y + xy + x2 − 1 4 y= r2 − 1 4 ∞ + k=2 c0 xr + r2 + 2r + 3 4 1 (k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r 4 = 0, which implies r2 − 1 = 4 c1 xr+1 r− 1 2 r2 + 2r + r+ 3 4 1 2 = 0, c1 = 0, and (k + r)2 − 1 ck + ck−2 = 0. 4 281 5.2 Solutions About Singular Points The indicial roots are r1 = 1/2 and r2 = −1/2, so c1 = 0. For r1 = 1/2 the recurrence relation is ck = − ck−2 , k(k + 1) k = 2, 3, 4, . . . , and 1 c0 3! c3 = c5 = c7 = · · · = 0 1 c4 = c0 5! (−1)n c2n = c0 . (2n + 1)! c2 = − For r2 = −1/2 the recurrence relation is ck = − ck−2 , k(k − 1) k = 2, 3, 4, . . . , and 1 c0 2! c3 = c5 = c7 = · · · = 0 1 c4 = c0 4! (−1)n c0 . c2n = (2n)! c2 = − The general solution on (0, ∞) is ∞ y = C1 x1/2 ∞ (−1)n 2n (−1)n 2n x + C2 x−1/2 x (2n + 1)! (2n)! n=0 n=0 = C1 x−1/2 ∞ ∞ (−1)n 2n+1 (−1)n 2n + C2 x−1/2 x x (2n + 1)! (2n)! n=0 n=0 = x−1/2 [C1 sin x + C2 cos x]. 27. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain ∞ xy − xy + y = r2 − r c0 xr−1 + [(k + r + 1)(k + r)ck+1 − (k + r)ck + ck ]xk+r = 0 k=0 which implies r2 − r = r(r − 1) = 0 and (k + r + 1)(k + r)ck+1 − (k + r − 1)ck = 0. The indicial roots are r1 = 1 and r2 = 0. For r1 = 1 the recurrence relation is ck+1 = kck , (k + 2)(k + 1) 282 k = 0, 1, 2, . . . , 5.2 Solutions About Singular Points and one solution is y1 = c0 x. A second solution is y2 = x e− −1 dx x2 ex dx = x x2 dx = x 1 x2 1 1 1 + x + x2 + x3 + · · · dx 2 3! 1 1 1 1 1 1 1 1 1 + + + x + x2 + · · · dx = x − + ln x + x + x2 + x3 + · · · 2 x x 2 3! 4! x 2 12 72 =x 1 1 1 = x ln x − 1 + x2 + x3 + x4 + · · · . 2 12 72 The general solution on (0, ∞) is y = C1 x + C2 y2 (x). 28. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain 3 y + y − 2y = r2 + 2r c0 xr−2 + r2 + 4r + 3 c1 xr−1 x ∞ [(k + r)(k + r − 1)ck + 3(k + r)ck − 2ck−2 ]xk+r−2 + k=2 = 0, which implies r2 + 2r = r(r + 2) = 0 r2 + 4r + 3 c1 = 0 (k + r)(k + r + 2)ck − 2ck−2 = 0. The indicial roots are r1 = 0 and r2 = −2, so c1 = 0. For r1 = 0 the recurrence relation is 2ck−2 ck = , k = 2, 3, 4, . . . , k(k + 2) and 1 c2 = c0 4 c3 = c5 = c7 = · · · = 0 1 c4 = c0 48 1 c0 . c6 = 1,152 The result is 1 1 1 y1 = c0 1 + x2 + x4 + x6 + · · · . 4 48 1,152 A second solution is y2 = y 1 = y1 = y1 e− (3/x)dx 2 y1 dx dx = y1 x3 1+ 1 2 4x dx x3 1 + 1 x2 + 2 5 4 48 x + 7 6 576 x + ··· + 1 4 48 x + ··· = y1 1 x3 2 1 7 19 6 x + · · · dx 1 − x2 + x4 + 2 48 576 1 1 19 4 1 1 7 19 3 7 − + x− x + · · · dx = y1 − 2 − ln x + x2 − x + ··· 3 x 2x 48 576 2x 2 96 2,304 1 1 7 19 4 = − y1 ln x + y − 2 + x2 − x + ··· . 2 2x 96 2,304 283 5.2 Solutions About Singular Points The general solution on (0, ∞) is y = C1 y1 (x) + C2 y2 (x). 29. Substituting y = ∞ n=0 cn xn+r into the differential equation and collecting terms, we obtain ∞ xy + (1 − x)y − y = r c0 x 2 r−1 [(k + r)(k + r − 1)ck + (k + r)ck − (k + r)ck−1 ]xk+r−1 = 0, + k=1 which implies r2 = 0 and (k + r)2 ck − (k + r)ck−1 = 0. The indicial roots are r1 = r2 = 0 and the recurrence relation is ck = ck−1 , k k = 1, 2, 3, . . . . One solution is 1 1 y1 = c0 1 + x + x2 + x3 + · · · 2 3! = c0 ex . A second solution is y2 = y 1 = ex e− (1/x−1)dx e2x 1 x ex /x dx = ex e2x dx = ex 1 −x e dx x 1 1 1 − x + x2 − x3 + · · · dx = ex 2 3! 1 1 1 − 1 + x − x2 + · · · dx x 2 3! ∞ = ex ln x − x + 1 2 1 3 (−1)n+1 n x − x + · · · = ex ln x − ex x . 2·2 3 · 3! n · n! n=1 The general solution on (0, ∞) is ∞ x x y = C1 e + C2 e 30. Substituting y = ∞ n=0 ln x − (−1)n+1 n x n · n! n=1 . cn xn+r into the differential equation and collecting terms, we obtain ∞ [(k + r)(k + r − 1)ck + (k + r)ck + ck−1 ]xk+r−1 = 0 xy + y + y = r2 c0 xr−1 + k=1 which implies r2 = 0 and (k + r)2 ck + ck−1 = 0. The indicial roots are r1 = r2 = 0 and the recurrence relation is ck = − ck−1 , k2 k = 1, 2, 3, . . . . One solution is y1 = c0 1 − x + 1 2 1 3 1 4 x − x + x − ··· 22 (3!)2 (4!)2 284 ∞ = c0 (−1)n n x . (n!)2 n=0 5.2 Solutions About Singular Points A second solution is e− y2 = y 1 (1/x)dx 2 y1 dx = y1 dx x 1−x+ = y1 dx x 1 − 2x + 3 x2 − 5 x3 + 2 9 = y1 1 x 35 4 288 x 1 2 4x − 1 3 36 x + ··· 2 − ··· 23 677 4 5 1 + 2x + x2 + x3 + x + · · · dx 2 9 288 1 677 3 5 23 + 2 + x + x2 + x + · · · dx x 2 9 288 = y1 5 23 677 4 = y1 ln x + 2x + x2 + x3 + x + ··· 4 27 1,152 5 23 677 4 = y1 ln x + y1 2x + x2 + x3 + x + ··· . 4 27 1,152 The general solution on (0, ∞) is y = C1 y1 (x) + C2 y2 (x). 31. Substituting y = ∞ n+r n=0 cn x into the differential equation and collecting terms, we obtain ∞ xy + (x − 6)y − 3y = (r2 − 7r)c0 xr−1 + (k + r)(k + r − 1)ck + (k + r − 1)ck−1 k=1 − 6(k + r)ck − 3ck−1 xk+r−1 = 0, which implies r2 − 7r = r(r − 7) = 0 and (k + r)(k + r − 7)ck + (k + r − 4)ck−1 = 0. The indicial roots are r1 = 7 and r2 = 0. For r1 = 7 the recurrence relation is (k + 7)kck + (k + 3)ck−1 = 0, k = 1, 2, 3, . . . , or ck = − k+3 ck−1 , k(k + 7) k = 1, 2, 3, . . . . Taking c0 = 0 we obtain 1 c1 = − c0 2 5 c0 c2 = 18 1 c3 = − c0 , 6 and so on. Thus, the indicial root r1 = 7 yields a single solution. Now, for r2 = 0 the recurrence relation is k(k − 7)ck + (k − 4)ck−1 = 0, 285 k = 1, 2, 3, . . . . 5.2 Solutions About Singular Points Then −6c1 − 3c0 = 0 −10c2 − 2c1 = 0 −12c3 − c2 = 0 −12c4 + 0c3 = 0 =⇒ c4 = 0 −10c5 + c4 = 0 =⇒ c5 = 0 −6c6 + 2c5 = 0 =⇒ c6 = 0 0c7 + 3c6 = 0 =⇒ c7 is arbitrary and ck = − k−4 ck−1 , k(k − 7) k = 8, 9, 10, . . . . Taking c0 = 0 and c7 = 0 we obtain 1 c1 = − c0 2 1 c2 = c0 10 1 c3 = − c0 120 c4 = c5 = c6 = · · · = 0. Taking c0 = 0 and c7 = 0 we obtain c1 = c2 = c3 = c4 = c5 = c6 = 0 1 c8 = − c7 2 5 c9 = c7 36 1 c10 = − c7 , 36 and so on. In this case we obtain the two solutions 1 1 1 3 y1 = 1 − x + x2 − x and 2 10 120 32. Substituting y = ∞ n=0 1 5 1 y2 = x7 − x8 + x9 − x10 + · · · . 2 36 36 cn xn+r into the differential equation and collecting terms, we obtain x(x − 1)y + 3y − 2y ∞ = 4r − r2 c0 xr−1 + [(k + r − 1)(k + r − 12)ck−1 − (k + r)(k + r − 1)ck + 3(k + r)ck − 2ck−1 ]xk+r−1 k=1 = 0, which implies 4r − r2 = r(4 − r) = 0 and −(k + r)(k + r − 4)ck + [(k + r − 1)(k + r − 2) − 2]ck−1 = 0. The indicial roots are r1 = 4 and r2 = 0. For r1 = 4 the recurrence relation is −(k + 4)kck + [(k + 3)(k + 2) − 2]ck−1 = 0 286 5.2 Solutions About Singular Points or k+1 ck−1 , k ck = k = 1, 2, 3, . . . . Taking c0 = 0 we obtain c1 = 2c0 c2 = 3c0 c3 = 4c0 , and so on. Thus, the indicial root r1 = 4 yields a single solution. For r2 = 0 the recurrence relation is −k(k − 4)ck + k(k − 3)ck−1 = 0, k = 1, 2, 3, . . . , −(k − 4)ck + (k − 3)ck−1 = 0, k = 1, 2, 3, . . . . or Then 3c1 − 2c0 = 0 2c2 − c1 = 0 c3 + 0c2 = 0 ⇒ c3 = 0 0c4 + c3 = 0 ⇒ c4 is arbitrary and ck = (k − 3)ck−1 , k−4 k = 5, 6, 7, . . . . Taking c0 = 0 and c4 = 0 we obtain 2 c0 3 1 c2 = c0 3 c3 = c4 = c5 = · · · = 0. c1 = Taking c0 = 0 and c4 = 0 we obtain c1 = c2 = c3 = 0 c5 = 2c4 c6 = 3c4 c7 = 4c4 , and so on. In this case we obtain the two solutions 2 1 y1 = 1 + x + x2 3 3 y2 = x4 + 2x5 + 3x6 + 4x7 + · · · . and 33. (a) From t = 1/x we have dt/dx = −1/x2 = −t2 . Then dy dy dy dt = = −t2 dx dt dx dt and d2 y d = 2 dx dx dy dx = d dx −t2 dy dt = −t2 dy d2 y dt − 2 dx dt dt 2t dt dx = t4 d2 y dy + 2t3 . 2 dt dt Now x4 d2 y 1 + λy = 4 dx2 t t4 d2 y dy + 2t3 dt2 dt 287 + λy = d2 y 2 dy + + λy = 0 dt2 t dt 5.2 Solutions About Singular Points becomes t (b) Substituting y = t ∞ n+r n=0 cn t d2 y dy +2 + λty = 0. 2 dt dt into the differential equation and collecting terms, we obtain d2 y dy +2 + λty = (r2 + r)c0 tr−1 + (r2 + 3r + 2)c1 tr 2 dt dt ∞ [(k + r)(k + r − 1)ck + 2(k + r)ck + λck−2 ]tk+r−1 + k=2 = 0, which implies r2 + r = r(r + 1) = 0, r2 + 3r + 2 c1 = 0, and (k + r)(k + r + 1)ck + λck−2 = 0. The indicial roots are r1 = 0 and r2 = −1, so c1 = 0. For r1 = 0 the recurrence relation is ck = − λck−2 , k(k + 1) k = 2, 3, 4, . . . , and λ c0 3! c3 = c5 = c7 = · · · = 0 c2 = − c4 = λ2 c0 5! . . . c2n = (−1)n λn c0 . (2n + 1)! For r2 = −1 the recurrence relation is ck = − λck−2 , k(k − 1) k = 2, 3, 4, . . . , and λ c0 2! c3 = c5 = c7 = · · · = 0 c2 = − c4 = λ2 c0 4! . . . c2n = (−1)n λn c0 . (2n)! 288 5.2 Solutions About Singular Points The general solution on (0, ∞) is ∞ (−1)n √ (−1)n √ ( λ t)2n + c2 t−1 ( λ t)2n (2n + 1)! (2n)! n=0 n=0 ∞ y(t) = c1 = ∞ ∞ (−1)n √ (−1)n √ 1 C1 ( λ t)2n+1 + C2 ( λ t)2n t (2n + 1)! (2n)! n=0 n=0 = √ √ 1 [C1 sin λ t + C2 cos λ t ]. t (c) Using t = 1/x, the solution of the original equation is √ √ λ λ y(x) = C1 x sin + C2 x cos . x x 34. (a) From the boundary conditions y(a) = 0, y(b) = 0 we find √ √ λ λ C1 sin + C2 cos =0 a a √ √ λ λ C1 sin + C2 cos = 0. b b Since this is a homogeneous system of linear equations, it will have nontrivial solutions for C1 and C2 if √ λ sin a √ λ sin b √ λ √ √ √ √ λ λ λ λ a cos − cos sin = sin √ a b a b λ cos b √ √ √ b−a λ λ = sin λ − = sin a b ab cos = 0. This will be the case if √ λ b−a ab = nπ or √ λ= nπab nπab = , n = 1, 2, . . . , b−a L or, if λn = n2 π 2 a2 b2 Pn b 4 = . 2 L EI √ √ The critical loads are then Pn = n2 π 2 (a/b)2 EI0 /L2 . Using C2 = −C1 sin( λ/a)/ cos( λ/a) we have √ √ √ λ λ sin( λ/a) √ y = C1 x sin cos − x x cos( λ/a) √ √ √ √ λ λ λ λ = C3 x sin cos − cos sin x a x a √ = C3 x sin λ 1 1 − x a , and yn (x) = C3 x sin nπab L 1 1 − x a = C3 x sin nπab La 289 nπab a a − 1 = C4 x sin 1− . x L x 5.2 Solutions About Singular Points (b) When n = 1, b = 11, and a = 1, we have, y for C4 = 1, 2 1 y1 (x) = x sin 1.1π 1 − x . 1 1 3 5 7 9 11 x 35. Express the differential equation in standard form: y + P (x)y + Q(x)y + R(x)y = 0. Suppose x0 is a singular point of the differential equation. Then we say that x0 is a regular singular point if (x − x0 )P (x), (x − x0 )2 Q(x), and (x − x0 )3 R(x) are analytic at x = x0 . 36. Substituting y = ∞ n+r n=0 cn x into the first differential equation and collecting terms, we obtain ∞ [ck + (k + r − 1)(k + r − 2)ck−1 ]xk+r = 0. x3 y + y = c0 xr + k=1 It follows that c0 = 0 and ck = −(k + r − 1)(k + r − 2)ck−1 . The only solution we obtain is y(x) = 0. Substituting y = ∞ n+r n=0 cn x into the second differential equation and collecting terms, we obtain ∞ x2 y + (3x − 1)y + y = −rc0 + [(k + r + 1)2 ck − (k + r + 1)ck+1 ]xk+r = 0, k=0 which implies −rc0 = 0 (k + r + 1) ck − (k + r + 1)ck+1 = 0. 2 If c0 = 0, then the solution of the differential equation is y = 0. Thus, we take r = 0, from which we obtain ck+1 = (k + 1)ck , k = 0, 1, 2, . . . . Letting c0 = 1 we get c1 = 2, c2 = 3!, c3 = 4!, and so on. The solution of the differential equation is then ∞ y = n=0 (n + 1)!xn , which converges only at x = 0. 37. We write the differential equation in the form x2 y + (b/a)xy + (c/a)y = 0 and identify a0 = b/a and b0 = c/a as in (12) in the text. Then the indicial equation is r(r − 1) + b c r+ =0 a a or ar2 + (b − a)r + c = 0, which is also the auxiliary equation of ax2 y + bxy + cy = 0. 290 5.3 Special Functions EXERCISES 5.3 Special Functions 1. Since ν 2 = 1/9 the general solution is y = c1 J1/3 (x) + c2 J−1/3 (x). 2. Since ν 2 = 1 the general solution is y = c1 J1 (x) + c2 Y1 (x). 3. Since ν 2 = 25/4 the general solution is y = c1 J5/2 (x) + c2 J−5/2 (x). 4. Since ν 2 = 1/16 the general solution is y = c1 J1/4 (x) + c2 J−1/4 (x). 5. Since ν 2 = 0 the general solution is y = c1 J0 (x) + c2 Y0 (x). 6. Since ν 2 = 4 the general solution is y = c1 J2 (x) + c2 Y2 (x). 7. We identify α = 3 and ν = 2. Then the general solution is y = c1 J2 (3x) + c2 Y2 (3x). 8. We identify α = 6 and ν = 1 2 . Then the general solution is y = c1 J1/2 (6x) + c2 J−1/2 (6x). 9. We identify α = 5 and ν = 2 . Then the general solution is y = c1 J2/3 (5x) + c2 J−2/3 (5x). 3 √ √ √ 10. We identify α = 2 and ν = 8. Then the general solution is y = c1 J8 ( 2x) + c2 Y8 ( 2x). 11. If y = x−1/2 v(x) then 1 y = x−1/2 v (x) − x−3/2 v(x), 2 3 y = x−1/2 v (x) − x−3/2 v (x) + x−5/2 v(x), 4 and 1 x2 y + 2xy + α2 x2 y = x3/2 v (x) + x1/2 v (x) + α2 x3/2 − x−1/2 v(x) = 0. 4 Multiplying by x1/2 we obtain x2 v (x) + xv (x) + α2 x2 − 1 4 v(x) = 0, whose solution is v = c1 J1/2 (αx) + c2 J−1/2 (αx). Then y = c1 x−1/2 J1/2 (αx) + c2 x−1/2 J−1/2 (αx). √ 12. If y = x v(x) then 1 y = x1/2 v (x) + x−1/2 v(x) 2 1 1/2 y = x v (x) + x−1/2 v (x) − x−3/2 v(x) 4 and x2 y + α2 x2 − ν 2 + 1 4 1 1 y = x5/2 v (x) + x3/2 v (x) − x1/2 v(x) + α2 x2 − ν 2 + 4 4 = x5/2 v (x) + x3/2 v (x) + (α2 x5/2 − ν 2 x1/2 )v(x) = 0. Multiplying by x−1/2 we obtain x2 v (x) + xv (x) + (α2 x2 − ν 2 )v(x) = 0, √ √ whose solution is v(x) = c1 Jν (αx) + c2 Yν (αx). Then y = c1 x Jν (αx) + c2 x Yν (αx). 291 x1/2 v(x) 5.3 Special Functions 13. Write the differential equation in the form y + (2/x)y + (4/x)y = 0. This is the form of (18) in the text with a = − 1 , c = 1 , b = 4, and p = 1, so, by (19) in the text, the general solution is 2 2 y = x−1/2 [c1 J1 (4x1/2 ) + c2 Y1 (4x1/2 )]. 14. Write the differential equation in the form y + (3/x)y + y = 0. This is the form of (18) in the text with a = −1, c = 1, b = 1, and p = 1, so, by (19) in the text, the general solution is y = x−1 [c1 J1 (x) + c2 Y1 (x)]. 15. Write the differential equation in the form y − (1/x)y + y = 0. This is the form of (18) in the text with a = 1, c = 1, b = 1, and p = 1, so, by (19) in the text, the general solution is y = x[c1 J1 (x) + c2 Y1 (x)]. 16. Write the differential equation in the form y − (5/x)y + y = 0. This is the form of (18) in the text with a = 3, c = 1, b = 1, and p = 2, so, by (19) in the text, the general solution is y = x3 [c1 J3 (x) + c2 Y3 (x)]. 17. Write the differential equation in the form y + (1 − 2/x2 )y = 0. This is the form of (18) in the text with a = c = 1, b = 1, and p = 3 , so, by (19) in the text, the general solution is 2 1 2 , y = x1/2 [c1 J3/2 (x) + c2 Y3/2 (x)] = x1/2 [C1 J3/2 (x) + C2 J−3/2 (x)]. 18. Write the differential equation in the form y + (4 + 1/4x2 )y = 0. This is the form of (18) in the text with a = 1 , c = 1, b = 2, and p = 0, so, by (19) in the text, the general solution is 2 y = x1/2 [c1 J0 (2x) + c2 Y0 (2x)]. 19. Write the differential equation in the form y + (3/x)y + x2 y = 0. This is the form of (18) in the text with a = −1, c = 2, b = 1 , and p = 1 , so, by (19) in the text, the general solution is 2 2 y = x−1 c1 J1/2 1 2 x 2 1 2 x 2 + c2 Y1/2 or y = x−1 C1 J1/2 1 2 x 2 + C2 J−1/2 1 2 x 2 . 20. Write the differential equation in the form y + (1/x)y + ( 1 x4 − 4/x2 )y = 0. This is the form of (18) in the 9 text with a = 0, c = 3, b = 1 9 , and p = 2 3 , so, by (19) in the text, the general solution is y = c1 J2/3 1 3 x 9 + c2 Y2/3 1 3 x 9 or y = C1 J2/3 1 3 x 9 + C2 J−2/3 292 1 3 x . 9 5.3 Special Functions 21. Using the fact that i2 = −1, along with the definition of Jν (x) in (7) in the text, we have Iν (x) = i−ν Jν (ix) = i−ν ∞ (−1)n n!Γ(1 + ν + n) n=0 ∞ = x (−1)n i2n+ν−ν n!Γ(1 + ν + n) 2 n=0 ∞ = x (−1)n (i2 )n n!Γ(1 + ν + n) 2 n=0 ∞ = (−1)2n n!Γ(1 + ν + n) n=0 ∞ = 1 n!Γ(1 + ν + n) n=0 x 2 2n+ν 2n+ν 2n+ν x 2 2n+ν ix 2 2n+ν , which is a real function. 22. (a) The differential equation has the form of (18) in the text with 1 2 2c − 2 = 2 =⇒ c = 2 1 1 b2 c2 = −β 2 c2 = −1 =⇒ β = and b = i 2 2 1 2 2 2 a − p c = 0 =⇒ p = . 4 1 − 2a = 0 =⇒ a = Then, by (19) in the text, y = x1/2 c1 J1/4 1 2 ix 2 + c2 J−1/4 1 2 ix 2 . In terms of real functions the general solution can be written y = x1/2 C1 I1/4 1 2 x 2 + C2 K1/4 1 2 x 2 . (b) Write the differential equation in the form y + (1/x)y − 7x2 y = 0. This is the form of (18) in the text with 1 − 2a = 1 =⇒ a = 0 2c − 2 = 2 =⇒ c = 2 1√ b2 c2 = −β 2 c2 = −7 =⇒ β = 7 2 a2 − p2 c2 = 0 =⇒ p = 0. and b = Then, by (19) in the text, y = c1 J0 1√ 7 ix2 2 + c2 Y0 1√ 7 ix2 . 2 In terms of real functions the general solution can be written y = C1 I0 1√ 2 7x 2 + C2 K0 293 1√ 2 7x . 2 1√ 7i 2 5.3 Special Functions 23. The differential equation has the form of (18) in the text with 1 2 2c − 2 = 0 =⇒ c = 1 1 − 2a = 0 =⇒ a = b2 c2 = 1 =⇒ b = 1 1 a2 − p2 c2 = 0 =⇒ p = . 2 Then, by (19) in the text, y = x1/2 [c1 J1/2 (x) + c2 J−1/2 (x)] = x1/2 c1 2 sin x + c2 πx 2 cos x = C1 sin x + C2 cos x. πx 24. Write the differential equation in the form y + (4/x)y + (1 + 2/x2 )y = 0. This is the form of (18) in the text with 3 1 − 2a = 4 =⇒ a = − 2 2c − 2 = 0 =⇒ c = 1 b2 c2 = 1 =⇒ b = 1 1 a2 − p2 c2 = 2 =⇒ p = . 2 Then, by (19), (23), and (24) in the text, y = x−3/2 [c1 J1/2 (x) + c2 J−1/2 (x)] = x−3/2 c1 2 sin x + c2 πx 2 1 1 cos x = C1 2 sin x + C2 2 cos x. πx x x 1 25. Write the differential equation in the form y + (2/x)y + ( 16 x2 − 3/4x2 )y = 0. This is the form of (18) in the text with 1 1 − 2a = 2 =⇒ a = − 2 2c − 2 = 2 =⇒ c = 2 1 1 b2 c2 = =⇒ b = 16 8 3 1 2 2 2 a − p c = − =⇒ p = . 4 2 Then, by (19) in the text, y = x−1/2 c1 J1/2 1 2 x 8 16 sin πx2 = x−1/2 c1 = C1 x−3/2 sin 1 2 x 8 + c2 J−1/2 1 2 x 8 1 2 x 8 16 cos πx2 + c2 + C2 x−3/2 cos 1 2 x 8 1 2 x . 8 26. Write the differential equation in the form y − (1/x)y + (4 + 3/4x2 )y = 0. This is the form of (18) in the text with 1 − 2a = −1 =⇒ a = 1 2c − 2 = 0 =⇒ c = 1 b2 c2 = 4 =⇒ b = 2 3 1 a2 − p2 c2 = =⇒ p = . 4 2 294 5.3 Special Functions Then, by (19) in the text, y = x[c1 J1/2 (2x) + c2 J−1/2 (2x)] = x c1 2 sin 2x + c2 π2x 2 cos 2x π2x = C1 x1/2 sin 2x + C2 x1/2 cos 2x. 27. (a) The recurrence relation follows from ∞ −νJν (x) + xJν−1 (x) = − (−1)n ν n!Γ(1 + ν + n) n=0 x 2 x 2 ∞ =− (−1)n ν n!Γ(1 + ν + n) n=0 ∞ = (−1)n (2n + ν) n!Γ(1 + ν + n) n=0 ∞ 2n+ν 2n+ν +x (−1)n n!Γ(ν + n) n=0 x 2 2n+ν−1 ∞ + (−1)n (ν + n) x ·2 n!Γ(1 + ν + n) 2 n=0 x 2 2n+ν−1 2n+ν x 2 = xJν (x). (b) The formula in part (a) is a linear first-order differential equation in Jν (x). An integrating factor for this equation is xν , so d ν [x Jν (x)] = xν Jν−1 (x). dx 28. Subtracting the formula in part (a) of Problem 27 from the formula in Example 5 we obtain 0 = 2νJν (x) − xJν+1 (x) − xJν−1 (x) or 2νJν (x) = xJν+1 (x) + xJν−1 (x). 29. Letting ν = 1 in (21) in the text we have xJ0 (x) = d [xJ1 (x)] dx x so r=x rJ0 (r) dr = rJ1 (r) 0 r=0 = xJ1 (x). 30. From (20) we obtain J0 (x) = −J1 (x), and from (21) we obtain J0 (x) = J−1 (x). Thus J0 (x) = J−1 (x) = −J1 (x). √ 31. Since Γ( 1 ) = π and 2 1 (2n − 1)! √ Γ 1− +n = π n = 1, 2, 3, . . . , 2 (n − 1)!22n−1 we obtain ∞ J−1/2 (x) = (−1)n n!Γ(1 − 1 + n) 2 n=0 1 =√ π x 2 2n−1/2 = 1 Γ( 1 ) 2 ∞ 2 (−1)n 21/2 x−1/2 2n √ x = + x n=1 2n(2n − 1)! π x 2 −1/2 2 + πx ∞ + (−1)n (n − 1)!22n−1 x2n−1/2 √ n!(2n − 1)!22n−1/2 π n=1 ∞ 2 (−1)n 2n x = πx n=1 (2n)! 32. (a) By Problem 28, with ν = 1/2, we obtain J1/2 (x) = xJ3/2 (x) + xJ−1/2 (x) so that J3/2 (x) = 2 πx sin x − cos x ; x with ν = −1/2 we obtain −J−1/2 (x) = xJ1/2 (x) + xJ−3/2 (x) so that 2 cos x + sin x ; πx x and with ν = 3/2 we obtain 3J3/2 (x) = xJ5/2 (x) + xJ1/2 (x) so that J−3/2 (x) = − J5/2 (x) = 2 πx 3 sin x 3 cos x − − sin x . x2 x 295 2 cos x. πx 5.3 Special Functions (b) y 1 0.5 y 1 0.5 ν = 1/2 5 10 ν = −1/2 20 x 15 5 -0.5 -1 y 1 0.5 10 20 x 15 5 -0.5 -1 -0.5 -1 y 1 0.5 y 1 0.5 ν = −3/2 5 10 ν = 5/2 20 x 15 5 -0.5 -1 10 20 x 15 -0.5 -1 33. Letting 2 α s= k −αt/2 , e m we have dx dx ds dx 2 = = dt ds dt dt α dx k α −αt/2 = − e m 2 ds k −αt/2 e m − and d2 x d = 2 dt dt dx dt = = dx ds α 2 k −αt/2 e m = dx ds α 2 k −αt/2 e m dx ds k −αt/2 e m + + d2 x ds ds2 dt − + α 2 d2 x ds2 d dt dx ds − k −αt/2 e m k −αt/2 e m k −αt . e m Then m d2 x d2 x mα + ke−αt x = ke−αt 2 + dt2 ds 2 k −αt/2 dx e + ke−αt x = 0. m ds Multiplying by 22 /α2 m we have 22 k −αt d2 x 2 e + 2 m 2 α ds α k −αt/2 dx 22 k −αt e + 2 e x=0 m ds α m or, since s = (2/α) k/m e−αt/2 , s2 34. Differentiating y = x1/2 w 2 3/2 3 αx d2 x dx +s + s2 x = 0. ds2 ds with respect to 2 αx3/2 we obtain 3 y = x1/2 w 2 3/2 1 αx1/2 + x−1/2 w αx 3 2 2 3/2 αx 3 and y = αxw 2 3/2 αx1/2 + αw αx 3 1 + αw 2 2 3/2 αx 3 2 3/2 αx 3 1 − x−3/2 w 4 2 3/2 . αx 3 Then, after combining terms and simplifying, we have 3 1 y + α2 xy = α αx3/2 w + w + αx3/2 − 2 4αx3/2 296 ν = 3/2 w = 0. 10 15 20 x 5.3 Special Functions Letting t = 2 αx3/2 or αx3/2 = 3 t this differential equation becomes 3 2 3 α 2 1 t w (t) + tw (t) + t2 − 2 t 9 w(t) = 0, t > 0. 35. (a) By Problem 34, a solution of Airy’s equation is y = x1/2 w( 2 αx3/2 ), where 3 w(t) = c1 J1/3 (t) + c2 J−1/3 (t) is a solution of Bessel’s equation of order 2 3/2 αx 3 y = x1/2 w 1 3 . Thus, the general solution of Airy’s equation for x > 0 is = c1 x1/2 J1/3 2 3/2 αx 3 2 3/2 . αx 3 + c2 x1/2 J−1/3 (b) Airy’s equation, y + α2 xy = 0, has the form of (18) in the text with 1 2 3 2c − 2 = 1 =⇒ c = 2 2 2 2 2 b c = α =⇒ b = α 3 1 2 2 2 a − p c = 0 =⇒ p = . 3 1 − 2a = 0 =⇒ a = Then, by (19) in the text, y = x1/2 c1 J1/3 2 3/2 αx 3 + c2 J−1/3 2 3/2 αx 3 . 36. The general solution of the differential equation is y(x) = c1 J0 (αx) + c2 Y0 (αx). In order to satisfy the conditions that limx→0+ y(x) and limx→0+ y (x) are finite we are forced to define c2 = 0. Thus, y(x) = c1 J0 (αx). The second boundary condition, y(2) = 0, implies c1 = 0 or J0 (2α) = 0. In order to have a nontrivial solution we require that J0 (2α) = 0. From Table 5.1, the first three positive zeros of J0 are found to be 2α1 = 2.4048, 2α2 = 5.5201, 2α3 = 8.6537 2 and so α1 = 1.2024, α2 = 2.7601, α3 = 4.3269. The eigenfunctions corresponding to the eigenvalues λ1 = α1 , 2 2 λ2 = α2 , λ3 = α3 are J0 (1.2024x), J0 (2.7601x), and J0 (4.3269x). 37. (a) The differential equation y + (λ/x)y = 0 has the form of (18) in the text with 1 2 1 2c − 2 = −1 =⇒ c = 2 √ 2 2 b c = λ =⇒ b = 2 λ 1 − 2a = 0 =⇒ a = a2 − p2 c2 = 0 =⇒ p = 1. Then, by (19) in the text, √ √ y = x1/2 [c1 J1 (2 λx ) + c2 Y1 (2 λx )]. (b) We first note that y = J1 (t) is a solution of Bessel’s equation, t2 y + ty + (t2 − 1)y = 0, with ν = 1. That is, t2 J1 (t) + tJ1 (t) + (t2 − 1)J1 (t) = 0, 297 5.3 Special Functions √ or, letting t = 2 x , Now, if y = √ √ xJ1 (2 x ), we have y = and √ √ √ √ 4xJ1 (2 x ) + 2 xJ1 (2 x ) + (4x − 1)J1 (2 x ) = 0. √ √ √ √ √ 1 1 1 x J1 (2 x ) √ + √ J1 (2 x ) = J1 (2 x ) + x−1/2 J1 (2 x ) 2 x 2 x √ √ √ 1 1 y = x−1/2 J1 (2 x ) + J1 (2 x ) − x−3/2 J1 (2 x ). 2x 4 Then √ √ √ √ √ 1 1 x J1 2 x + J1 (2 x ) − x−1/2 J1 (2 x ) + x J(2 x ) 2 4 √ √ √ √ √ 1 = √ [4xJ1 (2 x ) + 2 x J1 (2 x ) − J1 (2 x ) + 4xJ(2 x )] 4 x xy + y = √ = 0, √ and y = √ x J1 (2 x ) is a solution of Airy’s differential equation. 38. We see from the graphs below that the graphs of the modified Bessel functions are not oscillatory, while those of the Bessel functions, shown in Figures 5.3 and 5.4 in the text, are oscillatory. I0 I1 I2 20 20 20 15 15 15 10 10 10 5 5 5 1 2 3 4 5 x 1 2 3 5 x 4 1 K0 5 K1 5 4 3 2 1 1 2 3 4 5 x 2 1 5 x 3 2 4 4 3 3 K2 5 4 2 1 1 2 3 4 5 x 1 2 3 4 5 x 39. (a) We identify m = 4, k = 1, and α = 0.1. Then x(t) = c1 J0 (10e−0.05t ) + c2 Y0 (10e−0.05t ) and x (t) = −0.5c1 J0 (10e−0.05t ) − 0.5c2 Y0 (10e−0.05t ). Now x(0) = 1 and x (0) = −1/2 imply c1 J0 (10) + c2 Y0 (10) = 1 c1 J0 (10) + c2 Y0 (10) = 1. 298 5.3 Special Functions Using Cramer’s rule we obtain c1 = Y0 (10) − Y0 (10) J0 (10)Y0 (10) − J0 (10)Y0 (10) c2 = J0 (10) − J0 (10) . J0 (10)Y0 (10) − J0 (10)Y0 (10) and Using Y0 = −Y1 and J0 = −J1 and Table 5.2 we find c1 = −4.7860 and c2 = −3.1803. Thus x(t) = −4.7860J0 (10e−0.05t ) − 3.1803Y0 (10e−0.05t ). x (b) 10 5 t −5 50 40. (a) Identifying α = 1 2 150 100 200 , the general solution of x + 1 tx = 0 is 4 x(t) = c1 x1/2 J1/3 1 3/2 x 3 1 3/2 . x 3 + c2 x1/2 J−1/3 Solving the system x(0.1) = 1, x (0.1) = − 1 we find c1 = −0.809264 and c2 = 0.782397. 2 x (b) 1 t 50 −1 150 100 200 41. (a) Letting t = L − x, the boundary-value problem becomes d2 θ + α2 tθ = 0, dt2 θ (0) = 0, θ(L) = 0, where α2 = δg/EI. This is Airy’s differential equation, so by Problem 35 its solution is y = c1 t1/2 J1/3 2 3/2 αt 3 + c2 t1/2 J−1/3 2 3/2 αt 3 = c1 θ1 (t) + c2 θ2 (t). (b) Looking at the series forms of θ1 and θ2 we see that θ1 (0) = 0, while θ2 (0) = 0. Thus, the boundary condition θ (0) = 0 implies c1 = 0, and so √ θ(t) = c2 t J−1/3 From θ(L) = 0 we have √ c2 L J−1/3 2 3/2 . αt 3 2 3/2 αL 3 = 0, so either c2 = 0, in which case θ(t) = 0, or J−1/3 ( 2 αL3/2 ) = 0. The column will just start to bend when L 3 is the length corresponding to the smallest positive zero of J−1/3 . 299 5.3 Special Functions (c) Using Mathematica, the first positive root of J−1/3 (x) is x1 ≈ 1.86635. Thus 2 αL3/2 = 1.86635 implies 3 L= = 2/3 3(1.86635) 2α = 9EI (1.86635)2 4δg 9(2.6 × 107 )π(0.05)4 /4 (1.86635)2 4(0.28)π(0.05)2 1/3 1/3 ≈ 76.9 in. 42. (a) Writing the differential equation in the form xy + (P L/M )y = 0, we identify λ = P L/M . Problem 37 the solution of this differential equation is √ √ y = c1 x J1 2 P Lx/M + c2 x Y1 2 P Lx/M . Now J1 (0) = 0, so y(0) = 0 implies c2 = 0 and √ y = c1 x J1 2 P Lx/M From . √ (b) From y(L) = 0 we have y = J1 (2L P M ) = 0. The first positive zero of J1 is 3.8317 so, solving 2L P1 /M = 3.8317, we find P1 = 3.6705M/L2 . Therefore, √ y1 (x) = c1 x J1 (c) For c1 = 1 and L = 1 the graph of y1 = is shown. 3.6705x L 2 √ √ = c1 x J1 3.8317 √ √ x L √ x J1 (3.8317 x ) . y 0.3 0.2 0.1 x 0.2 0.4 0.6 0.8 1 43. (a) Since l = v, we integrate to obtain l(t) = vt + c. Now l(0) = l0 implies c = l0 , so l(t) = vt + l0 . Using sin θ ≈ θ in l d2 θ/dt2 + 2l dθ/dt + g sin θ = 0 gives (l0 + vt) d2 θ dθ + 2v + gθ = 0. dt2 dt (b) Dividing by v, the differential equation in part (a) becomes l0 + vt d2 θ dθ g +2 + θ = 0. 2 v dt dt v Letting x = (l0 + vt)/v = t + l0 /v we have dx/dt = 1, so dθ dθ dx dθ = = dt dx dt dx and d2 θ d(dθ/dx) dx d2 θ d(dθ/dt) = = 2. = dt2 dt dx dt dx Thus, the differential equation becomes x d2 θ dθ g +2 + θ=0 2 dx dx v or 300 2 dθ d2 θ g + + θ = 0. 2 dx x dx vx 5.3 Special Functions (c) The differential equation in part (b) has the form of (18) in the text with 1 − 2a = 2 =⇒ a = − 2c − 2 = −1 =⇒ c = b2 c2 = 1 2 1 2 g v g =⇒ b = 2 v a2 − p2 c2 = 0 =⇒ p = 1. Then, by (19) in the text, g 1/2 x v θ(x) = x−1/2 c1 J1 2 + c2 Y1 2 g 1/2 x v or θ(t) = v c1 J1 l0 + vt 2 v g(l0 + vt) + c2 Y1 2 v g(l0 + vt) . (d) To simplify calculations, let u= 2 v g 1/2 x , v g(l0 + vt) = 2 √ and at t = 0 let u0 = 2 gl0 /v. The general solution for θ(t) can then be written θ = C1 u−1 J1 (u) + C2 u−1 Y1 (u). (1) Before applying the initial conditions, note that dθ dθ du = dt du dt so when dθ/dt = 0 at t = 0 we have dθ/du = 0 at u = u0 . Also, dθ d −1 d −1 = C1 [u J1 (u)] + C2 [u Y1 (u)] du du du which, in view of (20) in the text, is the same as dθ = −C1 u−1 J2 (u) − C2 u−1 Y2 (u). du Now at t = 0, or u = u0 , (1) and (2) give the system C1 u−1 J1 (u0 ) + C2 u−1 Y1 (u0 ) = θ0 0 0 C1 u−1 J2 (u0 ) + C2 u−1 Y2 (u0 ) = 0 0 0 whose solution is easily obtained using Cramer’s rule: C1 = u0 θ0 Y2 (u0 ) , J1 (u0 )Y2 (u0 ) − J2 (u0 )Y1 (u0 ) C2 = −u0 θ0 J2 (u0 ) . J1 (u0 )Y2 (u0 ) − J2 (u0 )Y1 (u0 ) In view of the given identity these results simplify to π C1 = − u2 θ0 Y2 (u0 ) 2 0 and C2 = π 2 u θ0 J2 (u0 ). 2 0 The solution is then θ= π 2 J1 (u) Y1 (u) u θ0 −Y2 (u0 ) + J2 (u0 ) . 2 0 u u 301 (2) 5.3 Special Functions √ Returning to u = (2/v) g(l0 + vt) and u0 = (2/v) gl0 , we have 2 √ J1 g(l0 + vt) π gl0 θ0 2 2 v −Y2 √ θ(t) = gl0 + J2 v v v l0 + vt 2 g(l0 + vt) v √ l0 + vt Y1 gl0 . 1 radian, and v = 60 ft/s, the above function is √ √ J1 (480 2(1 + t/60)) Y1 (480 2(1 + t/60)) θ(t) = −1.69045 − 2.79381 . 1 + t/60 1 + t/60 (e) When l0 = 1 ft, θ0 = 1 10 The plots of θ(t) on [0, 10], [0, 30], and [0, 60] are Θ t 0.1 Θ t 0.1 Θ t 0.1 0.05 0.05 2 4 6 8 10 0.05 t 5 10 15 20 25 30 t 10 -0.05 -0.05 -0.1 30 40 50 60 t -0.05 -0.1 20 -0.1 (f ) The graphs indicate that θ(t) decreases as l increases. The Θ t 0.1 graph of θ(t) on [0, 300] is shown. 0.05 50 100 150 200 -0.05 -0.1 44. (a) From (26) in the text, we have 6·7 2 4·6·7·9 4 2 · 4 · 6 · 7 · 9 · 11 6 x + x = x , 2! 4! 6! P6 (x) = c0 1 − where c0 = (−1)3 1·3·5 5 =− . 2·4·6 16 Thus, P6 (x) = − 5 16 1 − 21x2 + 63x4 − 231 6 x 5 = 1 (231x6 − 315x4 + 105x2 − 5). 16 Also, from (26) in the text we have P7 (x) = c1 x − 6 · 9 3 4 · 6 · 9 · 11 5 2 · 4 · 6 · 9 · 11 · 13 7 x + x − x 3! 5! 7! where c1 = (−1)3 1·3·5·7 35 =− . 2·4·6 16 Thus P7 (x) = − 35 16 x − 9x3 + 99 5 429 7 x − x 5 35 = 1 (429x7 − 693x5 + 315x3 − 35x). 16 (b) P6 (x) satisfies 1 − x2 y − 2xy + 42y = 0 and P7 (x) satisfies 1 − x2 y − 2xy + 56y = 0. 302 250 300 t 5.3 Special Functions 45. The recurrence relation can be written Pk+1 (x) = 3 2 1 x − 2 2 5 3 2 1 k = 2: P3 (x) = x x − 3 2 2 2k + 1 k xPk (x) − Pk−1 (x), k+1 k+1 k = 2, 3, 4, . . . . k = 1: P2 (x) = 3 2 5 − x = x3 − x 3 2 2 k = 3: P4 (x) = 7 x 4 5 3 3 3 x − x − 2 2 4 k = 4: P5 (x) = 9 x 5 35 4 30 2 3 x − x + 8 8 8 k = 5: P6 (x) = 11 x 6 3 2 1 x − 2 2 − 4 5 = 35 4 30 2 3 x − x + 8 8 8 5 3 3 x − x 2 2 63 5 35 3 15 5 x − x + x − 8 4 8 6 = 63 5 35 3 15 x − x + x 8 4 8 35 4 30 2 3 x − x + 8 8 8 13 5 231 6 315 4 105 2 6 x x − x + x − − 7 16 16 16 16 7 429 7 693 5 315 3 35 = x − x + x − x 16 16 16 16 46. If x = cos θ then dy dy = − sin θ , dθ dx k = 6: P7 (x) = = 5 231 6 315 4 105 2 x − x + x − 16 16 16 16 63 5 35 3 15 x − x + x 8 4 8 d2 y d2 y dy = sin2 θ 2 − cos θ , 2 dθ dx dx and sin θ d2 y dy + cos θ + n(n + 1)(sin θ)y = sin θ dθ2 dθ 1 − cos2 θ d2 y dy − 2 cos θ + n(n + 1)y = 0. dx2 dx That is, 1 − x2 dy d2 y − 2x + n(n + 1)y = 0. 2 dx dx 47. The only solutions bounded on [−1, 1] are y = cPn (x), c a constant and n = 0, 1, 2, . . . . By (iv) of the properties of the Legendre polynomials, y(0) = 0 or Pn (0) = 0 implies n must be odd. Thus the first three positive eigenvalues correspond to n = 1, 3, and 5 or λ1 = 1 · 2, λ2 = 3 · 4 = 12, and λ3 = 5 · 6 = 30. We can take the eigenfunctions to be y1 = P1 (x), y2 = P3 (x), and y3 = P5 (x). 48. Using a CAS we find 1 d P1 (x) = (x2 − 1)1 = x 2 dx 1 d2 P2 (x) = 2 (x2 − 1)2 = 2 2! dx2 1 d3 P3 (x) = 3 (x2 − 1)3 = 2 3! dx3 1 d4 P4 (x) = 4 (x2 − 1)4 = 2 4! dx4 1 d5 P5 (x) = 5 (x2 − 1)5 = 2 5! dx5 1 d6 P6 (x) = 6 (x2 − 1)6 = 2 6! dx6 1 d7 P7 (x) = 7 (x2 − 1)7 = 2 7! dx7 1 (3x2 − 1) 2 1 (5x3 − 3x) 2 1 (35x4 − 30x2 + 3) 8 1 (63x5 − 70x3 + 15x) 8 1 (231x6 − 315x4 + 105x2 − 5) 16 1 (429x7 − 693x5 + 315x3 − 35x) 16 303 5.3 Special Functions P2 1 P1 1 49. 0.5 P3 1 0.5 -1 -0.5 0.5 1 x -1 -0.5 P4 1 0.5 0.5 1 x -1 -0.5 0.5 0.5 1 x -1 -0.5 -0.5 -0.5 -0.5 -1 -1 1 x -0.5 -1 0.5 -1 P5 1 P6 1 0.5 P7 1 0.5 -1 -0.5 0.5 1 x -1 -0.5 0.5 0.5 1 x -1 -0.5 -0.5 -0.5 -1 1 x -0.5 -1 0.5 -1 50. Zeros of Legendre polynomials for n ≥ 1 are P1 (x) : 0 P2 (x) : ±0.57735 P3 (x) : 0, ±0.77460 P4 (x) : ±0.33998, ±0.86115 P5 (x) : 0, ±0.53847, ±0.90618 P6 (x) : ±0.23862, ±0.66121, ±0.93247 P7 (x) : 0, ±0.40585, ±0.74153 , ±0.94911 P10 (x) : ±0.14887, ±0.43340, ±0.67941, ±0.86506, ±0.097391 The zeros of any Legendre polynomial are in the interval (−1, 1) and are symmetric with respect to 0. CHAPTER 5 REVIEW EXERCISES 1. False; J1 (x) and J−1 (x) are not linearly independent when ν is a positive integer. (In this case ν = 1). The general solution of x2 y + xy + (x2 − 1)y = 0 is y = c1 J1 (x) + c2 Y1 (x). 2. False; y = x is a solution that is analytic at x = 0. 3. x = −1 is the nearest singular point to the ordinary point x = 0. Theorem 5.1 guarantees the existence of two power series solutions y = −1 2 −1 2 ≤x≤ ≤x≤ 1 2 is 1 2 . ∞ n n=1 cn x of the differential equation that converge at least for −1 < x < 1. Since properly contained in −1 < x < 1, both power series must converge for all points contained in 304 CHAPTER 5 REVIEW EXERCISES 4. The easiest way to solve the system 2c2 + 2c1 + c0 = 0 6c3 + 4c2 + c1 = 0 1 12c4 + 6c3 − c1 + c2 = 0 3 2 20c5 + 8c4 − c2 + c3 = 0 3 is to choose, in turn, c0 = 0, c1 = 0 and c0 = 0, c1 = 0. Assuming that c0 = 0, c1 = 0, we have 1 c2 = − c0 2 2 1 c3 = − c2 = c0 3 3 1 1 1 c4 = − c3 − c2 = − c0 2 12 8 2 1 1 1 c5 = − c4 + c2 − c3 = c0 ; 5 30 20 60 whereas the assumption that c0 = 0, c1 = 0 implies c2 = −c1 2 c3 = − c2 − 3 1 c4 = − c3 + 2 2 c5 = − c4 + 5 1 1 c1 = c1 6 2 1 1 5 c1 − c2 = − c1 36 12 36 1 1 1 c2 − c3 = − c1 . 30 20 360 five terms of two power series solutions are then 1 1 1 1 y1 (x) = c0 1 − x2 + x3 − x4 + x5 + · · · 2 3 8 60 and 1 5 1 5 y2 (x) = c1 x − x2 + x3 − x4 − x + ··· . 2 36 360 5. The interval of convergence is centered at 4. Since the series converges at −2, it converges at least on the interval [−2, 10). Since it diverges at 13, it converges at most on the interval [−5, 13). Thus, at −7 it does not converge, at 0 and 7 it does converge, and at 10 and 11 it might converge. 6. We have x5 x3 + − ··· x− sin x 2x5 x3 6 120 f (x) = = + + ··· . =x+ 2 4 cos x 3 15 x x 1− + − ··· 2 24 7. The differential equation (x3 − x2 )y + y + y = 0 has a regular singular point at x = 1 and an irregular singular point at x = 0. 8. The differential equation (x − 1)(x + 3)y + y = 0 has regular singular points at x = 1 and x = −3. 9. Substituting y = ∞ n=0 cn xn+r into the differential equation we obtain ∞ 2xy + y + y = 2r2 − r c0 xr−1 + [2(k + r)(k + r − 1)ck + (k + r)ck + ck−1 ]xk+r−1 = 0 k=1 305 CHAPTER 5 REVIEW EXERCISES which implies 2r2 − r = r(2r − 1) = 0 and (k + r)(2k + 2r − 1)ck + ck−1 = 0. The indicial roots are r = 0 and r = 1/2. For r = 0 the recurrence relation is ck−1 ck = − , k = 1, 2, 3, . . . , k(2k − 1) so 1 1 c1 = −c0 , c2 = c0 , c3 = − c0 . 6 90 For r = 1/2 the recurrence relation is ck−1 ck = − , k = 1, 2, 3, . . . , k(2k + 1) so 1 1 1 c2 = c3 = − c1 = − c0 , c0 , c0 . 3 30 630 Two linearly independent solutions are 1 1 y1 = 1 − x + x2 − x3 + · · · 6 90 and 1 1 3 1 y2 = x1/2 1 − x + x2 − x + ··· . 3 30 630 10. Substituting y = ∞ n n=0 cn x into the differential equation we have ∞ ∞ y − xy − y = ∞ n(n − 1)cn xn−2 − n=2 ncn xn − n=1 k=n−2 n=0 k=n ∞ k=n ∞ (k + 2)(k + 1)ck+2 xk − = cn xn k=0 ∞ kck xk − k=1 ck xk k=0 ∞ = 2c2 − c0 + [(k + 2)(k + 1)ck+2 − (k + 1)ck ]xk = 0. k=1 Thus 2c2 − c0 = 0 (k + 2)(k + 1)ck+2 − (k + 1)ck = 0 and 1 c0 2 1 = ck , k+2 c2 = ck+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 2 c3 = c5 = c7 = · · · = 0 1 c4 = 8 1 c6 = 48 c2 = 306 CHAPTER 5 REVIEW EXERCISES and so on. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 1 c3 = 3 1 c5 = 15 1 c7 = 105 and so on. Thus, two solutions are 1 1 1 y1 = 1 + x2 + x4 + x6 + · · · 2 8 48 and 1 1 1 7 y2 = x + x3 + x5 + x + ···. 3 15 105 11. Substituting y = ∞ n=0 cn xn into the differential equation we obtain ∞ (x − 1)y + 3y = (−2c2 + 3c0 ) + [(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck ]xk = 0 k=1 which implies c2 = 3c0 /2 and ck+2 = (k + 1)kck+1 + 3ck , (k + 2)(k + 1) k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 3 , 2 c3 = 1 , 2 c4 = 5 8 c2 = 0, c3 = 1 , 2 c4 = 1 4 c2 = and so on. For c0 = 0 and c1 = 1 we obtain and so on. Thus, two solutions are 3 1 5 y1 = 1 + x2 + x3 + x4 + · · · 2 2 8 and 1 1 y2 = x + x3 + x4 + · · · . 2 4 12. Substituting y = ∞ n=0 cn xn into the differential equation we obtain ∞ y − x2 y + xy = 2c2 + (6c3 + c0 )x + [(k + 3)(k + 2)ck+3 − (k − 1)ck ]xk+1 = 0 k=1 which implies c2 = 0, c3 = −c0 /6, and ck+3 = k−1 ck , (k + 3)(k + 2) k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 6 c4 = c7 = c10 = · · · = 0 c3 = − c5 = c8 = c11 = · · · = 0 1 c6 = − 90 307 CHAPTER 5 REVIEW EXERCISES and so on. For c0 = 0 and c1 = 1 we obtain c3 = c6 = c9 = · · · = 0 c4 = c7 = c10 = · · · = 0 c5 = c8 = c11 = · · · = 0 and so on. Thus, two solutions are 1 1 y1 = 1 − x3 − x6 − · · · 6 90 13. Substituting y = ∞ n+r n=0 cn x and y2 = x. into the differential equation, we obtain ∞ xy − (x + 2)y + 2y = (r2 − 3r)c0 xr−1 + [(k + r)(k + r − 3)ck − (k + r − 3)ck−1 ]xk+r−1 = 0, k=1 which implies r2 − 3r = r(r − 3) = 0 and (k + r)(k + r − 3)ck − (k + r − 3)ck−1 = 0. The indicial roots are r1 = 3 and r2 = 0. For r2 = 0 the recurrence relation is k(k − 3)ck − (k − 3)ck−1 = 0, Then k = 1, 2, 3, . . . . c1 − c0 = 0 2c2 − c1 = 0 0c3 − 0c2 = 0 =⇒ c3 is arbitrary and ck = 1 ck−1 , k k = 4, 5, 6, . . . . Taking c0 = 0 and c3 = 0 we obtain c1 = c0 1 c2 = c0 2 c3 = c4 = c5 = · · · = 0. Taking c0 = 0 and c3 = 0 we obtain c0 = c1 = c2 = 0 1 6 c4 = c3 = c3 4 4! 1 6 c3 = c3 c5 = 5·4 5! 1 6 c6 = c3 = c3 , 6·5·4 6! and so on. In this case we obtain the two solutions and 1 y1 = 1 + x + x2 2 y2 = x3 + 6 4 6 6 1 x + x5 + x6 + · · · = 6ex − 6 1 + x + x2 . 4! 5! 6! 2 308 CHAPTER 5 REVIEW EXERCISES ∞ n n=0 cn x 14. Substituting y = into the differential equation we have ∞ (cos x)y + y = 1 1 1 6 x + · · · (2c2 + 6c3 x + 12c4 x2 + 20c5 x3 + 30c6 x4 + · · ·) + 1 − x2 + x4 − cn xn 2 24 720 n=0 = 2c2 + 6c3 x + (12c4 − c2 )x2 + (20c5 − 3c3 )x3 + 30c6 − 6c4 + 1 c2 x4 + · · · 12 + [c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + · · · ] = (c0 + 2c2 ) + (c1 + 6c3 )x + 12c4 x2 + (20c5 − 2c3 )x3 + 30c6 − 5c4 + 1 c2 x4 + · · · 12 = 0. Thus c0 + 2c2 = 0 c1 + 6c3 = 0 12c4 = 0 20c5 − 2c3 = 0 1 30c6 − 5c4 + c2 = 0 12 and 1 c2 = − c0 2 1 c3 = − c1 6 c4 = 0 1 c5 = c3 10 1 1 c6 = c4 − c2 . 6 360 Choosing c0 = 1 and c1 = 0 we find 1 1 c2 = − , c3 = 0, c4 = 0, c5 = 0, c6 = 2 720 and so on. For c0 = 0 and c1 = 1 we find 1 1 c2 = 0, c3 = − , c4 = 0, c5 = − , c6 = 0 6 60 and so on. Thus, two solutions are 1 1 6 y1 = 1 − x2 + x + ··· 2 720 15. Substituting y = ∞ n n=0 cn x 1 1 y2 = x − x3 − x5 + · · · . 6 60 and into the differential equation we have ∞ ∞ ∞ n(n − 1)cn xn−2 + y + xy + 2y = n=2 ncn xn + 2 n=1 k=n−2 n=0 k=n ∞ k=n ∞ (k + 2)(k + 1)ck+2 xk + = cn xn k=0 ∞ kck xk + 2 k=1 ck xk k=0 ∞ [(k + 2)(k + 1)ck+2 + (k + 2)ck ]xk = 0. = 2c2 + 2c0 + k=1 309 CHAPTER 5 REVIEW EXERCISES Thus 2c2 + 2c0 = 0 (k + 2)(k + 1)ck+2 + (k + 2)ck = 0 and c2 = −c0 ck+2 = − 1 ck , k+1 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find c2 = −1 c3 = c5 = c7 = · · · = 0 1 c4 = 3 1 c6 = − 15 and so on. For c0 = 0 and c1 = 1 we obtain c2 = c4 = c6 = · · · = 0 1 c3 = − 2 1 c5 = 8 1 c7 = − 48 and so on. Thus, the general solution is 1 1 1 1 1 y = C0 1 − x2 + x4 − x6 + · · · + C1 x − x3 + x5 − x7 + · · · 3 15 2 8 48 and 4 3 2 5 7 y = C0 −2x + x3 − x5 + · · · + C1 1 − x2 + x4 − x6 + · · · . 3 5 2 8 48 Setting y(0) = 3 and y (0) = −2 we find c0 = 3 and c1 = −2. Therefore, the solution of the initial-value problem is 1 1 1 y = 3 − 2x − 3x2 + x3 + x4 − x5 − x6 + x7 + · · · . 4 5 24 16. Substituting y = ∞ n n=0 cn x into the differential equation we have ∞ ∞ n(n − 1)cn xn−1 + 2 (x + 2)y + 3y = n=2 ∞ n(n − 1)cn xn−2 + 3 n=2 k=n−1 k=n ∞ (k + 1)kck+1 xk + 2 = n=0 k=n−2 ∞ cn xn k=1 ∞ (k + 2)(k + 1)ck+2 xk + 3 k=0 ck xk k=0 ∞ [(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + 3ck ]xk = 0. = 4c2 + 3c0 + k=1 Thus 4c2 + 3c0 = 0 (k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + 3ck = 0 310 CHAPTER 5 REVIEW EXERCISES and 3 c2 = − c0 4 ck+2 = − k 3 ck+1 − ck , 2(k + 2) 2(k + 2)(k + 1) k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find c2 = − 3 4 1 8 1 c4 = 16 c3 = c5 = − 9 320 and so on. For c0 = 0 and c1 = 1 we obtain c2 = 0 c3 = − 1 4 1 16 c5 = 0 c4 = and so on. Thus, the general solution is 3 1 1 1 9 5 1 y = C0 1 − x2 + x3 + x4 − x + · · · + C1 x − x3 + x4 + · · · 4 8 16 320 4 16 and 3 3 3 1 9 1 y = C0 − x + x2 + x3 − x4 + · · · + C1 1 − x2 + x3 + · · · . 2 8 4 64 4 4 Setting y(0) = 0 and y (0) = 1 we find c0 = 0 and c1 = 1. Therefore, the solution of the initial-value problem is 1 1 y = x − x3 + x4 + · · · . 4 16 17. The singular point of (1 − 2 sin x)y + xy = 0 closest to x = 0 is π/6. Hence a lower bound is π/6. 18. While we can find two solutions of the form y1 = c0 [1 + · · · ] and y2 = c1 [x + · · · ], the initial conditions at x = 1 give solutions for c0 and c1 in terms of infinite series. Letting t = x − 1 the initial-value problem becomes Substituting y = d2 y dy + (t + 1) + y = 0, y(0) = −6, y (0) = 3. dt2 dt ∞ n n=0 cn t into the differential equation, we have ∞ ∞ ∞ ∞ d2 y dy +y = + (t + 1) n(n − 1)cn tn−2 + ncn tn + ncn tn−1 + cn tn 2 dt dt n=2 n=1 n=1 n=0 k=n−2 k=n ∞ ∞ (k + 2)(k + 1)ck+2 tk + = k=n−1 k=0 ∞ kck tk + k=1 k=n ∞ (k + 1)ck+1 tk + k=0 ck tk k=0 ∞ [(k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck ]tk = 0. = 2c2 + c1 + c0 + k=1 311 CHAPTER 5 REVIEW EXERCISES 2c2 + c1 + c0 = 0 Thus (k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck = 0 c1 + c0 2 ck+1 + ck =− , k+2 c2 = − and ck+2 k = 1, 2, 3, . . . . Choosing c0 = 1 and c1 = 0 we find 1 c2 = − , 2 c3 = 1 , 6 c4 = 1 , 12 and so on. For c0 = 0 and c1 = 1 we find 1 c2 = − , 2 1 c3 = − , 6 c4 = 1 , 6 and so on. Thus, the general solution is 1 1 1 1 1 1 y = c0 1 − t2 + t3 + t4 + · · · + c1 t − t2 − t3 + t4 + · · · . 2 6 12 2 6 6 The initial conditions then imply c0 = −6 and c1 = 3. Thus the solution of the initial-value problem is 1 1 1 y = −6 1 − (x − 1)2 + (x − 1)3 + (x − 1)4 + · · · 2 6 12 1 1 1 + 3 (x − 1) − (x − 1)2 − (x − 1)3 + (x − 1)4 + · · · . 2 6 6 19. Writing the differential equation in the form y + 1 − cos x y + xy = 0, x and noting that 1 − cos x x x3 x5 = − + − ··· x 2 24 720 is analytic at x = 0, we conclude that x = 0 is an ordinary point of the differential equation. 20. Writing the differential equation in the form y + x y=0 ex − 1 − x and noting that x 2 2 x x2 = − + + − ··· ex − 1 − x x 3 18 270 we see that x = 0 is a singular point of the differential equation. Since x2 x ex − 1 − x = 2x − 2x2 x3 x4 + + − ··· , 3 18 270 we conclude that x = 0 is a regular singular point. 312 CHAPTER 5 REVIEW EXERCISES 21. Substituting y = ∞ n n=0 cn x into the differential equation we have ∞ ∞ ∞ n(n − 1)cn xn−2 + y + x2 y + 2xy = n=2 ncn xn+1 + 2 n=1 k=n−2 n=0 k=n+1 ∞ k=n+1 ∞ k=0 ∞ (k − 1)ck−1 xk + 2 (k + 2)(k + 1)ck+2 xk + = cn xn+1 k=2 ck−1 xk k=1 ∞ [(k + 2)(k + 1)ck+2 + (k + 1)ck−1 ]xk = 5 − 2x + 10x3 . = 2c2 + (6c3 + 2c0 )x + k=2 Thus, equating coefficients of like powers of x gives 2c2 = 5 6c3 + 2c0 = −2 12c4 + 3c1 = 0 20c5 + 4c2 = 10 (k + 2)(k + 1)ck+2 + (k + 1)ck−1 = 0, k = 4, 5, 6, . . . , and c2 = 5 2 1 1 c3 = − c0 − 3 3 1 c4 = − c1 4 1 1 1 1 c5 = − c2 = − 2 5 2 5 1 ck+2 = − ck−1 . k+2 5 2 =0 Using the recurrence relation, we find 1 1 1 1 c6 = − c3 = (c0 + 1) = 2 c0 + 2 6 3·6 3 · 2! 3 · 2! 1 1 c7 = − c4 = c1 7 4·7 c8 = c11 = c14 = · · · = 0 1 1 1 c9 = − c6 = − 3 c0 − 3 9 3 · 3! 3 · 3! 1 1 c10 = − c7 = − c1 10 4 · 7 · 10 1 1 1 c0 + 4 c12 = − c9 = 4 12 3 · 4! 3 · 4! 1 1 c13 = − c0 = c1 13 4 · 7 · 10 · 13 313 CHAPTER 5 REVIEW EXERCISES and so on. Thus 1 1 1 1 y = c0 1 − x3 + 2 x6 − 3 x9 + 4 x12 − · · · 3 3 · 2! 3 · 3! 3 · 4! 1 1 7 1 1 + c1 x − x4 + x − x10 + x13 − · · · 4 4·7 4 · 7 · 10 4 · 7 · 10 · 13 + 22. (a) From y = − 5 2 1 3 1 1 1 x − x + 2 x6 − 3 x9 + 4 x12 − · · · . 2 3 3 · 2! 3 · 3! 3 · 4! 1 du we obtain u dx dy 1 1 d2 u + 2 =− dx u dx2 u 2 du dx . Then dy/dx = x2 + y 2 becomes − 1 d2 u 1 + 2 2 u dx u du dx 2 = x2 + du dx 1 u2 2 , d2 u + x2 u = 0. dx2 so (b) The differential equation u + x2 u = 0 has the form (18) in the text with 1 2 2c − 2 = 2 =⇒ c = 2 1 b2 c2 = 1 =⇒ b = 2 1 a2 − p2 c2 = 0 =⇒ p = . 4 1 − 2a = 0 =⇒ a = Then, by (19) in the text, u = x1/2 c1 J1/4 1 2 x 2 + c2 J−1/4 1 2 x 2 . (c) We have y=− =− 1 du dw dt 1 d 1/2 1 1 x1/2 = − 1/2 x w(t) = − 1/2 + x−1/2 w u dx dt dx 2 x w(t) dx x w 1 dw dw 1 1 1 dw x3/2 + 1/2 w = − 2x2 +w =− 4t +w . dt 2xw dt 2xw dt x1/2 w 2x Now 4t dw d + w = 4t [c1 J1/4 (t) + c2 J−1/4 (t)] + c1 J1/4 (t) + c2 J−1/4 (t) dt dt = 4t c1 J−3/4 (t) − 1 1 J1/4 (t) + c2 − J−1/4 (t) − J3/4 (t) 4t 4t = 4c1 tJ−3/4 (t) − 4c2 tJ3/4 (t) = 2c1 x2 J−3/4 so y=− 1 2 x 2 − 2c2 x2 J3/4 + c1 J1/4 (t) + c2 J−1/4 (t) 1 2 x , 2 2c1 x2 J−3/4 ( 1 x2 ) − 2c2 x2 J3/4 ( 1 x2 ) −c1 J−3/4 ( 1 x2 ) + c2 J3/4 ( 1 x2 ) 2 2 2 2 =x . 2x[c1 J1/4 ( 1 x2 ) + c2 J−1/4 ( 1 x2 )] c1 J1/4 ( 1 x2 ) + c2 J−1/4 ( 1 x2 ) 2 2 2 2 314 CHAPTER 5 REVIEW EXERCISES Letting c = c1 /c2 we have y=x J3/4 ( 1 x2 ) − cJ−3/4 ( 1 x2 ) 2 2 . cJ1/4 ( 1 x2 ) + J−1/4 ( 1 x2 ) 2 2 23. (a) Equations (10) and (24) of Section 5.3 in the text imply Y1/2 (x) = cos π J1/2 (x) − J−1/2 (x) 2 = −J−1/2 (x) = − sin π 2 2 cos x. πx (b) From (15) of Section 5.3 in the text I1/2 (x) = i−1/2 J1/2 (ix) I−1/2 (x) = i1/2 J−1/2 (ix) and so ∞ I1/2 (x) = 2 1 x2n+1 = πx n=0 (2n + 1)! and ∞ I−1/2 (x) = 2 1 x2n = πx n=0 (2n)! 2 sinh x πx 2 cosh x. πx (c) Equation (16) of Section 5.3 in the text and part (b) imply K1/2 (x) = π I−1/2 (x) − I1/2 (x) π = 2 sin π 2 2 2 cosh x − πx π ex + e−x ex − e−x − = 2x 2 2 = 2 sinh x πx π −x e . 2x 24. (a) Using formula (5) of Section 3.2 in the text, we find that a second solution of (1 − x2 )y − 2xy = 0 is y2 (x) = 1 · e 2x dx/(1−x2 ) dx 1 = ln 2 1−x 2 = e− ln(1−x ) dx 2 dx = 12 1+x 1−x , where partial fractions was used to obtain the last integral. (b) Using formula (5) of Section 3.2 in the text, we find that a second solution of (1 − x2 )y − 2xy + 2y = 0 is y2 (x) = x · =x e 2x dx/(1−x2 ) x2 e− ln(1−x x2 2 dx = x dx 1 dx = x ln x2 (1 − x2 ) 2 1+x 1−x where partial fractions was used to obtain the last integral. 315 − ) dx 1 x = ln x 2 1+x 1−x − 1, CHAPTER 5 REVIEW EXERCISES y2 2 y2 2 1 (c) 1 1x -1 1x -1 -1 -2 y2 (x) = -1 -2 1 ln 2 1+x 1−x y2 = x ln 2 1+x 1−x −1 25. (a) By the binomial theorem we have 1 + t2 − 2xt −1/2 1 2 (−1/2)(−3/2) 2 (−1/2)(−3/2)(−5/2) 2 t − 2xt + (t − 2xt)2 + (t − 2xt)3 + · · · 2 2! 3! 1 3 5 = 1 − (t2 − 2xt) + (t2 − 2xt)2 − (t2 − 2xt)3 + · · · 2 8 16 ∞ 1 1 = 1 + xt + (3x2 − 1)t2 + (5x3 − 3x)t3 + · · · = Pn (x)tn . 2 2 n=0 =1− (b) Letting x = 1 in (1 − 2xt + t2 )−1/2 , we have (1 − 2t + t2 )−1/2 = (1 − t)−1 = From part (a) we have ∞ 1 = 1 + t + t2 + t 3 + . . . 1−t Pn (1)tn = (1 − 2t + t2 )−1/2 = n=0 ∞ tn . (|t| < 1) = n=0 ∞ tn . n=0 Equating the coefficients of corresponding terms in the two series, we see that Pn (1) = 1. Similarly, letting x = −1 we have (1 + 2t + t2 )−1/2 = (1 + t)−1 = 1 = 1 − t + t2 − 3t3 + . . . 1+t ∞ ∞ (−1)n tn = = n=0 Pn (−1)tn , n=0 n so that Pn (−1) = (−1) . 316 (|t| < 1) ...
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This note was uploaded on 05/05/2008 for the course MATH 240 taught by Professor Storm during the Spring '08 term at UPenn.

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