張衡工數-第 7 次測驗 範圍:級數解(Bessel、Legendre 方程式) 學號: 就讀學校科系: 姓名: 目標: 1. Find the first five nonzero terms of the power series solution of the initial value problem, about the point where the initial conditions are given: y e x y x2 , y(0) 4 . (15%) 【99 成大光電&交大機械丁(17%)】 3 解: y ( x) 4 4 x x 2. Let 1 4 7 5 x x 12 60 y be a real function of x . Find two linearly independent Frobenius solutions of the following differential equation at x 0 : 2x2 y x( x 3) y 3 y 0 (20%) 【99 台聯大(A)、(C) 、(D) (20%)】 1 1 1 y c1 x1 x x 2 c2 x 2 1 x x 2 3 8 2 3 解: 3. Use the power series method to solve the following equation (1 x 2 ) where n is a nonnegative integer. 1 (20%) 2 d y dy 2 x n(n 1) y 0 2 dx dx 【98 中山電機乙&台科化工(15%)】 1 2 4 解: y c0 1 2 n(n 1) x 4! n(n 2)(n 1)(n 3) x 1 1 c1 x (n 1)(n 2) x3 (n 1)(n 3)(n 2)(n 4) x 5 3! 5! 4. Bessel equation: x2 y xy ( x2 2 ) y 0 ; is real and non-negative; (20%) (1) Are x 0 and x 1 ordinary, regular singular or irregular singular point? (5%) (2) If n (integer), using Frobenius method to prove Bessel equation’s solution y : (15%) y J n ( x) x n m0 (3) If (1) m x 2 m 22 m n m !(m n)! n (integer), Prove: J n ( x) (1)n J n ( x) (5%) 【98 暨南電機電子(25%)】 1 張衡工數-第 7 次測驗 範圍:級數解(Bessel、Legendre 方程式) 學號: 就讀學校科系: 姓名: 目標: 解:(1) x 0 為 regular singular point, x 1 為 ordinary point (1)n a0(n 1) 2n 2 m (1)n a0n!2n n x x 2m n x2m (2) y1 x 2 m n m!(m n 1) m!(m n 1) m 0 2 m 0 2 n (1) n x 2 m AJ n ( x) (其中 A a 0 n!2 n ) 2m n m!(m n)! m 0 2 Ax n (3) Gamma函數( )之 0 (1) m x J n ( x) m n m! ( m n 1) 2 (1) k n x k 0 ( k m)! k! 2 2mn 2k n (1) m x m n m!( m n)! 2 2m n (1) k x (1) k 0 ( k m)! k! 2 , 令mk n 2k n n (1) k x (1) k 0 k!( k n 1) 2 2k n (1) n J n ( x) n 即 J n ( x) (1) n J n ( x) 得證 5. Let’s denote J v ( x) and Yv ( x) to be the Bessel functions of 1st kind and of 2nd kind, respectively. I recall you that x2 y xy ( x2 v2 ) y 0 is called Bessel’s equation where v is a real and nonnegative number. Find a general solution for the ordinary differential equation xy 11y xy 0 in terms of J v ( x) and Yv ( x) . [Hint: use the substitution y x5u in your derivation.] 【99 清大工科(15%)】 (1) n x 2n ) n ! ( n 1 ) n 0 解: y x 5 c1 J 5 ( x) c2Y5 ( x) (其中 J ( x) 6. Find the power series solutions about 0 for the differential equation. ( x 2 4) y xy x 2 (15%) 【98 台大化工(15%)】 解: y C 0 (1 1 3 1 5 1 4 1 1 3 1 4 1 5 x x ) C1 ( x x ) ( x 2 x x x ) 24 320 48 4 24 96 160 2
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