2
解 8x u''+ 6xu'+ (x-1)u = 0
我的解法: 用method of Frobenius
2
x u''+ (3x/4)u'+ (x-1)u/8 = 0
P_0(x) = 3/4 => b_0 = 3/4 otherwise b_n = 0
P_1(x) = (x-1)/8 => c_0 = -1/8, c_1 = 1/8 otherwise c_n = 0
r ∞ n
則解 u(x) = x Σ (a_n)x , a_n ≠ 0
n=0
indicial eq : F(r) = r(r-1) + (3/4)r - (1/8) = 0 => r = 1/2 or -1/4
n
解遞迴 F(n+r)a_n + Σ [G_n(k)][a_(n-k)] = 0, (n = 1,2,3,...) ………(*)
k=1
2
F(n+(1/2)) = (n+(1/2))(n-(1/2)) + (3/4)(n+(1/2)) - (1/8) = n + (3/4)n
G_n(k) = (n-k+r)b_k + c_k
=> G_n(1) = (n-1+(1/2))*b_1 + c_1 = c_1 = 1/8, otherwise G_n(k) = 0
(*) 2
===> (n +(3/4)n)a_n + (1/8)a_(n-1) = 0
n
-1 (-1) a_0
=> a_n = ────── a_(n-1) => a_n = ───────，令 a_0 = 1
8n^2 + 6n n 2
Π (8i + 6i)
i=1
n
(1/2) ∞ (-1) n (1/2)
所以一解為 u_1(x) = x Σ ─────── x + x
n=1 n 2
Π (8i + 6i)
i=1
因為(1/2)-(-1/4) = 3/4，非整數，所以同理
n
(-1/4) ∞ (-1) n (-1/4)
另外一解為 u_2(x) = x Σ ─────── x + x
n=1 n 2
Π (8i - 6i)
i=1
則 u(x) = Au_1 + Bu_2
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我用mathematica解，他給我Bessel和Gama function的型態
我不知道該怎麼轉換，還是我算錯了呢?
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