Re: [工數]一題卡很久的高階ODE...
※ 引述《storange (我是誰?)》之銘言：
: y''- ay/x^2 = 0
: 其中a是常數項
: 邊界條件分別是 x=kR y=0
: x=R y=w
: 我試過令 y = exp(mx) 去化簡
: 可是分母的平方項讓我整個卡住了orz
: 希望有神人能幫我解出答案~~謝謝!!!
令t=x^(1/2)
dt 1 1
--- = --- x^(-1/2) = ----
dx 2 2t
dy dy dt 1 dy
--- = --- --- = ---- ---
dx dt dx 2t dt
d^2y d 1 dy dt 1 d^2y 1 dy 1
---- = --- ( ---- --- ) --- = (---- ---- - ---- --- ) ----
dx^2 dt 2t dt dx 2t dt^2 2t^2 dt 2t
1 d^2y 1 dy
= ----- ---- - ----- ---
4t^2 dt^2 4t^3 dt
代回原方程式得
1 d^2y 1 dy a
----- ---- - ----- --- - --- y = 0 同乘4t^4
4t^2 dt^2 4t^3 dt t^4
d^2y dy
t^2 ---- - t --- - 4ay = 0 為Cauchy–Euler equation
dt^2 dt
令 y = t^m 代入
m(m-1) - m - 4a = 0 => m^2 - 2m -4a = 0, m = 1 + √(1+4a), 1-√(1+4a)
y = C1 t^[1 + √(1+4a)] + C2 t^[1 - √(1+4a)]
= C1 x^[(1 + √(1+4a))/2] + C2 x^[(1 - √(1+4a))/2]
y(kR) = 0, y(R) = w
(kR)^[(1 + √(1+4a))/2] (kR)^[(1 - √(1+4a))/2] C1 0
=> [ ] [ ] = [ ]
R^[(1 + √(1+4a))/2] R^[(1 - √(1+4a))/2] C2 w
=> A X = b
det(A) = k^[(1 + √(1+4a))/2]*R - k^[(1 - √(1+4a))/2]*R
X = inv(A) * b, C1 = -(kR)^[(1 - √(1+4a))/2]*w / det(A)
C2 = (kR)^[(1 + √(1+4a))/2]*w / det(A)
真不是普通的醜...
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