推 zendla:我知道自己哪裡算錯了= = 感謝你的詳細解題過程 11/17 12:17
※ 引述《zendla (夏夜薄荷)》之銘言：
: 2.Use the transformation z = sinx to solve [成大土木]
: 2
: d y dy 2
: ─── + (tanx) ── + (cos x)y = 0
: dx^2 dx
: ans: y = c1cos(sinx) + c2sin(sinx)
: 這題我無法轉成常係數類型的，那應該怎麼做?
令 z = sinx, dz/dx = cosx = sqrt(1-z^2)
dy/dx = (dy/dz)(dz/dx) = sqrt(1-z^2) (dy/dz)
d^2y/dx = d/dx (sqrt(1-z^2) (dy/dz))
= d/dz (sqrt(1-z^2) (dy/dz)) (dz/dx)
= (sqrt(1-z^2) (d^2y/dz^2) + 1/sqrt(1-z^2)/2*(-2z) dy/dz) sqrt(1-z^2)
= (1-z^2) d^2y/dz^2 -2z dy/dz
cosx = sqrt(1-z^2), tanx = z/sqrt(1-z^2)
原式可整理為
(1-z^2)(d^2y/dz^2) + (1-z^2)y = 0
若 1-z^2 = cosx ~=0
=> d^2y/dz^2 + y = 0
=> y = C1 cosz + C2 sinz
= C1 cos(sinx) + C2 sin(sinx)
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