Re: [理工] [工數] ODE
※ 引述《justwantyou (饅頭)》之銘言：
: 這題解到一半卡住...
: solve xp^2 - 2yp + 4x =0 p=y'
: xp 2x
: 我移項成 y=－ + －
: 2 p
: 然後一起微分 p= p/2 + xdp/2dx + 2/p - (2x/p^2)dp/dx
: 整理>>> 0=(p/2 - 2/p)( 1+ xdp/dx)
這邊我整理的似乎跟你有些出入
1 2 dp
[ --- - --- ][p - x*----] =0
2 p^2 dx
____________ ______________
↓ ↘
解得p = ±2 = y' ↘
p =c2*x = y'
xp 2x 1 2
帶入y= --- + --- y = c2*---*x + c3
2 p 2
負不合,故取正 帶入原方程式
x(y')^2 - 2yy' +4x =0
y = 2x + c1
可解得c2*c3= 2
2
帶入原方程式 配合你的答案取c2 = ----
x(y')^2 - 2yy' +4x =0 c3
1 2
得c1 = 0 y = ---*x + c3
y= 2x 為一組特解 c3
────────# 2 2
c3*y = x + c3 c3 is constant = c
2 2
c*y = x + c
──────#
: 然後我就不會了~"~...想請問接著要怎麼處理
: 答案是 cy=x^2 + c^2
: 另外一個是解 y" + 4y = sin2t y(0)=y'(0)=0
: 這題我算出來的答案怪怪的 想問一下解題的方法
: 謝謝大家~
這題看下來用Laplace最容易
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