Re: [理工] [工數]-ODE 矩陣
※ 引述《NewFighterZR (NFT-ZR)》之銘言：
: 3
: 1. xy'' = y' + (y')
: 這題應該是令 P=y' P'=y''去解吧0.0?
: 但解到後面完全卡死..不會整理成y的表示式 請高手幫忙列詳細過程
: Ans: y = ±√(c1 - x^2) + c2
3
xy'' = y' + (y') ...(*)
Let u=y' ...(1)
du
u' = ---- = y'' ...(2)
dx
substituting (1) (2) into (*)
xu' = u + u^3
du
x ---- = u + u^3 → by separable method
dx
du dx
---------- = ----
u(1 + u^2) x
1 u dx
[--- - ------- ] du = ----
u 1 + u^2 x
1
ln|u| - ---ln(1+u^2) = ln(x) + c
2
u
ln(--------------) = ln(c1 * x)
(1+u^2)^(1/2)
u
------------- = c1x
(1+u^2)^(1/2)
u = c1 * x * ((1+u^2)^(1/2))
u^2 = (c1*x)^2 * (1+u^2)
(u^2)(1- (c1*x)^2) = (c1*x)^2
(c1*x)^2
u^2 = -------------
(1- (c1*x)^2)
(c1*x) dy
u = ± --------------- = ----
√(1- (c1*x)^2) dx
(c1*x)
dy = ±--------------- dx
√(1- (c1*x)^2)
y = ±√(c1 - x^2) + c2
小弟淺見，有錯還煩請高手不吝指正 <(_ _)>
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 140.114.123.85
推
03/04 15:21, , 1F
03/04 15:21, 1F
討論串 (同標題文章)