Re: [微積] 一些很簡單的微分方程式
※ 引述《dkcheng (電磁霸主)》之銘言：
: 小弟不才
: 以下幾題ODE真的想不到要怎麼解
: 煩請板上各位高手了
: 1. Solve (sinycosy+xcos^2y)dx+xdy=0
: y^2+2y
: 2. Solve y'= --------------
: y^4+2xy+4x
: 3. Solve (2xcosy+4x^2)dx=xsinydy
: 4. Solve the ODE and find an integrating factor
: (3y^2+x+1)dx+2y(x+1)dy=0
1. 同除cos^2y 得
tany dx + xdx + x/(cos^2y) dy = 0
tany dx + xdx + xd(tany) = 0
d(xtany) + xdx = 0
xtany + 0.5x^2 = c
2. 先乘開 得
(y^4+2xy+4x) dy = (y^2+2y)dx
(y+2)(ydx-2xdy) = y^4dy
y(y+2) d(xy^-2) = y^4dy
d(xy^-2) = { y^4/[y(y+1)] }dy
兩邊積分
3. 同除x 得
2cosydx + 4xdx = sinydy
2cosy dx + d(cosy) = -4x dx
積分因子 I(x) = e^2x
d(cosy*e^2x) = -4x*e^2x dx
兩邊積分
4.
(3y^2+x+1)dx+2y(x+1)dy=0
3y^2 dx + (x+1)dx + (x+1)d(y^2) = 0
同除 x+1 得
d(y^2) + [3/(x+1)]y^2 dx + 1dx = 0
同乘 e^[3ln(x+1)] = (x+1)^3 後正合
===> I(x) = (x+1)^2
d[ y^2*(x+1)^3 ] + (x+1)^3dx = 0
兩邊積分
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