Question

求微分方程的通解…要过程

Answer 1

设u=y/x du/dx=(y'x-y)/x^2
y'=xdu/dx+y/x=xdu/dx+u................1
dy/dx=-(x^2+2xy-y^2)/(y^2+2xy-x^2)
dy/dx=(u^2-2u-1)/(u^2+2u-1)
将1式代入上式：
xdu/dx+u=(u^2-2u-1)/(u^2+2u-1)
xdu/dx=[(u^2-2u-1)/(u^2+2u-1)-u]
1/xdx=1/[(u^2-2u-1)/(u^2+2u-1)-u]
1/xdx=(u^3+2u^2-u)/(u^3-2u^2-u-u^3-2u^2+u)du
1/xdx=(u^3+2u^2-u)/(-4u^2)du
1/xdx=(u^2+2u-1)/(-4u)du=-1/4(u+2-1/u)du
两边积分：
lnx=-1/4[1/2u^2+2u-lnu]+C
lnx=-1/8u^2-1/2u+1/4 lnu+C
8lnx=-(y/x)^2-1/2(y/x)+1/4ln(y/x)+C
x=1, y=1
0=-1-1/2+0+c
c=3/2
8lnx=-(y/x)^2-1/2(y/x)+1/4ln(y/x)+3/2

追问

能不能写在纸上？

追答

。错了，等等，在改
-1/xdx=(u^2+2u-1)/[(u^2+1)(u+1)]du=[(2u/(u^2+1)-1/(u+1)]du
积分；-lnx=ln[(u^2+1)/(u+1)]+c
1/x=C(u^2+1)/(u+1)
y+x=C(y^2+x^2)

Answer 2