微分方程dx/dy=(2xy-y^2)/(x^2-2xy)满足y(1)=-2的特解是?

dx/dy=(2xy-y²)/(x²-2xy)
dy/dx=(x²-2xy)/(2xy-y²) 分子分母同时除以x²
dy/dx=(1-1y/x)/[2y/x-(y/x)²]
设y/x=u
u+xu'=(1-2u)/(2u-u²)
xdu/dx=(u³-2u²-2u+1)/(2u-u²)
分离变量得
(2u-u²)/[(u+1)(u²-3u+1)]du=dx/x
-1/5[(2u-3)/(u²-3u+1)+3/(u+1)]du=dx/x
两边积分
-1/5∫(2u-3)/(u²-3u+1)]dx-3/5∫1/(u+1)du=∫1/xdx
-1/5∫1/(u²-3u+1)d(u²-3u+1)-3/5∫1/(u+1)d(u+1)=∫1/xdx
-1/5ln(u²-3u+1)-3/5ln(u+1)=lnx+C1
(u²-3u+1)(u+1)³=C/x^5
即[(y/x)²-3(y/x)+1][(y/x)+1]³x^5=C
(y²-3xy+x²)(y+x)³=C
将y(1)=-2代入通解中
得C=-11
特解为(x²-3xy+y²)(x+y)³=-11