求(x-2xy-y^2)y'+y^2=0的通解

求(x-2xy-y^2)y'+y^2=0的通解
求(x-2xy-y^2)y'+y^2=0的通解

求(x-2xy-y^2)y'+y^2=0的通解
(x-2xy-y^2)y'+y^2=0
dx/dy=-(x-2xy-y^2)/y^2=-x/y^2+2x/y+2
令x/y=p
x=py
x'=p+p'y
代入得
p+p'y=1/y*p+2p+2
你确定前面是x-2xy-y^2吗?