令y/x=z,y=zx,y'=z'x+z
原方程化为
(x+y)y'=x
(1+y/x)y'=1
(1+z)(z'x+z)=1
z'x=1/(1+z)-z=(1-z-z^2)/(1-z)
(1-z)/(1-z-z^2)dz=dx/x
两边积分就可以啦