求y'=y/(y-x)

∵令y=xt,则y'=xt'+t
代入原方程,得xt'+t=t/(t-1)
==>xt'=(2t-t^2)/(t-1)
==>(t-1)dt/(2t-t^2)=dx/x
==>2dx/x+[1/t+1/(t-2)]dt=0
==>2ln│x│+ln│t│+ln│t-2│=ln│C│ (C是常数)
==>x^2t(t-2)=C
==>x^2(y/x)(y/x-2)=C
==>y(y-2x)=C
∴原方程的通解是y(y-2x)=C.