解法一：∵y'=y/(y-x) ==>(y-x)y'=y
==>(y-x)dy=ydx
==>ydy=ydx+xdy
==>d(y²)=2d(xy)
==>y²=2xy+C (C是積分常數(shù))
∴原方程的通解是y²=2xy+C (C是積分常數(shù)).
解法二：令y=xt,則y'=xt'+t
代入原方程,得xt'+t=t/(t-1)
==>xt'=t(2-t)/(t-1)
==>(t-1)dt/[t(2-t)]=dx/x
==>[1/(2-t)-1/t]dt=2dx/x
==>ln│t-2│+ln│t│=-2ln│x│+ln│C│ (C是積分常數(shù))
==>t(t-2)=C/x²
==>(y/x)(y/x-2)=C/x²
==>y(y-2x)=C
==>y²-2xy=C
==>y²=2xy+C
故原方程的通解是y²=2xy+C (C是積分常數(shù)).