微分方程y'=y^2/(xy-x^2)

网友回答

令y/x=t,则y'=xt'+t
代入原方程,得
y'=(y/x)²/((y/x)-1)
==>xt'+t=t²/(t-1)
==>xt'=t/(t-1)
==>dx/x=(1-1/t)dt
==>ln│x│=t-ln│t│+ln│C│ (C是积分常数)
==>xt=Ce^t
==>x(y/x)=Ce^(y/x)
==>y=Ce^(y/x)
故原方程的通解是y=Ce^(y/x) (C是积分常数).