令y=xt,则dy=xdt+tdx
代入原方程,化简得dx/x=-3t^2dt/(t^3+1)
==>ln│x│=ln│C│-ln│t^3+1│ (C是积分常数)
==>x=C/(t^3+1)
==>x=C/((y/x)^3+1)
==>1=Cx^2/(x^3+y^3)
==>x^3+y^3=Cx^2
故 原方程的通解是x^3+y^3=Cx^2.