dy/dx=(x³+y³)/3xy²=(1/3)[(x/y)²+(y/x)]=(1/3)[1/(y/x)²+(y/x)]
令y/x=u,則y=ux,dy/dx=u+x(du/dx),代入上式得：
u+x(du/dx)=(1/3)[(1/u²)+u]
故有x(du/dx)=1/(3u²)-(2/3)u=(1-2u³)/(3u²)
分離變量得x/dx=(1-2u³)/(3u²du)
取倒數(shù)得(1/x)dx=3u²du/(1-2u³)=-(1/2)[d(1-2u³)]/(1-2u³)
兩邊取積分得lnx=-(1/2)ln(1-2u³)+lnC₁=ln[C₁/√(1-2u³)]
故得x=C₁/√(1-2u³)],將u=y/x代入得x=C₁/√[1-2(y/x)³)]=C₁x(√x)/√(x³-2y³)
于是得√(x³-2y³)=C₁√x
平方去根號(hào)便得原方程的通解為：x³-2y³=Cx,其中C=C²₁