(x^3+y^3)dx-3xy^2dy=0, 齐次方程的通解?

(x³+y³)dx-3xy²dy=0,齐次方程的通解?dy/dx=(x³+y³)/3xy²=(1/3)[(x/y)²+(y/x)]=(1/3)[1/(y/x)²+(y/x)]令y/x=u,则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)分离变量得x/dx=(1-2u³)/(3u²du)取倒数得(1/x)dx=3u²du/(1-2u³)=-(1/2)[d(1-2u³)]/(1-2u³)两边取积分得lnx=-(1/2)ln(1-2u³)+lnC₁=ln[C₁/√(1-2u³)]故得x=C₁/√(1-2u³)],将u=y/x代入得x=C₁/√[1-2(y/x)³)]=C₁x(√x)/√(x³-2y³)于是得√(x³-2y³)=C₁√x平方去根号便得原方程的通解为：x³-2y³=Cx,其中C=C²₁