设z=1/y³,则dz/dx=(-3/y^4)(dy/dx) ==>dy/dx=(-y^4/3)(dz/dx)
代入原方程,得(-y^4/3)(dz/dx)+2xy+xy^4=0
==>dz/dx-6x/y³-3x=0
==>dz/dx-6xz-3x=0
==>dz/dx=3x(2z+1)
==>2dz/(2z+1)=6xdx
==>ln│2z+1│=3x²+ln│C│ (C是积分常数)
==>2z+1=Ce^(3x²)
==>2/y³=Ce^(3x²)-1
==>2=[Ce^(3x²)-1]y³
故原方程的通解是[Ce^(3x²)-1]y³=2 （C是积分常数）.