設(shè)z=1/y³,則y³=1/z,dy=-y^4dz/3
代入原方程得-xy^4dz/3+(y+x²y^4)dx=0
==>-xy³dz/3+(1+x²y³)dx=0
==>-xdz/(3z)+(1+x²/z)dx=0
==>xdz-(z+x²)dx=0
==>dz/dx-z/x=x.(1)
方程(1)是一階線性方程,可用常數(shù)變易法求解.
先求齊次方程dz/dx-z/x=0的通解
∵dz/dx-z/x=0 ==>dz/z=dx/x
==>ln│z│=ln│x│+ln│C│ (C是積分常數(shù))
==>z=Cx
∴齊次方程dz/dx-z/x=0的通解z=Cx (C是積分常數(shù))
于是,設(shè)方程(1)的通解為z=C(x)x (C(x)表示關(guān)于x的函數(shù))
∵dz/dx=C'(x)x+C(x)
代入方程(1)得C'(x)x+C(x)-C(x)=x
==>C'(x)=1
==>C(x)=x+C (C是積分常數(shù))
∴方程(1)的通解是z=x(x+C)
==>1/y³=x(x+C)
==>y³=1/[x(x+C)]
故原微分方程通解是y³=1/[x(x+C)] (C是積分常數(shù)).