求dy/dx+2xy+xy^4的通解
1个回答

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).