∵dy/dx-(1/x)y=2x^2
==>dy-ydx/x=2x^2dx
==>dy/x-ydx/x^2=2xdx (等式两端同除x)
==>d(y/x)=d(x^2)
==>y/x=x^2+C (C是常数)
==>y=x^3+Cx
∴原方程的通解是y=x^3+Cx.
求dy/dx-(1/x)y=2x^2的通解
∵dy/dx-(1/x)y=2x^2
==>dy-ydx/x=2x^2dx
==>dy/x-ydx/x^2=2xdx (等式两端同除x)
==>d(y/x)=d(x^2)
==>y/x=x^2+C (C是常数)
==>y=x^3+Cx
∴原方程的通解是y=x^3+Cx.