求微分方程dy/dx+2xy=4x的通解 求大神指教啊

dy/dx+2xy=4xdy/dx=4x-2xy=2x(2-y)dy/(2-y)=2xdx-d(2-y)/(2-y)=dx^2-dln(2-y)=dx^2dln[1/(2-y)]=dx^2ln[1/(2-y)]=x^2+C1/(2-y)=e^(x^2+C)2-y=e^(-x^-C)y=2-e^(-x-C)=2-Ce^(-x)