Solve: dy/dx − 2xy = x, with y(0) = 0.

Answer:[tex]1 + 2y = x^2[/tex]Step-by-step explanation:Given differential equation,[tex]\frac{dy}{dx}-2xy=x[/tex][tex]\frac{dy}{dx}=x+2xy[/tex][tex]\frac{dy}{dx}=x(1+2y)[/tex][tex]\frac{dy}{1+2y}=xdx[/tex]Integrating both sides,[tex]\int \frac{dy}{1+2y}=\int xdx----(1)[/tex]Put 1 + 2y = u Differentiating both sides,2dy = du[tex]\implies dy=\frac{du}{2}[/tex]From equation (1),[tex]\frac{1}{2} \int \frac{du}{u}=\int xdx[/tex][tex]\frac{1}{2}\log u = \log x + C[/tex][tex]\frac{1}{2} \log (1+2y)=\log x+C---(2)[/tex]If x = 0, y = 0[tex]\implies C=0[/tex]From equation (2),[tex]\frac{1}{2} \log(1+2y)=log x[/tex][tex]\log (1+2y) = 2\log x[/tex][tex]\log (1+2y) = \log x^2[/tex][tex]\implies 1 + 2y = x^2[/tex]