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]
