Rearranging the terms in the differential equation, we get :
$\frac{dy}{1-4y}=xdx$
Integrating on both sides, we get:
$\int \frac{dy}{1-4y}=\int xdx$
$\Rightarrow$ $\frac{1}{-4}\times ln(1-4y)=\frac{x^{2}}{2}+c$
Now, using the condition given in the question i.e. $y(0)=0$, we get $c=0$
$\therefore$ We get the equation as $\frac{1}{-4}\times ln(1-4y)=\frac{x^{2}}{2}$
$\Rightarrow$ $ln(1-4y)=-2x^{2}$ $\Rightarrow$ $e^{-2x^{2}}=1-4y$
$\Rightarrow$ $y(x)=\frac{1}{4}(1-e^{-2x^{2}})$
Option B is the correct answer.