Finding an explicit solution of a differential equation 
Find an explicit solution of an initial value problem $$\frac{dy}{dx}=\frac{2x}{1+2y}, \; y(2)=0$$

Attempt: I have no problem finding the general solution which is 
$y+y^2=x^2+C$
Then, I find the implict solution which I think is correct, but I am not sure  
$y+y^2=x^2-4$
Now, how should I go about finding an explicit solution? Thanks for your help.

Complete the square on the right hand side i.e.
\begin{align}
y+y^2 & = x^2-4\\
y^2 + y + \dfrac14 & = x^2-4 + \dfrac14\\
\left( y + \dfrac12 \right)^2 & = x^2-\dfrac{15}4\\
y + \dfrac12 & = \pm \sqrt{x^2 - \dfrac{15}4}\\
y & = - \dfrac12 \pm \dfrac{\sqrt{4x^2-15}}2
\end{align}
Further, we need $y(2) = 0$ and hence $y = - \dfrac12 + \dfrac{\sqrt{4x^2-15}}2$
