I would leave it as $\displaystyle \frac{1}{4}\log\left|4y+1\right|=\frac{x^2}{2}+C$
Now, multiply both sides by 4 and then get rid of the log:
$\displaystyle \log|4y+1|=2x^2+C\implies e^{\log|4y+1|}=e^{2x^2+C}\implies 4y+1=Ce^{2x^2}$
Now solve for y. Then apply the initial condition.
Can you take it from here?
--Chris