If $\frac{dy}{dx}=e^{-2y}$ and $y=0$ when $x=5,$ the value of x for $y=3$, is

We have,

$\frac{dy}{dx}=e^{-2y}\Rightarrow e^{2y}dy=dx$

On integrating, we obtain

$\frac{1}{2}{e}^{2y}=x+C$

Putting $x =5, y = 0$, we obtain

$\begin{array}{l}\frac{1}{2}=5+C\Rightarrow C=-\frac{9}{2}\\ \therefore \frac{1}{2}{e}^{2y}=x-\frac{9}{2}\Rightarrow e^{2y}=2x-9\end{array}$

Putting $y = 3$, we obtain 

$e^6=2x-9\Rightarrow x=\frac{1}{2}\left(e^6+9\right)$