Given $ \large\frac{dy}{dx}=e^{-2y}$ and y=0 when x=5
We have to find the value of x when y=3
Let us seperate the variables,
$\Large\frac{dy}{e^{-2y}}$$=dx$
=>$ e^{2y}dy=dx$
On integrating both the sides we get,
$\int e^{2y}dy=\int dx$
=>$\large\frac{1}{2}$$e^{2y}=x+c$
When $y=0$ and $x=5$
$\large\frac{1}{2}$$ e^{2 \times 0}=5+c$
But $e^0=1$
Therefore $\large\frac{1}{2} $$=5+c$
$-c=\large\frac{9}{2}$
When $y=3,$ let us find the value of x
$\large\frac{1}{2} $$e^{2 \times 3}=x+\large\frac{9}{2}$
$=\Large\frac{e^6}{2}-\large\frac{9}{2}=$$x$
$=\Large\frac{e^6-\large9}{\large2}$
Hence the value is $\Large\frac{e^6-\large9}{\large2}$