If $\frac{d y}{d x}=e^{-2 y}$ and y = 0 when x = 5, then value of x where y = 3 is given by :

$e^5$ $\frac{e^6+9}{2}$ $e^6+1$ $\log _e 6$

$\frac{e^6+9}{2}$

$e^{2 y} dy=dx \Rightarrow \frac{e^{2 y}}{2}+c=x \Rightarrow c=5-\frac{1}{2}=\frac{9}{2}$ At. $y=3 \frac{e^6}{2}+\frac{9}{2}=x \Rightarrow x=\frac{e^6+9}{2}$ Hence (2) is the correct answer.