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

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

$\Rightarrow$ $\int {{e^{2y}}} dy = \int d x \Rightarrow \frac{{{e^{2y}}}}{2} = x + C$ …….(i)

When $x = 5$ and $y = 0,$ then

substituting these values in Eq. (i),

we get

$\frac{{{e^0}}}{2} = 5 + C$

$\Rightarrow$ $\frac{1}{2} = 5 + C \Rightarrow C = \frac{1}{2} - 5 = - \frac{9}{2}$

Eq.(i) becomes ${e^{2y}} = 2x - 9$

When $y = 3$,

then ${e^6} = 2x - 9 \Rightarrow 2x = {e^6} + 9$

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