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

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

The correct answer is Option (1) → $\frac{e^6 + 9}{2}$ ##

Given that, $\frac{dy}{dx} = e^{-2y} \Rightarrow \frac{dy}{e^{-2y}} = dx \quad$ [applying variable separation]

$\Rightarrow \int e^{2y} dy = \int dx \Rightarrow \frac{e^{2y}}{2} = x + C \quad \dots (i)$

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 $e^{2y} = 2x - 9$

This is the required equation.

When $y = 3$, then $e^{6} = 2x - 9 \Rightarrow 2x = e^{6} + 9$

$∴x = \frac{(e^{6} + 9)}{2}$ or $\frac{1}{2}(e^{6} + 9)$