Question

Given that $\frac{dy}{dx}=e^{-2y}$ and $y=0$ when $x=5$.Find the value of $x$ when $y=3$.

Answer 1

Given, $\frac{dy}{dx}=e^{-2y}$
$\Rightarrow \frac{dy}{e^{-2y}}=dx\Rightarrow e^{2y}dy=dx$
Integrating on both sides, we get
$\Rightarrow \int e^{2y}dy=\int dx$
$\Rightarrow \frac{e^{2y}}{2}=x+c....\left(1\right)$
$(\because \int e^{ax}dx=\frac{e^{ax}}{a}+c)$
Since $y=0$ when $x=5$
From equation (1)
$\frac{e^0}{2}=5+c$
$\Rightarrow \frac{1}{2}=5+c\Rightarrow c=-\frac{9}{2}$
Putting back the value of $c$
$\frac{1}{2}{e}^{2y}=x-\frac{9}{2}$
Now putting $y=3$, we get
$\Rightarrow \frac{1}{2}{e}^6=x-\frac{9}{2}$
$\Rightarrow x=\frac{9}{2}+\frac{1}{2}{e}^6=\frac{1}{2}(9+e^6)$
Hence, the required value of
$x=\frac{1}{2}(9+e^6)$ at $y=3$.