Given Differential Equation is :-
$\frac{dy}{dx} + xe^{-y} + 2x = 0$
$\Rightarrow \frac{dy}{dx}= -x\left ( \frac{1}{e^{y}}+2 \right )$
$\Rightarrow \frac{-dy}{\left ( \frac{1}{e^{y}}+2 \right )}= xdx$
Now, separating terms in $y$ and $dy$ one side and terms in $x$ and $dx$ another side,
$ -(\frac{e^{y}}{1 + 2e^{y}})dy = xdx$
On Integrating both sides,
$\int (\frac{-e^{y}}{1+2e^{y}})dy = \int xdx$
$ \frac{-log|(1 + 2e^{y})|}{2} = \frac{x^{2}}{2} + c$
When $x=0$ then $y=0$
So, $c= -\frac{log 3}{2}$ ----------- Equation $(1)$
After putting value of '$c$' in equation $(1)$, we get
$ \frac{-log|(1 + 2e^{y})|}{2} = \frac{x^{2}}{2} – \frac{log 3}{2}$
$\Rightarrow log|1+2e^{y}|=-x^{2}+log3$
Now, When $x=1,$
$\Rightarrow log|1+2e^{y}|=-1+log3$
$\Rightarrow log|1+2e^{y}|-log3=-1$
$\Rightarrow log\frac{|1+2e^{y}|}{3}= -1$
$\Rightarrow 1+2e^{y} = 3e^{-1}$
$\Rightarrow e^{y} = \frac{3e^{-1}-1}{2}$
$y = ln(\frac{3e^{-1}-1}{2})$
$y = ln(\frac{3}{2e}-\frac{1}{2})$
option A will be correct