Answer
$e^y=-\dfrac{e^{-x}}{e^2}+1+\dfrac{1}{e^2}$
Work Step by Step
Given: $e^{-x-y-2}=(e^{-y})(e^{-x-2})$
Re-arrange as: $\dfrac{dy}{dx}=(e^{-y})(e^{-x-2})$
Separate the differentials and integrate.
$\int e^y dy=\int \dfrac{e^{-x}}{e^2}dx$
This implies that $e^2e^y=-e^{-x}+ce^2$
Applying the initial conditions $y(0)=0$, we get
Thus, $e^{0}=-\dfrac{e^{-0}}{e^2}+c \implies c=1+\dfrac{1}{e^2}$
Hence, $e^y=-\dfrac{e^{-x}}{e^2}+1+\dfrac{1}{e^2}$
