## University Calculus: Early Transcendentals (3rd Edition)

$e^y=-\dfrac{e^{-x}}{e^2}+1+\dfrac{1}{e^2}$
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, $e^y=-\dfrac{e^{-x}}{e^2}+1+\dfrac{1}{e^2}$