Answer
$$y = \ln \left( {2{e^{ - 2}} - {e^{ - x - 2}}} \right)$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {e^{ - x - y - 2}},\,\,\,\,y\left( 0 \right) = - 2 \cr
& {\text{Use exponential properties}} \cr
& \frac{{dy}}{{dx}} = {e^{ - x - 2}}{e^{ - y}} \cr
& {\text{separating variables gives}} \cr
& \frac{{dy}}{{{e^{ - y}}}} = {e^{ - x - 2}}dx \cr
& {e^y}dy = {e^{ - x - 2}}dx \cr
& {\text{Integrating gives}} \cr
& \int {{e^y}} dy = \int {{e^{ - x - 2}}} dx \cr
& {e^y} = - {e^{ - x - 2}} + C \cr
& {\text{Find }}C{\text{ by using the initial value }}y\left( 0 \right) = - 2 \cr
& {e^{ - 2}} = - {e^{ - 0 - 2}} + C \cr
& C = 2{e^{ - 2}} \cr
& {\text{then}} \cr
& {e^y} = - {e^{ - x - 2}} + 2{e^{ - 2}} \cr
& y = \ln \left( {2{e^{ - 2}} - {e^{ - x - 2}}} \right) \cr} $$
