Answer
$$y = - \ln \left( {1 - \frac{{{x^2}}}{2} - 2x} \right)$$
Work Step by Step
$$\eqalign{
& y' - x{e^y} = 2{e^y},\,\,\,\,\,\,\,\,y\left( 0 \right) = 0 \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} - x{e^y} = 2{e^y} \cr
& \cr
& {\text{separating the variables}} \cr
& \frac{{dy}}{{dx}} = 2{e^y} + x{e^y} \cr
& \frac{{dy}}{{dx}} = \left( {2 + x} \right){e^y} \cr
& {e^{ - y}}dy = \left( {x + 2} \right)dx \cr
& {\text{integrate both sides of the equation}} \cr
& \int {{e^{ - y}}dy} = \int {\left( {x + 2} \right)} dx \cr
& - {e^{ - y}} = \frac{{{x^2}}}{2} + 2x + C \cr
& \cr
& {\text{use the initial condition }}y\left( 0 \right) = 0 \cr
& - {e^{ - 0}} = \frac{{{{\left( 0 \right)}^2}}}{2} + 2\left( 0 \right) + C \cr
& - 1 = + C \cr
& C = - 1 \cr
& \cr
& Then,{\text{ the particular solution of the differential equation is}} \cr
& - {e^{ - y}} = \frac{{{x^2}}}{2} + 2x - 1 \cr
& {\text{solve for }}y \cr
& {e^{ - y}} = - \frac{{{x^2}}}{2} - 2x + 1 \cr
& \ln \left( {{e^{ - y}}} \right) = \ln \left( {1 - \frac{{{x^2}}}{2} - 2x} \right) \cr
& - y = \ln \left( {1 - \frac{{{x^2}}}{2} - 2x} \right) \cr
& y = - \ln \left( {1 - \frac{{{x^2}}}{2} - 2x} \right) \cr} $$
