Answer
$$y = - 4x - 4 + {e^{x + 1}}$$
Work Step by Step
$$\eqalign{
& y' = y + 4x,{\text{ }}\left( { - 1,1} \right) \cr
& \frac{{dy}}{{dx}} = y + 4x \cr
& \frac{{dy}}{{dx}} - y = 4x \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = - 1{\text{ and }}Q\left( x \right) = 4x \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{ - \int {dx} }} = {e^{ - x}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{ - x}}\frac{{dy}}{{dx}} - {e^{ - x}}y = 4x{e^{ - x}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {y{e^{ - x}}} \right] = 4x{e^{ - x}} \cr
& {\text{Integrate both sides}} \cr
& y{e^{ - x}} = \int {4x{e^{ - x}}} dx \cr
& y{e^{ - x}} = - 4x{e^{ - x}} - 4{e^{ - x}} + C \cr
& {\text{Solve for }}y \cr
& y = - 4x - 4 + C{e^x}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( { - 1,1} \right) \cr
& 1 = - 4\left( { - 1} \right) - 4 + C{e^{ - 1}} \cr
& 1 = C{e^{ - 1}} \cr
& C = e \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = - 4x - 4 + {e^{x + 1}} \cr
& \cr
& {\text{Graph}} \cr} $$
