Answer
$$y = 1 + C{e^{ - 2{x^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} + 4xy = 4x \cr
& {\text{this equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{comparing }}\frac{{dy}}{{dx}} + 4xy = 4x{\text{ with }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ we can note that }}P\left( x \right){\text{ is }}4x{\text{ }} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {4x} dx}} = {e^{2{x^2}}} \cr
& {\text{multiplying both sides of the differential equation by }}{e^{{x^2}}} \cr
& {e^{2{x^2}}}\frac{{dy}}{{dx}} + 4xy{e^{2{x^2}}} = 4x{e^{2{x^2}}} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {{e^{2{x^2}}}y} \right] = 4x{e^{2{x^2}}} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& {e^{2{x^2}}}y = \int {4x{e^{2{x^2}}}} dx \cr
& {e^{2{x^2}}}y = \int {{e^{2{x^2}}}} \left( {4x} \right)dx \cr
& {e^{2{x^2}}}y = {e^{2{x^2}}} + C \cr
& y = 1 + \frac{C}{{{e^{2{x^2}}}}} \cr
& {\text{or}} \cr
& y = 1 + C{e^{ - 2{x^2}}} \cr} $$
