Answer
$y = \frac{1}{4}{x^2}{e^{{x^2}}} - \frac{1}{8}{e^{{x^2}}} + \frac{C}{{{e^{{x^2}}}}}$
Work Step by Step
$$\eqalign{
& y' + 2xy = {x^3}{e^{{x^2}}} \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + 2xy = {x^3}{e^{{x^2}}} \cr
& {\text{The differential equation is in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = 2x{\text{ and }}Q\left( x \right) = {x^3}{e^{{x^2}}} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {2xdx} }} = {e^{{x^2}}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{{x^2}}}\frac{{dy}}{{dx}} + 2x{e^{{x^2}}}y = {x^3}{e^{{x^2}}}{e^{{x^2}}} \cr
& {e^{{x^2}}}\frac{{dy}}{{dx}} + 2x{e^{{x^2}}}y = {x^3}{e^{2{x^2}}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{e^{{x^2}}}y} \right] = {x^3}{e^{2{x^2}}} \cr
& d\left[ {{e^{{x^2}}}y} \right] = {x^3}{e^{2{x^2}}}dx \cr
& {\text{Integrate both sides}} \cr
& {e^{{x^2}}}y = \underbrace {\int {{x^3}{e^{2{x^2}}}} dx}_{{\text{Integrate by parts}}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrating }}\int {{x^3}{e^{2{x^2}}}} dx \cr
& {\text{Let }}u = {x^2} \to du = 2xdx \cr
& dv = x{e^{2{x^2}}}dx \to v = \frac{1}{4}{e^{2{x^2}}} \cr
& \int {udv} = uv - \int {vdu} \cr
& = \frac{1}{4}{x^2}{e^{2{x^2}}} - \int {\left( {\frac{1}{4}{e^{2{x^2}}}} \right)\left( {2x} \right)} dx \cr
& = \frac{1}{4}{x^2}{e^{2{x^2}}} - \frac{1}{4}\int {{e^{2{x^2}}}\left( {2x} \right)} dx \cr
& = \frac{1}{4}{x^2}{e^{2{x^2}}} - \frac{1}{8}{e^{2{x^2}}} + C \cr
& {\text{Substitute the previous result into }}\left( {\bf{1}} \right) \cr
& {e^{{x^2}}}y = \frac{1}{4}{x^2}{e^{2{x^2}}} - \frac{1}{8}{e^{2{x^2}}} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{1}{{4{e^{{x^2}}}}}{x^2}{e^{2{x^2}}} - \frac{1}{{8{e^{{x^2}}}}}{e^{2{x^2}}} + \frac{C}{{{e^{{x^2}}}}} \cr
& y = \frac{1}{4}{x^2}{e^{{x^2}}} - \frac{1}{8}{e^{{x^2}}} + \frac{C}{{{e^{{x^2}}}}} \cr} $$
