Answer
$$y = \frac{1}{2}{\left( {x + 3} \right)^2} + \frac{C}{{{{\left( {x + 3} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& \left( {x + 3} \right)y' + 2y = 2{\left( {x + 3} \right)^2} \cr
& {\text{Divide both sides of the equation by }}x + 3 \cr
& y' + \frac{2}{{x + 3}}y = 2\left( {x + 3} \right) \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} + \frac{2}{{x + 3}}y = 2\left( {x + 3} \right) \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \frac{2}{{x + 3}}{\text{, }}Q\left( x \right) = 2\left( {x + 3} \right) \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{2}{{x + 3}}dx} }} = {e^{2\ln \left| {x + 3} \right|}} = {\left( {x + 3} \right)^2} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {\left( {x + 3} \right)^2}\frac{{dy}}{{dx}} + \frac{2}{{x + 3}}{\left( {x + 3} \right)^2}y = 2\left( {x + 3} \right){\left( {x + 3} \right)^2} \cr
& {\left( {x + 3} \right)^2}\frac{{dy}}{{dx}} + 2\left( {x + 3} \right)y = 2{\left( {x + 3} \right)^3} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{{\left( {x + 3} \right)}^2}y} \right] = 2{\left( {x + 3} \right)^3} \cr
& d\left[ {{{\left( {x + 3} \right)}^2}y} \right] = 2{\left( {x + 3} \right)^3}dx \cr
& {\text{Integrate both sides}} \cr
& {\left( {x + 3} \right)^2}y = \int {2{{\left( {x + 3} \right)}^3}} dx \cr
& {\left( {x + 3} \right)^2}y = \frac{1}{2}{\left( {x + 3} \right)^4} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{1}{2}\frac{{{{\left( {x + 3} \right)}^4}}}{{{{\left( {x + 3} \right)}^2}}} + \frac{C}{{{{\left( {x + 3} \right)}^2}}} \cr
& y = \frac{1}{2}{\left( {x + 3} \right)^2} + \frac{C}{{{{\left( {x + 3} \right)}^2}}} \cr} $$
