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