Answer
$$y = \frac{1}{5}{x^3} - x - \frac{{17}}{5}{x^{1/2}}$$
Work Step by Step
$$\eqalign{
& 2xy' - y = {x^3} - x,{\text{ }}y\left( 4 \right) = 2 \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& 2x\frac{{dy}}{{dx}} - y = {x^3} - x \cr
& \frac{{dy}}{{dx}} - \frac{1}{{2x}}y = \frac{{{x^3} - x}}{{2x}} \cr
& \frac{{dy}}{{dx}} - \frac{1}{{2x}}y = \frac{1}{2}{x^2} - \frac{1}{2} \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{1}{{2x}}{\text{ }}Q\left( x \right) = \frac{1}{2}{x^2} - \frac{1}{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 {\frac{1}{{2x}}dx} }} = {e^{ - \frac{1}{2}\ln \left| x \right|}} = {x^{ - 1/2}} = \frac{1}{{{x^{1/2}}}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& \frac{1}{{{x^{1/2}}}}\left( {\frac{{dy}}{{dx}} - \frac{1}{{2x}}y} \right) = \frac{1}{{{x^{1/2}}}}\left( {\frac{1}{2}{x^2} - \frac{1}{2}} \right) \cr
& \frac{1}{{{x^{1/2}}}}\frac{{dy}}{{dx}} - \frac{1}{{2{x^{3/2}}}}y = \frac{1}{2}{x^{3/2}} - \frac{1}{{2{x^{1/2}}}} \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^{1/2}}}}} \right] = \frac{1}{2}{x^{3/2}} - \frac{1}{{2{x^{1/2}}}} \cr
& d\left[ {\frac{y}{{{x^{1/2}}}}} \right] = \left( {\frac{1}{2}{x^{3/2}} - \frac{1}{{2{x^{1/2}}}}} \right)dx \cr
& {\text{Integrate both sides}} \cr
& \frac{y}{{{x^{1/2}}}} = \int {\left( {\frac{1}{2}{x^{3/2}} - \frac{1}{{2{x^{1/2}}}}} \right)} dx \cr
& \frac{y}{{{x^{1/2}}}} = \frac{1}{5}{x^{5/2}} - {x^{1/2}} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{1}{5}{x^3} - x + C{x^{1/2}}{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 4 \right) = 2 \cr
& 2 = \frac{1}{5}{\left( 4 \right)^3} - 4 + C{\left( 4 \right)^{1/2}} \cr
& 2 = \frac{1}{5}{\left( 4 \right)^3} - 4 + C{\left( 4 \right)^{1/2}} \cr
& 2 = \frac{{64}}{5} - 4 + 2C \cr
& - \frac{{34}}{5} = 2C \cr
& C = - \frac{{17}}{5} \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{1}{5}{x^3} - x - \frac{{17}}{5}{x^{1/2}} \cr} $$
