Answer
$$y = \frac{{\left( {x - 1} \right)}}{{{x^2}}}{e^x} + C{x^{ - 2}}$$
Work Step by Step
$$\eqalign{
& \left( {2y - {e^x}} \right)dx + xdy = 0 \cr
& {\text{Divide by }}xdx \cr
& \frac{{\left( {2y - {e^x}} \right)}}{{xdx}}dx + \frac{{xdy}}{{xdx}} = 0 \cr
& \frac{{2y}}{x} - \frac{{{e^x}}}{x} + \frac{{dy}}{{dx}} = 0 \cr
& \frac{{dy}}{{dx}} + \frac{{2y}}{x} - \frac{{{e^x}}}{x} = 0 \cr
& \frac{{dy}}{{dx}} + \frac{2}{x}y = \frac{{{e^x}}}{x} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \frac{2}{x}{\text{, }}Q\left( x \right) = \frac{{{e^x}}}{x} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{2}{x}dx} }} = {e^{2\ln x}} = {x^2} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {x^2}\left( {\frac{{dy}}{{dx}} + \frac{2}{x}y} \right) = {x^2}\left( {\frac{{{e^x}}}{x}} \right) \cr
& {x^2}\frac{{dy}}{{dx}} + 2xy = x{e^x} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{x^2}y} \right] = x{e^x} \cr
& {\text{Integrate both sides}} \cr
& {x^2}y = \underbrace {\int {x{e^x}} dx}_{{\text{Int by parts}}} \cr
& {x^2}y = x{e^x} - {e^x} + C \cr
& {x^2}y = \left( {x - 1} \right){e^x} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{\left( {x - 1} \right){e^x}}}{{{x^2}}} + \frac{C}{{{x^2}}} \cr
& y = \frac{{\left( {x - 1} \right)}}{{{x^2}}}{e^x} + C{x^{ - 2}} \cr} $$
