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