Answer
$$y = x\ln x + Cx$$
Work Step by Step
$$\eqalign{
& x\frac{{dy}}{{dx}} - y - x = 0,\,\,\,\,\,\,\,\,x > 0 \cr
& {\text{this equation is not written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right){\text{ }} \cr
& {\text{adding }}x{\text{ to both sides we obtain}} \cr
& x\frac{{dy}}{{dx}} - y = x \cr
& {\text{now}}{\text{, divide both sides of the equation by }}x \cr
& \frac{{dy}}{{dx}} - \frac{1}{x}y = 1 \cr
& {\text{the equation is already written in the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{ we can note that }}P\left( x \right){\text{ is }} - \frac{1}{x} \cr
& {\text{The integrating factor is }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{ - \int {\frac{1}{x}} dx}} = {e^{\ln {x^{ - 1}}}} = {x^{ - 1}} = \frac{1}{x} \cr
& {\text{multiplying both sides of the differential equation }}\frac{{dy}}{{dx}} - \frac{1}{x}y = 1{\text{ by }}\frac{1}{x} \cr
& \frac{1}{x}\frac{{dy}}{{dx}} - \frac{1}{{{x^2}}}y = \frac{1}{x} \cr
& {\text{Write the terms on the left in the form }}{D_x}\left[ {I\left( x \right)y} \right] \cr
& {D_x}\left[ {\frac{y}{x}} \right] = \frac{1}{x} \cr
& {\text{solve for }}y{\text{ integrating both sides}} \cr
& \frac{y}{x} = \int {\frac{1}{x}} dx \cr
& \frac{y}{x} = \ln \left| x \right| + C \cr
& y = x\ln \left| x \right| + Cx \cr
& {\text{we know that }}x > 0,{\text{ then removing the aboslute value}} \cr
& y = x\ln x + Cx \cr} $$
