Answer
$$y = Cx{e^x}$$
Work Step by Step
$$\eqalign{
& xy' - \left( {x + 1} \right)y = 0 \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& x\frac{{dy}}{{dx}} - \left( {x + 1} \right)y = 0 \cr
& x\frac{{dy}}{{dx}} = \left( {x + 1} \right)y \cr
& {\text{Separate the variables}} \cr
& \frac{{dy}}{y} = \frac{{x + 1}}{x}dx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{{dy}}{y}} = \int {\frac{{x + 1}}{x}} dx \cr
& \ln \left| y \right| = \int {\left( {1 + \frac{1}{x}} \right)} dx \cr
& \ln \left| y \right| = x + \ln \left| x \right| + C \cr
& {\text{Solve for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{x + \ln \left| x \right| + C}} \cr
& y = {e^x}{e^{\ln \left| x \right|}}{e^C} \cr
& y = Cx{e^x} \cr} $$
