Answer
$$\frac{{{y^3}}}{3} - \frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + C$$
Work Step by Step
$$\eqalign{
& \left( {{y^2} - y} \right)\frac{{dy}}{{dx}} = x \cr
& {\text{Separating variables leads to}} \cr
& \left( {{y^2} - y} \right)dy = xdx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {\left( {{y^2} - y} \right)} = \int x dx \cr
& {\text{integrating by using the power rule we obtain}} \cr
& \frac{{{y^3}}}{3} - \frac{{{y^2}}}{2} = \frac{{{x^2}}}{2} + C \cr
& \cr
& {\text{verifying that the solution satisfies the original differential equation}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{y^3}}}{3} - \frac{{{y^2}}}{2}} \right] = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{2} + C} \right] \cr
& {y^2}\frac{{dy}}{{dx}} - y\frac{{dy}}{{dx}} = x \cr
& \frac{{dy}}{{dx}} = \frac{x}{{{y^2} - y}} \cr
& {\text{substitute }}\frac{{dy}}{{dx}} = \frac{x}{{{y^2} - y}}{\text{ into }}\left( {{y^2} - y} \right)\frac{{dy}}{{dx}} = x \cr
& \left( {{y^2} - y} \right)\left( {\frac{x}{{{y^2} - y}}} \right) = x \cr
& x = x \cr
& {\text{The general solution is verified}} \cr} $$
