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