Answer
$$y = \frac{2}{3}{x^3} - \frac{1}{2}{x^2} + 2$$
Work Step by Step
$$\eqalign{
& y' = 2{x^2} - x,{\text{ }}\left( {0,2} \right) \cr
& \frac{{dy}}{{dx}} = 2{x^2} - x \cr
& {\text{Separate the variables}} \cr
& dy = \left( {2{x^2} - x} \right)dx \cr
& {\text{Integrate both sides}} \cr
& y = \int {\left( {2{x^2} - x} \right)dx} \cr
& y = \frac{2}{3}{x^3} - \frac{1}{2}{x^2} + C \cr
& {\text{Use the initial condition }}\left( {0,2} \right) \cr
& 2 = \frac{2}{3}{\left( 0 \right)^3} - \frac{1}{2}{\left( 0 \right)^2} + C \cr
& C = 2 \cr
& \cr
& y = \frac{2}{3}{x^3} - \frac{1}{2}{x^2} + 2 \cr
& \cr
& {\text{Graph}} \cr} $$
