Answer
$$y = \frac{{{x^3}}}{3} - 3{x^2} + 3$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {x^2} - 6x;\,\,\,\,\,\,\,\,\,y\left( 0 \right) = 3 \cr
& {\text{Separating variables leads to}} \cr
& dy = \left( {{x^2} - 6x} \right)dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {dy} = \int {\left( {{x^2} - 6x} \right)dx} \cr
& {\text{integrating by using the power rule }} \cr
& y = \frac{{{x^3}}}{3} - 3{x^2} + C\,\,\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{we can find the constant }}C{\text{ using the initial value problem}}\,\,y\left( 0 \right) = 3 \cr
& y\left( 0 \right) = 3{\text{ implies that }}y = 3{\text{ when }}x = 0 \cr
& {\text{substituting these values into }}\left( {\bf{1}} \right) \cr
& 3 = \frac{{{{\left( 0 \right)}^3}}}{3} - 3{\left( 0 \right)^2} + C \cr
& C = 3 \cr
& {\text{substitute }}C = 3{\text{ into }}\left( {\bf{1}} \right){\text{ to find the particular solution}} \cr
& y = \frac{{{x^3}}}{3} - 3{x^2} + 3 \cr} $$
