Calculus with Applications (10th Edition)

$$y = \frac{{{x^3}}}{3} - 3{x^2} + 3$$
\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}