## Calculus with Applications (10th Edition)

$$y = \frac{{{x^3}}}{3} - \frac{2}{3}x + C$$
\eqalign{ & 3{x^2} - 3\frac{{dy}}{{dx}} = 2 \cr & {\text{This differential equation is not in the proper form}}{\text{, but we can easily fix this }}\,{\text{by }} \cr & {\text{subtracting }}3{x^2}{\text{ to both sides of the equation}}{\text{. That is}} \cr & 3{x^2} - 3\frac{{dy}}{{dx}} - 3{x^2} = 2 - 3{x^2} \cr & - 3\frac{{dy}}{{dx}} = 2 - 3{x^2} \cr & {\text{multiplying both sides by }} - \frac{1}{3} \cr & \frac{{dy}}{{dx}} = {x^2} - \frac{2}{3} \cr & {\text{Separating variables leads to}} \cr & dy = \left( {{x^2} - \frac{2}{3}} \right)dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {dy} = \int {\left( {{x^2} - \frac{2}{3}} \right)} dx \cr & {\text{integrate using the power rule}} \cr & \boxed{y = \frac{{{x^3}}}{3} - \frac{2}{3}x + C} \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^3}}}{3} - \frac{2}{3}x + C} \right] \cr & \frac{{dy}}{{dx}} = {x^2} - \frac{2}{3} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = {x^2} - \frac{2}{3}{\text{ into }}3{x^2} - 3\frac{{dy}}{{dx}} = 2 \cr & 3{x^2} - 3\left( {{x^2} - \frac{2}{3}} \right) = 2 \cr & 3{x^2} - 3{x^2} + 2 = 2 \cr & 2 = 2 \cr & {\text{The general solution is verified}} \cr}