## Calculus with Applications (10th Edition)

$$y = \frac{{{x^4}}}{2} + C$$
\eqalign{ & 4{x^3} - 2\frac{{dy}}{{dx}} = 0 \cr & {\text{This differential equation is not in the proper form}}{\text{, but we can easily fix this }}\,{\text{by }} \cr & {\text{subtracting }}4{x^3}{\text{ to both sides of the equation}}{\text{. That is}} \cr & 4{x^3} - 2\frac{{dy}}{{dx}} - 4{x^3} = - 4{x^3} \cr & - 2\frac{{dy}}{{dx}} = - 4{x^3} \cr & {\text{multiplying both sides by }} - \frac{1}{2} \cr & \frac{{dy}}{{dx}} = 2{x^3} \cr & {\text{Separating variables leads to}} \cr & dy = 2{x^3}dx \cr & {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr & \int {dy} = \int {2{x^3}} dx \cr & {\text{integrate using the power rule}} \cr & y = 2\left( {\frac{{{x^4}}}{4}} \right) + C \cr & y = \frac{{{x^4}}}{2} + C \cr & \cr & {\text{verifying that the solution satisfies the original differential equation}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{2} + C} \right] \cr & \frac{{dy}}{{dx}} = \frac{{4{x^3}}}{2} + 0 \cr & \frac{{dy}}{{dx}} = 2{x^3} \cr & {\text{substitute }}\frac{{dy}}{{dx}} = 2{x^3}{\text{ into }}4{x^3} - 2\frac{{dy}}{{dx}} = 0 \cr & 4{x^3} - 2\left( {2{x^3}} \right) = 0 \cr & 4{x^3} - 4{x^3} = 0 \cr & 0 = 0 \cr & {\text{The general solution is verified}} \cr}