## Calculus 10th Edition

$$\frac{dy}{dx}=\frac{6-x^2}{2y^3}$$ $$y=\sqrt[4]{12x-\frac{2}{3}x^3+C}$$
We first need to gather the $y$ terms on one side, and our $x$ terms on the other side. The easiest way of doing this is by multiplying both sides by $y^3$. We get $y^3\frac{dy}{dx}=\frac{6-x^2}{2}$. Integrating both sides with respect to $x$ we get $\int y^3\frac{dy}{dx}dx=\int \frac{6-x^2}{2}dx$. Evaluating these two integrals we get $\frac{y^4}{4}=3x-\frac{x^3}{6}+C$. Finally by solving for $y$ we are left with $y=\sqrt[4]{12x-\frac{2}{3}x^3+C}$