## Calculus (3rd Edition)

$y= \frac{1}{2x^{2}-C}$
We have $\frac{dy}{dx}+4xy^{2}=0$ $\frac{dy}{dx}= -4xy^{2}$ Separating the variables, we get $\frac{dy}{y^{2}}= -4xdx$ Integrating both sides, we get $\int \frac{dy}{y^{2}}=\int -4xdx$ ⇒ $-\frac{1}{y}= -4\times\frac{x^{2}}{2}+C$ ⇒ $-\frac{1}{y}= -2x^{2}+C$ or $\frac{1}{y}= 2x^{2}-C$ or $y= \frac{1}{2x^{2}-C}$ where C is an arbitrary constant.