Answer
$ y $ = $ -\frac{1}{x^{2}} + C $
Work Step by Step
$\frac{dy}{dx} $ = $ 2x^{-3}$
To get the original equation, we have to integrate the aforementioned differential equation.
$y=\int dy=\int 2x^{-3} dx$
= $\frac{2x^{-3+1}}{-3+1} + C $
= $ \frac{2x^{-2}}{-2} + C $
= $ \frac{2}{-2}\frac{x^{-2}}{1} + C $
= $ -\frac{1}{1}x^{-2} + C $
= $ -x^{-2} + C $
= $ -\frac{1}{x^{2}} + C $
Hence, $ y $ = $ -\frac{1}{x^{2}} + C$.
The result checks out.
