## Use partial fractions to solve the separable equation dy/dx=-2y/x^2-1

Jul 4, 2018

I’ll interpret the equation as $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 y}{x} ^ 2 - 1$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} + \frac{2 y}{x} ^ 2 = - 1$

We now can see the equation is in the form $\frac{\mathrm{dy}}{\mathrm{dx}} + f \left(x\right) y = g \left(x\right)$, a linear differential equation. The first step here is to find the integrating factor which will always be given by ${e}^{\int f \left(x\right) \mathrm{dx}}$.

It follows that $I = {e}^{\int \frac{2}{x} ^ 2 \mathrm{dx}} = {e}^{- \frac{2}{x}}$

Multiply both sides of the equation by ${e}^{- \frac{2}{x}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} {e}^{- \frac{2}{x}} + {e}^{- \frac{2}{x}} \frac{2 y}{x} ^ 2 = - {e}^{- \frac{2}{x}}$

(ye^(-2/x))’ = e^(-2/x)

$y {e}^{- \frac{2}{x}} = \int {e}^{- \frac{2}{x}} \mathrm{dx} + C$

However, $\int {e}^{- \frac{2}{x}} \mathrm{dx}$ can’t be integrated in terms of elementary functions so it is completely allowable for us to leave it as this in the final answer.

$y = \frac{\int {e}^{- \frac{2}{x}} \mathrm{dx} + C}{e} ^ \left(- \frac{2}{x}\right)$

Hopefully this helps!