What is the general solution of the differential equation (y^2-1)xdx + (x+2)ydy = 0?

Mar 11, 2018

the general (implicit) solution is:

${y}^{2} = 1 + A {\left(x + 2\right)}^{4} {e}^{- 2 x}$

Explanation:

We have:

$\left({y}^{2} - 1\right) x \mathrm{dx} + \left(x + 2\right) y \mathrm{dy} = 0$

Which if we write in the form

$\left(x + 2\right) y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(1 - {y}^{2}\right) x$
$\therefore \frac{y}{1 - {y}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{x + 2}$

Which is separable, so we can "separate the variables" to get:

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

Then we integrate (steps omitted) to get:

$- \frac{1}{2} \ln | {y}^{2} - 1 | = x - 2 \ln | x + 2 | + C$

$\therefore \ln | {y}^{2} - 1 | = 4 \ln | x + 2 | - 2 x - 2 C$

$\therefore \ln | {y}^{2} - 1 | = \ln {\left(x + 2\right)}^{4} - 2 x - 2 C$

$\therefore {y}^{2} - 1 = {e}^{\ln {\left(x + 2\right)}^{4} - 2 x - 2 C}$

$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{\ln {\left(x + 2\right)}^{4}} {e}^{- 2 x} {e}^{- 2 C}$

$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = A {\left(x + 2\right)}^{4} {e}^{- 2 x}$

Leading to the general (implicit) solution:

${y}^{2} = 1 + A {\left(x + 2\right)}^{4} {e}^{- 2 x}$