# What is a general solution to the differential equation y'=y/(x^2-1)?

Dec 13, 2016

The answer is $y = K \sqrt{\frac{x - 1}{x + 1}}$

#### Explanation:

Rewrite the equation as

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

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

${x}^{2} - 1 = \left(x - 1\right) \left(x + 1\right)$

$\frac{1}{{x}^{2} - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}$

$= \frac{A \left(x + 1\right) + B \left(x - 1\right)}{\left(x - 1\right) x + 1 \left(\right)}$

$\therefore 1 = A \left(x + 1\right) + B \left(x - 1\right)$

Let $x = 1$, $\implies$, $1 = 2 A$, $\implies$, $A = \frac{1}{2}$

Let $x = - 1$, $\implies$, $1 = - 2 B$, $\implies$, $B = - \frac{1}{2}$

So,

$\frac{1}{{x}^{2} - 1} = \frac{\frac{1}{2}}{x - 1} - \frac{\frac{1}{2}}{x + 1}$

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

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

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

$\ln y = \frac{1}{2} \left(\ln \left(x - 1\right) - \ln \left(x + 1\right)\right) + C$

$\ln y = \frac{1}{2} \ln \left(\frac{x - 1}{x + 1}\right) + C$

$y = K \sqrt{\frac{x - 1}{x + 1}}$