Question

Solve the following differential equation: $(x^2-1)\frac{dy}{dx}+2xy=\frac{2}{x^2-1}.$

Answer 1

$(x^2-1)\frac{dy}{dx}+2xy=\frac{2}{x^2-1}$

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

It is in the form of linear differential equation.

$\frac{dy}{dx}+Py=Q$

Where $P=\frac{2x}{x^2-1},Q=\frac{2}{x^2-1}$

Integrating factor, IF = $e^{\int Pdx}=e^{\int \frac{2x}{x^2-1}dx}=e^{log|x^2-1|}=x^2-1$

Therefore, $y(x^2-1)=\int \frac{2}{(x^2-1{)}^2}(x^2-1)dx+C$

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