Find a particular solution to the equation
satisfying y(0)=1y(0)=1.
We have to be slightly careful here, as this equation is not in the form dydx+p(x)y=q(x)\frac{dy}{dx}+p(x)y=q(x). We first divide through by (x2-1)(x^{2}-1) to obtain
Now the integrating factor is eF(x)e^{F(x)} where F(x)=∫2dxx2-1F(x)=\int\frac{2\,dx}{x^{2}-1}.