The integrating factor of the differential equation \(\left(1-x^2\right) \frac{d y}{d x}+x y=a x\), \(-1<x<1\), is…

\(\left(1-x^2\right) \frac{d y}{d x}+x y=a x\)

\(\Rightarrow \frac{d y}{d x}+\frac{x}{1-x^2} y=\frac{a x}{1-x^2}\)

\(P(x)=\frac{x}{1-x^2}\)

\(\int P(x) d x=\int \frac{x}{1-x^2} d x\)

Let \(u=1-x^2, d u=-2 x d x\)

\(\int \frac{x}{1-x^2} d x=-\frac{1}{2} \int \frac{d u}{u}=-\frac{1}{2} \ln \left(1-x^2\right)\)

I.F. \(=e^{\int P d x}=e^{-\frac{1}{2} \ln \left(1-x^2\right)}=\left(1-x^2\right)^{-1 / 2}\)