Find the two linearly independent power series solution of $(1-x^2)y^{\prime\prime}+2xy^{\prime}-2y=0$
There are other methods also to solve this problem. From inspection we can see $y_1(x)=x$ is a solution of this D.E, then $y_2 = x \int \frac{1}{x^2} e^{-\int \frac{2x}{1-x^2}}dx$ $=1+x^2$. We get the same solutions from power series method, but when I try to calculate the radius of convergence then I get $R=1$. $p(x)=\frac{2x}{1-x^2}$ and $q(x)=\frac{-2}{1-x^2}$ , both of which are analytic for $x \in \mathbb{R} - \{ \pm 1\}$ , $0$ is a regular point. Therefore we can start by assuming $y=\sum_{n=0}^{\infty} a_nx^n$ , and we can calulate radius of convergence of the solutions as $1$ (which is the distance from the center $0$ to the singularity $1$). So, the interval of convergence of the solutions is $(−1,1)$. But our solutions are valid for all $x\in \mathbb{R}$. What am i doing wrong ?
You aren't doing anything wrong. It's just that $(1-x^2)y^{\prime\prime}+2xy^{\prime}-2y=0$ and $y^{\prime\prime}+\frac{2x}{1-x^2}y^{\prime}-\frac{2}{1-x^2}y=0$ are not identical on the whole of $\mathbb{R}$. They are identical except at $x=\pm 1$.
So, the radius of convergence you have found is for the $2nd$ equation, for which indeed can be at most $1$, as at $\pm 1$, the equation itself fails to exist.