Series solution of a 2nd order ODE

Question

Is the ODE

$(1-x^2)y''+y'+y=0$ solvable by simple power series (not Frobenius) method? The reason I am asking this, is because if the eq were $(1-x^2)y''+xy'+y=0$, it could have been easy, since all the individual series could have been synched to $\sum A_nx^n$, i.e in terms of a single power series. Is there any workaround to it, without using the generalised method, i.e the Frobenius method?

Answer 1

Let $x=t-1$ ,

Then $(1-(t-1)^2)y''+y'+y=0$

$(1-t^2+2t-1)y''+y'+y=0$

$(-t^2+2t)y''+y'+y=0$

$t(t-2)y''-y'-y=0$

Which is similar as the Gaussian hypergeometric ODE and can solve easily by Frobenius method