Checking the radius of convergence of the PS solution of $(1+x^2) y'' + y = 0$

Question

In the book of Int. Ordinary Differential Equation by Coddington, at page 130, it is asked to find the radius of convergence of the solution of $$(1+x^2) y'' + y = 0.$$

However, when I compute the PS solution I get

$$a_{k+2} = \frac{ (-1) (k^3 +1) a_k}{(k+2) (k+1)^2 },$$ which leads to

$$a_{2k} = \frac{ (-1)^k (k+1) a_0}{(2k)! },$$ but for odd terms $$a_3 = \frac{ (-1)a_1 }{3! } \& a_5 = \frac{(-1)^2 7 a_1}{5! } $$ and does not give me a sensible series which I can check it radius of convergence.

So my question is what am I missing in here ?

Any help of hint is appreciated.

Note that, the book also suggests me to put $y= \sum_k a_k x^k$ into the equation and find $r >0$ for that series.

Answer 1

Apply the quotient formula for the radius of convergence. As $\frac{a_{k+2}}{a_k}$ has a limit $-1$ you get $R^2=1$ so that also $R=1$. Which is also conform with the ODE having a singularity at $x=\pm i$.