Why does the answer to $(x^2+1)y''+xy'-y=0$ converge for $|x|<1$?

Question

For $$(x^2+1)y''+xy'-y=0,$$ the answer is $$y=C_1x +C_0\left(1+\frac{x^2}{2} - \frac{x^4}{2^2\cdot2!}+ ......\right)$$ According to my textbook, this power series only converges for $|x|<1$ because $x= ±i$ is the questions singular point. Can anyone explain why this is the case?

Answer 1

The power series solution is not only valid for real arguments, but also for complex ones. As such the interval of convergence of the real series is the intersection of the disk of convergence of the (same) complex series with the real line. The disk of convergence of the complex power series is restricted by the singularities in the complex plane. As you found, $x=\pm i$ are such obstacles, as the ODE is singular there, the order of the ODE collapses by one, which in general leads to divergent solutions.

As a consequence it is to be expected that for the coefficients of the power series $\limsup \sqrt[n]{|a_n|}\to1$ leading to the divergence of the power series for all $|x|>1$.

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The recursion for the power series coefficients is $$ [(n+2)(n+1)a_{n+2}+n(n-1)a_n]+na_n-a_n=0 \\\iff (n+2)(n+1)a_{n+2}+(n+1)(n-1)a_n=0 $$ As $n\ge 0$ the factor $(n+1)$ cancels without problem, $a_{2n+1}=0$ for $n>0$ follows immediatly, for the others one gets $$ \frac{a_{2m}}{a_{2(m+1)}}=-\frac{2m+2}{2m-1}\xrightarrow{~m\to\infty~}-1 $$ so that also from the coefficients the quotient test gives convergence for $|x|^2<1$, confirming the radius $1$ of convergence.

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This is not so readily visible from the explicit form of the coefficients, \begin{align} a_{2m}&=(-1)^m\frac{(2m-3)(2m-5)...1\cdot(-1)}{2m(2m-2)...2}a_0 =(-1)^{m+1}\frac{(2m-2)!}{m!(m-1)!2^{2m-1}}a_0 \end{align} but the coefficient recursion can be read also as \begin{align} a_{2m+2}&=\frac{\frac12-m}{m+1}a_m\\[1em] \implies a_{2m}&=\frac{\frac12(\frac12-1)(\frac12-2)...(\frac12-m+1)}{m!}=\binom{1/2}{m} \end{align} which gives the even part of the solution as $$a_0\sum_{k=0}\infty \binom{1/2}{k}x^{2k}=a_0\sqrt{1+x^2}.$$

Plotting some partial sums demonstrates the very rapid divergence outside $[-1,1]$ from a seemingly nice numerical solution curve.