How to find the radius of convergence of series solution of an ODE? Recently, I watched a video https://www.youtube.com/watch?v=8lHAMZWDQHI which taught me how to find the radius of convergence of power series solution of an ODE. For example, if we want to find the radius of convergence of power series solution $\displaystyle y=\sum \limits _{i=0}^\infty c_n(x-1)^n$ of ODE $(x^2+2x+5)y''+xy'+y=0$, just finding the roots of $x^2+2x+5$ then computing the minimum distance between $1$ and one of the roots. Generally, we have an ODE as follows:$$y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots +p_1(x)y'+p_0(x)y=0,$$where $p_i(x)$ is analytic in the complex plane except for some poles, and $\rho _1,\rho _2,\ldots ,\rho _k$ are regular singular points. Assume $\displaystyle y(x)=\sum c_kx^n$ is a nonzero formal power series solution. Is it true that the radius of convergence of $y(x)$ equals $\displaystyle d=\min \limits _{1\leq i\leq k}\rho _i$? Is it possible that there exists $|z_0|>d$ making $y(z_0)$ converge?
Thanks very much for your precious help.
 A: The radius of convergence is at least $d=\min_{1\leq i\leq k}\{|\rho_k|\}$.  In general, ODEs have infinitely many solutions.  Typically (whatever that means) almost all of the solutions have that exact radius of convergence, but there is one (or more) solution with larger radius of convergence.
I like the presentation of this in the textbook
Coddington, Earl A., An introduction to ordinary differential equations, Englewood Cliffs, N.J.: Prentice-Hall, Inc. XI, 292 p. (1961). ZBL0123.27301.

Example.  Differential equation
$$
y''(x) + \frac{1}{x-1}\,y'(x) - \frac{1}{x-1}\,y(x) = 0
$$
has general solution
$$
y(x) = A(x-1) + \frac{B}{x-1}
\tag1$$
where $A,B$ are two arbitrary constants.
Consider the series expansion for $(1)$ at $x=0$,
$$
y(x) = \sum_{k=0}^\infty a_k x^k
\tag2$$
What is its radius of convergence?
If $B \ne 0$, then $(1)$ has a pole at $x=1$, so $(2)$ has radius of convergence $1$.  But if $B = 0$, then $(1)$ is a polynomial $A(x-1)$, so $(2)$ has infinite radius of convergence.
A: There can be (isolated) solutions that have a larger radius of convergence, most trivially the zero solution.
However, a basic fact about power series is that if it converges for $z=z_0$, then it also converges for all $|z|<|z_0|$, or in other words, the radius of convergence is $|z_0|$ or larger. So no, the scenario of your question is not possible.
