Singularities implies divergence of a Taylor series at that point?

Question

Does a power series solution to an ODE of variable coefficients with no singularities converges everywhere?. Taylors series have a radius of convergence at least to the nearest singularity, at this point the series diverges. Can we regard them as the same?

Answer 1

A Taylor series of infinite radius of convergence corresponds to an entire function, i.e. a function analytic on all $\mathbb C$. A linear differential equation whose coefficients are analytic in a region of the complex plane, and whose leading coefficient is $1$, has solutions that are analytic in that region. In particular if the coefficients are entire and the leading coefficient is $1$, the solutions are entire, and their Taylor series converge everywhere.

You do need that restriction on the leading coefficient: e.g. $(x-p) y'(x) + y(x) = 0$ has solutions $y = c/(x-p)$ that have a pole at $x=p$.

It is not true that the series always diverges at a singularity. For example, a homogeneous linear differential equation always has at least one entire solution, $0$.