Without solving the following ODE, determine the minimum radius of convergence original image

Without solving the following ODE, determine the minimum radius of convergence $R$ of its power series solution about $x=3$:
  $$
(x^2+16)y''+xy'+y=0
$$

I actually got stuck after I got the general term involves $x^k$.
I got 
$$
C_0+2 C_1(x)+32 C_2 +96 C_3 (x) + \sum_{k=2} [(16(k+2)(k+1)C_{k+2} + (k(k-1) C_k) + (k C_k)+ C_k] (x)^k =0.
$$
Yet again, I might be wrong. 
 A: i would expect the radius of convergence to be $5$ because $z^2 + 4$ has zeros at $z =\pm 4i$ and $y'' + \frac{x}{x^2 + 4} y' + \frac 1{x^2+4} y = 0$ is singular at these points.
A: You are looking for the singular points of the differential equation.  The minimum radius of convergence of a Taylor series solution to a differential equation is the distance from the center of the expansion to the nearest singular point.  Singular points are points at which any of the coefficient functions diverge (are undefined) or the leading coefficient vanishes.  Taylor series can only converge on $disks$ in the complex plane, so we cannot ignore complex numbers when looking at a radius of convergence.  This particular differential equation is guaranteed to have well-defined solutions for arbitrary initial conditions specified over the whole range of $real$ $x$, but the Taylor series must contend with the singular points at $\pm 4i$.  This means that the radius of convergence must be at least 5, the distance from the center at 3 to either of the singular points.  
It may be the case that one or both of the homogeneous solutions to the equation actually has a Taylor series with a larger radius of convergence, but this is not guaranteed (it is, instead, understood as a sort of 'accident' associated with the interplay between the coefficients of this particular equation).  We are, in any case, guaranteed a radius of convergence $at\space least$ as great as the distance to the nearest singular point.
