Frobenius series Consider the differential equation
\begin{equation}
(1+x^3)y''+4xy'+y =0
\end{equation}
I need to find a lower bound on the radius of convergence for above equation at $x=0\ \&\ x=2$.
I wrote the solution in Frobenius form
\begin{equation}
y(x) =\sum_{k=0}^{\infty}a_k x^{r+k}
\end{equation}
Substituting back into the equation I got the following equations:
\begin{align}
a_0 r(r-1) &= 0\\
a_1 r(r+1 &= 0\\
a_2 (r+2)(r+1) +a_0(4r+1)  &=0\\
a_{k+3}(k+r+3)(k+r+2) + a_k (k+r-1)(k+r) + a_{k+1}(4k+4r+5) &=0 \quad \forall k\ge0 
\end{align}
I try to use the fact that the necessary criterion for convergence of Frobenius series is
\begin{equation}
\left|\frac{xa_{k}}{a_{k-1}}\right|<1 \implies \left|x\right|<\left|\frac{a_{k-1}}{a_{k}}\right|
\end{equation}
For $r=0$, i get the recurrence relations as above. But can proceed on the bound.
Same thing for $r=1$, but I find only one parameter arbitrary instead of two even though this a second order ODE.
For $r=-1$, the solution is trivial. I'm stuck.
 A: The critical piece of information is the first coefficient, as its roots are the points where the ODE is singular. 
It is a well known fact that an explicit analytical ODE (with leading coefficient $1$) has an analytical or holomorphic solution in any disk where the coefficient functions are analytical. This means that you get a converging power series on any disk that has not the roots of $(1+x^3)$ as interior points, as the the coefficient functions $\frac{4x}{1+x^3}$ and $\frac{1}{1+x^3}$ of the normalized equation are analytical there.
The roots are $-1$, $\frac{1\pm\sqrt{3}i}2$. The the minimal distance from $z=0$ t these roots is $r=1$. For $z=2$ the minimal distance is $|2-\frac{1\pm\sqrt{3}i}2|=\frac{\sqrt{9+3}}2=\sqrt3$.
A: Since $a_0$ is the first non-zero term, then we must have $r=0$ or $r=1.$ The second equation then means that $r=0$ or $a_1=0,$ and does not imply that $r=-1$ is a possibility. In particular, $a_1=0$ in the $r=1$ case, which is why only one of the parameters is arbitrary. See here for another example where this happens, as well as an explanation of why we can (and should) assume that $a_0$ is the first non-zero term.
A: Let $x=t-1$ ,
Then $(1+(t-1)^3)y''+4(t-1)y'+y=0$
$(1+t^3-3t^2+3t-1)y''+4(t-1)y'+y=0$
$(t^3-3t^2+3t)y''+4(t-1)y'+y=0$
$t(t^2-3t+3)y''+4(t-1)y'+y=0$
Not that
this belongs to a Heun's Equation and don't expect simple form of series solution.
