# Differential equation (2nd order) with divergent coefficients.

I have this equation:

$$x(x-1)y''+6x^2y'+3y=0$$

I try to get the series for the solution around $x=0$, using Frobenius (however it's written). the first solution must be of the form:

$$y_1=\sum_{n=0}^\infty c_nx^{n+1}$$

If I try to get coefficients, I will get a divergent series of coefficients, in which I can't see any sense. The first one is $c_1=\frac{3}{2}c_0$, I calculated the first 1000 by computer and the never go down, although I think you can deduce that from the expression for the coefficients: $$a_{n+2}=\frac{[(n+2)(n+1)+3]a_{n+1}+6(n+1)a_{n}}{(n+3)(n+2)}$$ Since the upper thing goes "like" a factorial with the $a_{n-1,2}$.

So... how would you get the solutions by series for this? What does it mean that the coefficients are divergent?

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Without looking at the actual form of your specific coefficients, the fact that the coefficients of a power series go to infinity is not a "curse" — it is possible, albeit it implies that the radius of convergence will be at most $1$ (if it is $0$, then it is a problem).
For instance, consider $a_n = n$: the radius of convergence is $R=1$, as $\sum_{n=1}^\infty nx^n$ is well-defined for all $|x|<1$ (a characterization is actually $R=\sup \{r \geq 0: a_n r^n \to 0\}$).
Even more striking, $a_n = 2^n$: the radius of convergence is $R=\frac{1}{2}$.
Yes... Is there a relatively simple way to get the radius of convergence without knowing the general expression $a_n(n)$, from that recurrence relation? –  MyUserIsThis May 18 '13 at 20:55
I'm ashamed to say that I don't know this theorem... I cannot help in that regard, I tend to do it "manually". On a different level, I do not find the same recurrence relation as you do -- I might be wrong, though; what I get is $a_0=0$, $a_2=\frac{3}{2} a_1$ and, for $n\geq 3$, $$a_n = \frac{(n^2-3n+5)a_{n-1}+6(n-2)a_{n-2}}{n(n-1)}$$ –  Clement C. May 18 '13 at 21:08
I'm sorry, the coefficients $a_k$ were wrong, I've corrected them, although I still get something different than you... still the same i had before. Anyway I already got my answer, so I have accepted yours. Thanks. –  MyUserIsThis May 18 '13 at 21:21
By the way (it is not a proof, but a hunch backed by substituting in the recurrence relation), I have the hazy feeling that $a_n\operatorname*{\sim}_{n\to\infty} n$. If you can prove it (provided it is correct), then you get the radius of convergence. –  Clement C. May 18 '13 at 21:22