Irregular singular point & Frobenius form I have got the following differential equation. 
$$z^3{\rm y}′′(z) + z\,{\rm y}′(z) + \lambda\,{\rm y} (z)=0$$
where $\lambda$ is a constant. 
I think I proved that at $z=0,$ the ODE has an irregular singular point. I always learned that you can't use Frobenius method now. But the question in the textbook says: By considering a solution of the Frobenius form:
$$y=\sum_{n=0}^\infty a_nx^{n+r}$$
show that only one value of r is possible. 
Can somebody please help me. Because I thought it was not possible. When I use the indicial equation, I can't get p(0) and q(0), so how do I have to continue?
 A: You need to fall back to the raw form of the power series approach, insert the power series in the DE and compare coefficients,
$$
\sum_{n\ge0}a_n(n+r)(n+r-1)z^{n+r-2+3}+\sum_{n\ge 0}a_n(n+r)z^{n+r-1+1}+λ\sum_{n\ge 0}a_nz^{n+r}=0
\\\iff\\
\sum_{n\ge1}a_{n-1}(n+r-1)(n+r-2)z^{n}+\sum_{n\ge 0}a_n(n+r)z^{n}+λ\sum_{n\ge 0}a_nz^{n}=0
\\\iff\\
a_0(r+λ)+\sum_{n\ge1}\Bigl[a_{n-1}(n+r-1)(n+r-2)+a_n(n+r+λ)\Bigr]z^{n}=0
$$
It could become a problem to prove that the resulting power series has a positive radius of convergence.

The constant term gives the indicial condition $r=-λ$, from the other coefficients one reads the recursion 
$$
a_{n+1}=-\frac{(n-λ)(n-λ-1)}{n+1}
$$
If $a_n=0$ at some index, then all the higher order coefficients are also zero and the solution is a polynomial of degree $n-1$ or less. If that is not the case, then 
$$
\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|
=\lim_{n\to\infty}n\,\frac{(1-\frac{λ}{n})(1-\frac{λ+1}{n})}{1+\frac{1}{n}}
=\infty
$$
giving no positive radius of convergence so that there is no solution via power series.
