Frobenius method. Problem finding a recurrence relation The equation is $2z^2w''+3zw'-w=0$
$z_0=0$ is a regular singular point, so $w(z)=\sum_{n=0}^{\infty} a_nz^{n+r}$
then $w'(z)=\sum_{n=0}^{\infty} (n+r)a_nz^{n+r-1}$ and $w''(z)=\sum_{n=0}^{\infty} (n+r)(n+r-1)a_nz^{n+r-2}$
Replacing in the equation:
$2z^2\sum_{n=0}^{\infty} (n+r)(n+r-1)a_nz^{n+r-2}+3z\sum_{n=0}^{\infty} (n+r)a_nz^{n+r-1}-\sum_{n=0}^{\infty} a_nz^{n+r}=0$
$\sum_{n=0}^{\infty} 2(n+r)(n+r-1)a_nz^{n+r}+\sum_{n=0}^{\infty} 3(n+r)a_nz^{n+r}-\sum_{n=0}^{\infty} a_nz^{n+r}=0$
$\sum_{n=0}^{\infty} [2(n+r)(n+r-1)a_nz^{n+r}+ 3(n+r)a_nz^{n+r}- a_nz^{n+r}]=0$
$2(n+r)(n+r-1)a_n+ 3(n+r)a_n- a_n=0$
$(2(n+r)(n+r-1)+3(n+r)-1) a_n=0$
At this point, how can I construct a recurrence relation since I only got $a_n$?, I'd need $a_{n+1}$, right?
I already found the indicial equation and its roots, which are $r_1=1/2$ and $r_2=-1$
Your last equation is all you get. There's no recurrence, and you don't need one.
You necessarily have $\color{blue}{a_n=0}$ for $n&gt;0$. Otherwise, if $a_n\neq 0$ for some $n&gt;0$, you would have both $r$ and $n+r$ be roots of $2r^2+r-1=0$ (the indical equation). In our case, this is clearly impossible.
Notice that, for this problem, Frobelius method is not required. Let $w(z)=z u(z)$ to end with
$$6u''+13u'=0$$ Just reduction of order and a quite simple problem.