If I'm not mistaken, the Frobenius Series is given by $$ y = a_{0}x^{r} + a_{1}x^{r+1} + a_{2}x^{r+2} + \dots + a_{n}x^{r+n} + \dots $$ and so we also have $$ y' = ra_{0}x^{r-1} + (r+1)a_{1}x^{r} + (r+2)a_{2}x^{r+1} + \dots + (r+n)a_{n}x^{r+n-1} + \dots $$ and $$ y'' = r(r-1)a_{0}x^{r-2} + (r+1)ra_{1}x^{r-1} + (r+2)(r+1)a_{2}x^{r} + \dots + (r+n)(r+n-1)a_{n}x^{r+n-2} + \dots $$
To then use the Frobenius method to solve an ODE of the form $p(x) y'' + q(x) y' + r(x) y = 0$, we must substitute in the values of $y'', y', y$ given by the Frobenius series, and simplify the resulting equation.
What is the next step in this method?
To specify: the Frobenius method is a method to find a solution of the ODE around a regular singular point of that ODE. For simplicity, assume that $x = 0$ is a regular singular point: that means that we can write \begin{align} p(x) &= x^2 P(x),\\ q(x) &= x Q(x), \end{align} where $P$, $Q$ and $r$ are not zero when $x=0$. The ODE can then be rewritten as \begin{equation} y'' + \frac{1}{x} \frac{Q(x)}{P(x)}\,y' + \frac{1}{x^2} \frac{r(x)}{P(x)} y= 0. \end{equation} If you now substitute the series expansions you found (where you already implicitly assumed that indeed $x=0$ was the regular singular point you're expanding about), you obtain \begin{equation} [r(r-1) + \frac{Q(x)}{P(x)} r + \frac{r(x)}{P(x)}]a_0 x^r + \sum_{n=1}^\infty [(n+r-1)(n+r) + \frac{Q(x)}{P(x)}(n+r) + \frac{r(x)}{P(x)}]a_n x^{n+r} = 0. \end{equation} This equation should hold for all $x$, so in particular for the regular singular point $x=0$. This gives you the so-called indicial equation \begin{equation} r(r-1) + \frac{Q(0)}{P(0)} r+ \frac{r(0)}{P(0)} = 0. \end{equation} This equation is a quadratic equation for $r$: its solutions give you two $r$-values for which the ODE can be solved using this series assumption. If the difference of those two roots of the indicial equation is not an integer, you're basically done; if the difference is equal to an integer, you have to adjust the series Ansatz a bit, see here for more information.