# Frobenius Method for indicial equations but my powers aren't the same

I've been doing some work on the frobenius method and I've been able to successfully use it to obtain indicial equations and roots.

However, in the this question I can't seem to make the powers equal when I plug my $$y$$ values back into my $$ODE$$, here's my process so far:

I want the indicial equation of the following $$ODE$$:

The $$ODE$$: $$xy'' + y' - y$$

$$xy'' +y' -y$$ => $$y''+(1/x)y' - (1/x)y$$

$$x$$ is a regular-singular point so we can use frobenius

$$y = \sum_{n=0}^\infty a_nx^{n+r}$$, $$y'= \sum_{n=0}^\infty (n+r)a_nx^{n+r-1}$$, $$y''=\sum_{n=0}^\infty(n+r-1)(n+r)a_nx^{n+r-2}$$

Plug these values into the $$ODE$$:

$$\sum_{n=0}^\infty(n+r-1)(n+r)a_nx^{n+r-2} + (1/x)*\sum_{n=0}^\infty (n+r)a_nx^{n+r-1} -(1/x)*\sum_{n=0}^\infty a_nx^{n+r}$$

From here I'm struggling to get the same powers, my $$y''$$ and $$y'$$ term will be to the power of $$(n+r-2)$$ but my $$y$$ power will only be $$(n+r-1)$$

If anyone could show me how I'd make the powers equal or how I deal with different powers I would really appreciate it, or if I've made mistake please show me.

Thanks in advance

## 1 Answer

Combine the factor $$1/x$$ with the power in the series and shift the index in the last term, $$\sum_{n=0}^\infty(n+r-1)(n+r)a_nx^{n+r-2} + \sum_{n=0}^\infty (n+r)a_nx^{n+r-2} -\sum_{n=0}^\infty a_nx^{n+r-1} \\ =\sum_{n=0}^\infty (n+r)^2a_nx^{n+r-2} - \sum_{n=1}^\infty a_{n-1}x^{n+r-2}$$

• what happened to the $(n+r-1)$ term? shouldn't that be with the $(n+r)^2$ term? May 29, 2021 at 13:11
• And then from here how would I obtain the indicial equation? should I set $n=1$ for both sums and then simplify to find an equation? May 29, 2021 at 13:16
• The indicial equation is the coefficient of $a_0$, that is, $r^2=0$. As that is a double root, you need the reduction-of-order step, almost certainly resulting in a logarithm term, to find the second basis solution. May 29, 2021 at 14:03