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 1


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

  • $\begingroup$ what happened to the $(n+r-1)$ term? shouldn't that be with the $(n+r)^2$ term? $\endgroup$
    – Charlie P
    May 29, 2021 at 13:11
  • $\begingroup$ 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? $\endgroup$
    – Charlie P
    May 29, 2021 at 13:16
  • $\begingroup$ 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. $\endgroup$ May 29, 2021 at 14:03

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