Frobenius Method When Initial Value of A Sum is not 1

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1. Dec 15, 2015

PatsyTy

1. The problem statement, all variables and given/known data
Solve

\begin{equation*}
36x^2y''+(5-9x^2)y=0
\end{equation*}

using the Frobenius method

2. Relevant equations

Assume a solution of the form

\begin{equation*}
y=\sum_{n=0}^{\infty}{a_nx^{n+s}}
\end{equation*}

then

\begin{equation*}
y''=\sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s-2}}
\end{equation*}

3. The attempt at a solution

I substitute the expanded forms of $y$ and $y''$ into the D.E

\begin{equation*}
36x^2 \sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s-2}}+5 \sum_{n=0}^{\infty}{a_nx^{n+s}} -9x^2 \sum_{n=0}^{\infty}{a_nx^{n+s}} =0
\end{equation*}

and put the terms of $x$ into the sums

\begin{equation*}
36 \sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s}}+5 \sum_{n=0}^{\infty}{a_nx^{n+s}} -9\sum_{n=0}^{\infty}{a_nx^{n+s+2}} =0
\end{equation*}

I then modify the last term so that the $x$ factor is $x^{n+s}$

\begin{equation*}
36 \sum_{n=0}^{\infty}{(n+s)(n+s-1)a_nx^{n+s}}+5 \sum_{n=0}^{\infty}{a_nx^{n+s}} -9\sum_{n=2}^{\infty}{a_{n-2}x^{n+s}} =0
\end{equation*}

This is where I get confused, the last term starts at $n=2$, all the examples I have seen have their highest starting value for a sum equal to $n=1$, in those cases I would comine the terms with the sum starting at $n=0$ and take out the first term of $n=0$ resulting in the indicial equation multiplied by $a_o$ which I would use to solve the D.E. Now however I have an extra term, just to be clear below is how I have found the indicial equation:

\begin{equation*}
\sum_{n=0}^{\infty}{[36(n+s)(n+s-1)+5]a_nx^{n+s}}-9\sum_{n=2}^{\infty}{a_{n-2}x^{n+s}}=0 \\
[36(s)(s-1)+5]a_0x^s+\sum_{n=1}^{\infty}{[36(n+s)(n+s-1)+5]a_nx^{n+s}}-9\sum_{n=2}^{\infty}{a_{n-2}x^{n+s}}=0
\end{equation*}

So I have my indicial equation but I can't combine the second and third terms together because their initial values don't match. I'm not sure if I should pull out the $n=1$ term on the second term so I can create a recursion relation to use however I am not sure what to do with that term I pull out.

Also a somewhat unrelated question, in this course we are also working with the Legendre D.Es and Bessel's equations, would I be able to solve both of these using the forbenius method? Or do I need some other techniques?

Any help would be appreciated!

Last edited: Dec 15, 2015
2. Dec 16, 2015

haruspex

Yes.
What would you do with the n=0 term?

3. Dec 17, 2015

PatsyTy

Thanks for the reply. For the n=0 term since we can't assume $a_0=0$ then we have to find a value for $s$ that eliminates the $a_0$ term. I get $s=\frac{5}{6}$ and $s=\frac{1}{6}$. I assume then I would do the same for the $n=1$ term. Below I have my equation with the $n=1$ taken out

\begin{equation*}
(36s(s-1)+5)a_0x^s+(36s(s+1)+5)a_1x^{s+1}+\sum_{n=2}^{\infty}{\big[(36(n+s)(n+s-1)+5)a_nx^{n+s}-9a_{n-2}x^{n+s}\big]}
\end{equation*}

Thinking about this wouldn't this mean that the only possible values for $s$ would be common roots of the $n=0$ term and the $n=1$ term? This is the only way we could garuntee that the terms including $a_0$ and $a_1$ would equal zero. The issue now is I don't have a common $s$ value for the $n=0$ and $n=1$ terms. For $n=0$ I have $s=\frac{5}{6}$ and $s=\frac{1}{6}$ and for $n=1$ I have $s=-\frac{5}{6}$ and $s=-\frac{1}{6}$. Does this mean I have made an error somewhere in my calculation or is there something I don't understand?

4. Dec 17, 2015

haruspex

Right, because of the way s is defined.
Why would the same restriction apply to that term?

5. Dec 17, 2015

PatsyTy

I didn't write it out in the equation above but that entire equation equals zero. From my understanding all the $a_n$ terms are unknown constants so the only way we can garuntee that this equation equals zero is if the $s$ portion of the terms equals zero.

Or do we define $a_0$ as zero and for any other $a_n$ where $n>0$ we can choose if it equals zero? Is there somewhere else in the equations that indicates that $a_1$ is zero? Would finding the recursion relations using $s=\frac{5}{6}$ and $s={1}{6}$ give an $a_1=0$ maybe? I'll try this out in the morning if it would help.

Reading my textbook it mentions that since $a_0x^s$ is the first term in the series we assume it is not zero but it doesn't say why, is there somewhere I could read up on why?

Sorry for all the questions, seems like once I figure something out five more questions arise.

Thanks again for the help, I really appreciate it.

6. Dec 17, 2015

haruspex

In the assumed form of the equation, s is defined to be the lowest power with a nonzero coefficient. Hence a0 is not zero. But any other coefficient can be zero.