# Frobenius method and Euler equations

Hi,

I'm having trouble with this one.

## Homework Statement

Find a particular solution of the second-order homogeneous lineal differential equation

$x^2y'' + xy' - y = 0$

taking in account that $x = 0$ is a regular singular point and performing a power series expansion.

## Homework Equations

$x^2y'' + xy' - y = 0$

## The Attempt at a Solution

I see that the equation given is an Euler equation, but the question asks for a power series solution, so i tried with the Frobenius method. Assuming there is at least one solution with the form $y = x^\sigma\sum{a_nx^n}$.

First, I divide the whole differential equation by $x^2$. Then I substitute the expression above so I get

$\sum{(n+\sigma)(n+\sigma-1)a_nx^{n+\sigma-2}} + \frac{1}{x}\sum{(n+\sigma)a_nx^{n+\sigma-1}}-\frac{1}{x^2}\sum{a_nx^{n+\sigma}}$

And now, dividing by $x^{\sigma-2}$, I get

$\sum{((n+\sigma)(n+\sigma-1)+(n+\sigma) - 1})a_nx^n$

Now I don't know how to find the recurrence relation I'm looking for in order to find the form of $a_n$. In all the examples I've been able to find, in the last expression one always finds terms of $a_{n-1}$, for example, but here I don't know how to continue.

Did I do something wrong? I tried to follow the steps given in my textbook. I'm confused because I believe the equation given fits the requeriments needed in order to the Frobenius method to be applicable, but this happens to me every time I try to solve an Euler equation using it.

Thank you very much in advance.

Related Calculus and Beyond Homework Help News on Phys.org
HallsofIvy
Homework Helper
You have overlooked an important step. You haven't solved for $\sigma$. A general power series can start at any power of n but if that power is larger than 0, we can always factor out a power of x, $x^{\sigma}$ so that our sum does start with a non-zero $a_n$ term. Taking the very first term, the n= 0 term, we get the "indicial equation" to solve for $\sigma$. What must $\sigma$ b in order that $a_0\ne 0$?

You have overlooked an important step. You haven't solved for $\sigma$. A general power series can start at any power of n but if that power is larger than 0, we can always factor out a power of x, $x^{\sigma}$ so that our sum does start with a non-zero $a_n$ term. Taking the very first term, the n= 0 term, we get the "indicial equation" to solve for $\sigma$. What must $\sigma$ b in order that $a_0\ne 0$?
So the indicial equation is

$\sigma^2 - 1 = 0$

For which I get $\sigma = \pm 1$

So one of the solutions would have the form $y = x\sum{a_nx^n}$. But I still don't know how to find an appropriate expression for $a_n$.

Ray Vickson
Homework Helper
Dearly Missed
So the indicial equation is

$\sigma^2 - 1 = 0$

For which I get $\sigma = \pm 1$

So one of the solutions would have the form $y = x\sum{a_nx^n}$. But I still don't know how to find an appropriate expression for $a_n$.
For ##n \geq 1## you need
$$((n+\sigma)(n+\sigma-1)+(n+\sigma) - 1)a_n = 0$$
What does this tell you?

1 person
For ##n \geq 1## you need
$$((n+\sigma)(n+\sigma-1)+(n+\sigma) - 1)a_n = 0$$
What does this tell you?
Oh, I think I see it. For $n>1$ the left term is not going to vanish, so, would it be $a_n = 0$ for all $n>1$?