Frobenius method and Euler equations

[Pablo815]

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.

 

[HallsofIvy]

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?

 

[Pablo815]

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]

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?

 

[Pablo815]

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?