Frobenius method and Euler equations

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Hi,

I'm having trouble with this one.

Homework Statement



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

[itex] x^2y'' + xy' - y = 0[/itex]

taking in account that [itex] x = 0 [/itex] is a regular singular point and performing a power series expansion.

Homework Equations



[itex] x^2y'' + xy' - y = 0[/itex]

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 [itex] y = x^\sigma\sum{a_nx^n} [/itex].

First, I divide the whole differential equation by [itex] x^2[/itex]. Then I substitute the expression above so I get

[itex]\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}} [/itex]

And now, dividing by [itex] x^{\sigma-2} [/itex], I get

[itex] \sum{((n+\sigma)(n+\sigma-1)+(n+\sigma) - 1})a_nx^n [/itex]

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

Answers and Replies

  • #2
HallsofIvy
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You have overlooked an important step. You haven't solved for [itex]\sigma[/itex]. 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, [itex]x^{\sigma}[/itex] so that our sum does start with a non-zero [itex]a_n[/itex] term. Taking the very first term, the n= 0 term, we get the "indicial equation" to solve for [itex]\sigma[/itex]. What must [itex]\sigma[/itex] b in order that [itex]a_0\ne 0[/itex]?
 
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You have overlooked an important step. You haven't solved for [itex]\sigma[/itex]. 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, [itex]x^{\sigma}[/itex] so that our sum does start with a non-zero [itex]a_n[/itex] term. Taking the very first term, the n= 0 term, we get the "indicial equation" to solve for [itex]\sigma[/itex]. What must [itex]\sigma[/itex] b in order that [itex]a_0\ne 0[/itex]?
So the indicial equation is

[itex] \sigma^2 - 1 = 0 [/itex]

For which I get [itex]\sigma = \pm 1[/itex]

So one of the solutions would have the form [itex] y = x\sum{a_nx^n} [/itex]. But I still don't know how to find an appropriate expression for [itex]a_n[/itex].
 
  • #4
Ray Vickson
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So the indicial equation is

[itex] \sigma^2 - 1 = 0 [/itex]

For which I get [itex]\sigma = \pm 1[/itex]

So one of the solutions would have the form [itex] y = x\sum{a_nx^n} [/itex]. But I still don't know how to find an appropriate expression for [itex]a_n[/itex].
For ##n \geq 1## you need
[tex]((n+\sigma)(n+\sigma-1)+(n+\sigma) - 1)a_n = 0 [/tex]
What does this tell you?
 
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  • #5
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For ##n \geq 1## you need
[tex]((n+\sigma)(n+\sigma-1)+(n+\sigma) - 1)a_n = 0 [/tex]
What does this tell you?
Oh, I think I see it. For [itex] n>1 [/itex] the left term is not going to vanish, so, would it be [itex] a_n = 0 [/itex] for all [itex] n>1 [/itex]?
 

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