Power series solution to a differential equation

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Homework Help Overview

The discussion revolves around finding a power series solution to the differential equation (x^2-1)y" + 8xy' + 12y = 0. The original poster expresses uncertainty regarding the choice of the expansion point, x0, since the problem does not specify one.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the concept of ordinary points and singular points in the context of power series solutions. The original poster questions the implications of not having a specified x0 and whether any point can be chosen for the series expansion.

Discussion Status

Some participants provide guidance on selecting an ordinary point for the power series expansion, suggesting that x0 = 0 is a viable option. There is an ongoing exploration of the implications of choosing different points, with no explicit consensus reached on the best approach.

Contextual Notes

Participants note that the singular points of the equation are -1 and 1, which influences the choice of x0. There is a recognition of the freedom to choose any ordinary point, but concerns about the impact of this choice on the solution are raised.

swtlilsoni
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Homework Statement



Using a power series solution, what is the solution to:
(x^2-1)y" + 8xy' + 12y = 0

Homework Equations



Normally these questions specify (about x0=0) but this one doesn't specify about which point. So if I use the power series equation, what am I supposed to plug in for x0?
 
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For your x_0, I assume you are talking about an ordinary point about which to expand your series? If so, in this case you can just choose one. I think you will find that x_0 = 0 works. Then just plug in a nice power series. You know the requirements for an ordinary point if you are not given one?
 
Last edited:
Right given an equation p(x)y" + q(x)y' + r(x)y=0,
an ordinary point is any point at which p(x0) does not equal 0.

Thus in this equation, the singular points are -1, and 1.
Does that mean I can choose any number that works as an ordinary point and plug it in for x0? I can choose 9? And plug that in as:
y(x)=\Sigmaan(x-9)n ?
 
swtlilsoni said:
Right given an equation p(x)y" + q(x)y' + r(x)y=0,
an ordinary point is any point at which p(x0) does not equal 0.

Thus in this equation, the singular points are -1, and 1.
Does that mean I can choose any number that works as an ordinary point and plug it in for x0? I can choose 9? And plug that in as:
y(x)=\Sigmaan(x-9)n ?

Yes, you can chose 9. But why? You have x's in the equation. It's going to be a bit easier just to choose x0=0.
 
Okay thank you. I did not know I had the freedom to choose any point. I thought the answer varied depending on the chosen point so I thought it needed to be specified.
 

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