Finding Series Solutions Using Method of Frobenius

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Homework Help Overview

The discussion revolves around finding a series solution to a differential equation using the method of Frobenius. The equation presented is a second-order linear differential equation with variable coefficients.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the method of Frobenius by substituting a power series into the differential equation. They express concerns about the implications of their findings, particularly regarding the values of s and the behavior of coefficients.
  • Some participants question the formulation of the differential equation, suggesting a need to clarify the notation used for derivatives.
  • Others express uncertainty about the implications of certain values of s, particularly s = -3, and its effect on the coefficients of the series.

Discussion Status

The discussion is ongoing, with participants exploring the implications of their findings and questioning the assumptions made in the original poster's approach. Some guidance has been offered regarding the formulation of the differential equation, but there is no explicit consensus on the interpretation of the results.

Contextual Notes

There are indications of potential trivial errors in the original formulation, and the participants are navigating the complexities of the method of Frobenius, particularly in relation to the nature of the solutions derived from the series expansion.

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Homework Statement



Using method of Frobenius, find a series solution to the following differential equation:

<br /> x^2\frac{d^2y(x)}{dx^2} + 4x\frac{dy(x)}{dx} + xy(x) = 0 <br />

Homework Equations



y(x) = \sum_{n = 0}^\infty C_{n} x^{n + s}


The Attempt at a Solution



<br /> y(x) = \sum_{n = 0}^\infty C_{n} x^{n + s}<br />
<br /> \frac{dy(x)}{dx} = \sum_{n = 0}^\infty C_{n} (n + s) x^{n + s - 1}<br />
<br /> \frac{d^2 y(x)}{dx^2} = \sum_{n = 0}^\infty C_{n} (n + s) (n + s - 1) x^{n + s - 2}<br />

Therefore, by substituting, I get:

<br /> x^2\frac{d^2y(x)}{dx^2} = \sum_{n = 0}^\infty C_{n} (n + s) (n + s - 1) x^{n + s}<br />
<br /> 4x\frac{dy(x)}{dx} = \sum_{n = 0}^\infty 4C_{n} (n + s) x^{n + s}<br />
<br /> xy(x) = \sum_{n = 0}^\infty C_{n} x^{n + s + 1} = \sum_{n = 1}^\infty C_{n - 1} x^{n + s} \rightarrow n + 1 = m \leftrightarrow n = m - 1, n \geq 0, m \geq 1<br />

Combining all terms, I get:

<br /> C_{0}((s + 0) (s + 0 - 1) + 4(s + 0))x^s + \sum_{n = 1}^\infty [C_{n} (n + s) (n + s + 3) + C_{n - 1}] x^{n + s}<br />

Assuming C_{0} is not 0, I get:

<br /> C_{0}(s(s + 3)) = 0<br />

and...

<br /> C_{n} (n + s) (n + s + 3) + C_{n - 1} = 0<br />

Now, with the assumption is that C_{0} is not 0, I conclude that:

<br /> <br /> s(s + 3) = 0, s = 0 , -3<br /> <br />

Now... So far, so good. The problem is within the generating terms.

<br /> C_{n} (n + s) (n + s + 3) + C_{n - 1} = 0<br />

This has to be zero at all times, meaning:

<br /> C_{n} (n + s) (n + s + 3) = - C_{n - 1}<br />

Therefore:

<br /> C_{n} = - \frac{C_{n - 1}}{(n + s) (n + s + 3)}<br />

So what's the problem? You see, if we assume s = -3, and C_{0} is not 0, then we got a problem at n = 3, s = -3 as that will mean the whole equation will explode. This means C_{0}, C_{1}, C_{2} are all zero, with no information about C_{3}

Am I doing it right? I am having my doubts.
 
Last edited:
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Shouldn't your DE be
x^2\frac{d^2 y(x)}{dx^2} + 4x\frac{d y(x)}{d x} + xy(x) = 0
?

IOW, the 2nd derivative in the first term, and derivatives intead of partial derivatives?

Take a look at this wikipedia article - http://en.wikipedia.org/wiki/Frobenius_method
 
Mark44 said:
Shouldn't your DE be
x^2\frac{d^2 y(x)}{dx^2} + 4x\frac{d y(x)}{d x} + xy(x) = 0
?

IOW, the 2nd derivative in the first term, and derivatives intead of partial derivatives?

Take a look at this wikipedia article - http://en.wikipedia.org/wiki/Frobenius_method

I wasn't done typing the problem, and my attempt at it.

Trivial errors are all fixed by now. That aside...

I still don't see how much sense I can get out of the situation above.
 
I guess this problem doesn't need any more attention.

My understanding is that Method of Frobenius may be of help to find a solution to the DEQ, but it may not be able to provide all the solutions.

In this case, s = -3 doesn't provide anything useful, for instance. s = 0 is the only sensible choice, in other words.
 

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