Finding Series Solutions Using Method of Frobenius

In summary, using the Method of Frobenius, a series solution to the given differential equation was found by substituting the series expansion of y(x) and its derivatives. The resulting equation was solved for the coefficients Cn, with the assumption that C0 is not equal to 0. However, the solution for s resulted in two values, 0 and -3, with only s = 0 providing meaningful information. This indicates that the Method of Frobenius may not always provide all solutions to a differential equation.
  • #1
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Homework Statement



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

[tex]
x^2\frac{d^2y(x)}{dx^2} + 4x\frac{dy(x)}{dx} + xy(x) = 0
[/tex]

Homework Equations



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


The Attempt at a Solution



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

Therefore, by substituting, I get:

[tex]
x^2\frac{d^2y(x)}{dx^2} = \sum_{n = 0}^\infty C_{n} (n + s) (n + s - 1) x^{n + s}
[/tex]
[tex]
4x\frac{dy(x)}{dx} = \sum_{n = 0}^\infty 4C_{n} (n + s) x^{n + s}
[/tex]
[tex]
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
[/tex]

Combining all terms, I get:

[tex]
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}
[/tex]

Assuming [itex] C_{0} [/itex] is not 0, I get:

[tex]
C_{0}(s(s + 3)) = 0
[/tex]

and...

[tex]
C_{n} (n + s) (n + s + 3) + C_{n - 1} = 0
[/tex]

Now, with the assumption is that [itex] C_{0} [/itex] is not 0, I conclude that:

[tex]

s(s + 3) = 0, s = 0 , -3

[/tex]

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

[tex]
C_{n} (n + s) (n + s + 3) + C_{n - 1} = 0
[/tex]

This has to be zero at all times, meaning:

[tex]
C_{n} (n + s) (n + s + 3) = - C_{n - 1}
[/tex]

Therefore:

[tex]
C_{n} = - \frac{C_{n - 1}}{(n + s) (n + s + 3)}
[/tex]

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

Am I doing it right? I am having my doubts.
 
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  • #3
Mark44 said:
Shouldn't your DE be
[tex]x^2\frac{d^2 y(x)}{dx^2} + 4x\frac{d y(x)}{d x} + xy(x) = 0 [/tex]
?

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.
 
  • #4
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.
 

FAQ: Finding Series Solutions Using Method of Frobenius

What is the Method of Frobenius?

The Method of Frobenius, also known as the Frobenius method, is a technique used to find solutions to differential equations with variable coefficients. It is particularly useful for solving ordinary differential equations with polynomial coefficients.

How does the Method of Frobenius work?

The Method of Frobenius involves assuming a solution in the form of a power series and then solving for the unknown coefficients. This allows for an infinite number of solutions to be found, which can then be combined to form a general solution to the differential equation.

When is the Method of Frobenius typically used?

The Method of Frobenius is typically used when other methods, such as separation of variables or variation of parameters, are not applicable. It is also useful for solving differential equations with singular points.

What are the advantages of using the Method of Frobenius?

One advantage of using the Method of Frobenius is that it allows for the solution of a wide range of differential equations, including those with variable coefficients. Additionally, it can often provide a simpler and more elegant solution compared to other methods.

Are there any limitations to the Method of Frobenius?

While the Method of Frobenius is a powerful technique, it does have some limitations. It is not always applicable to all types of differential equations, and it may not always yield a closed-form solution. Additionally, it can be a time-consuming process and may require advanced mathematical skills.

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