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Method of Frobenius (I)

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    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]

    2. Relevant equations

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


    3. 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.
     
    Last edited: Feb 1, 2009
  2. jcsd
  3. Feb 1, 2009 #2

    Mark44

    Staff: Mentor

    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
     
  4. Feb 1, 2009 #3
    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.
     
  5. Feb 1, 2009 #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.
     
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