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Frobenius series without recurrence relation

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider

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

    where n is a constant.

    a) find two linearly independent solutions in the form of a Frobenius series, initially keeping at least the first 3 terms. Can you find the solution to all orders?

    b) for n=1 you shouild find only one linearly independent solution. If y_1(x) and y_2(x) are two independent solution for n=/= 1, us the fact that

    [itex]\tilde{y} = [1/(n-1)][y_1(x) - y_2(x)][/itex]

    is a solution and take the [itex] n \rightarrow 1[/itex] of it to find another solution.

    2. Relevant equations



    3. The attempt at a solution

    I did the normal method of frobenius by taking the maclaurin expansion of y(x) and plugging it back in and got

    [itex]y(z) = \sum_{k = 0}^{ \infty} a_kx^{(k+s)}[n^2-(k+s)+(k+s)(k+s-1)] = 0[/itex]

    and by asserting that a_0 does not equal 0 i solved for s and got

    [itex]s = 1 \pm \sqrt{1-n^2}[/itex]

    But all the examples i have looked at have had another summation that starts at a different value of k in order to get a recurrence relationship between the coefficients. I only have one summation so i don't know what to do.
     
    Last edited: Oct 28, 2013
  2. jcsd
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