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Math Help - Power series and Frobenius' method

  1. #1
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    Power series and Frobenius' method

    I dont work with power seris that much and i have a problem:

    (x-1)y''-xy'+y=0

    I just need to know what happens to the x-1 when i multiply it into the sumation; since multiplying the x just takes the n on the X to n+1...(basically im having a mathmatical brain fart)

    Anywhos it...i could also use help with a power series in which i have to find two linealy indepentdent series about x=0 (using Frobenius' method) on the porblem 2xy''-y'+2y=0

    I really appreciate any help, thanks.
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  2. #2
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    The point x=0 is an ordinary point, there is no need to use Frobenius Method here.

    Look for a solution of the form,
    y=\sum_{n=0}^{\infty}a_nx^n
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  3. #3
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    I was told to use the method to get two independant linear sets.
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  4. #4
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    never mind, figured it out, but i was wondering if anyone could tel me about the x+1...

    im guessing it would basically become (X^(n+1)) +1 ?
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  5. #5
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    Quote Originally Posted by neven87 View Post
    I dont work with power seris that much and i have a problem:

    (x-1)y''-xy'+y=0

    I just need to know what happens to the x-1 when i multiply it into the sumation; since multiplying the x just takes the n on the X to n+1...(basically im having a mathmatical brain fart)
    Say that you are going to guess a solution of the form:
    y = \sum_{n = 0}^{\infty}a_nx^n

    Then
    y^{\prime} = \sum_{n = 1}^{\infty}na_nx^{n-1}
    and
    y^{\prime \prime} = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}

    Put this into your differential equation:
    (x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0

    The first term becomes:
    (x - 1) \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} = x \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - 1 \cdot \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}

    = \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2}

    The second term becomes:
    x \cdot \sum_{n = 1}^{\infty}na_nx^{n-1} = \sum_{n = 1}^{\infty}na_nx^n

    So your differential equation becomes:
    \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 1} - \sum_{n = 2}^{\infty}n(n - 1)a_nx^{n - 2} - \sum_{n = 1}^{\infty}na_nx^n + \sum_{n = 0}^{\infty}a_nx^n = 0

    -Dan
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