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A harder Frobenius Differential

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data

    xu'' + 2y' + xy = 0

    2. Relevant equations


    3. The attempt at a solution

    Ok, so i have managed the Frobenius method in the past, but this seems harder...

    \sum_{n=0}^{\infty} a_n(n+c)(n+c-1)x^{\ n+c-2} + 2\sum_{n=0}^{\infty} a_n(n+c)x^{\ n+c-2} + \sum_{n=0}^{\infty} a_nx^{\ n+c} = 0

    \sum_{n=0}^{\infty} a_n(n+c)(n+c-1)x^{\ n+c-2} + 2\sum_{n=0}^{\infty} a_n(n+c)x^{\ n+c-2} + \sum_{n=2}^{\infty} a_{\ n-2}x^{\ n+c-2} = 0

    i cannot simply remove n=0 as i would if there were only a difference of one between the sumation limits, here i would have to remove n=0 and n=1 which means that i would not have a quadratic multiplied by [tex]a_0[/tex]to solve, to find the c values. I would have one quadratic multiplied [tex]a_0[/tex] by and one by [tex]a_1[/tex]...
  2. jcsd
  3. Feb 14, 2009 #2
    I meant xy'' + 2y' + xy = 0...

    To clarify, from the wikipedia article and looking at the example,

    'We can take one element out of the sums that start with k=0 to obtain the sums starting at the same index.'
    ee64e892092bd35e2c05c3b869ca5aae.png e226e4a00adde6d2b49ce1d66cc39134.png
    'We obtain one linearly independent solution by solving the indicial polynomial r(r-1)-r+1 = r2-2r+1 =0 which gives a double root of 1'

    I have two eqautions if i remove n=0 and n=1, what shall i do to get my indicial polynomial?
  4. Feb 14, 2009 #3
    Here is the question:
  5. Feb 14, 2009 #4
    and my solution:
  6. Feb 14, 2009 #5
    am i right? is the question wrong? please can someone clarify this.
  7. Feb 16, 2009 #6
  8. Feb 16, 2009 #7
    .... :(
  9. Feb 18, 2009 #8
    f uckyall bitchez - i solved it.
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