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Power series solution to degree 2 ODE

  1. Mar 11, 2015 #1
    1. The problem statement, all variables and given/known data
    (x+1)y'' - (x-1)y' - y = 0
    centred around x=1
    y(1) = 2, y'(1) = 3

    3. The attempt at a solution
    I know I am supposed to get two power series, one with a0 and one with a1 but when I am trying to figure out a pattern, I keep getting both a0 and a1 in all of my terms.

    So I end up with a2 = (1/4)*a0
    and an+2 = (an*n - n*(n+1)*an+1)/(2*(n+1)*(n+2))
    for n = 1 I got a3 = (2a1-(1/2)*a0)/12

    for n = 2 I got a4 = ((3/4)*a0-6*((2a1-(1/2)*a0)/12)/24

    I did this for n=3 and n=4 and n=5 and I still can't come up with a pattern for two different series, one with a0 and one with a1
  2. jcsd
  3. Mar 11, 2015 #2
    My recurrence relation is a little different. Did you replace (x+1) with [(x-1) + 2] in the first term? Then my recurrence relation turned out to be ##a_{n+2} = \frac{a_{n} - na_{n+1}}{2(n+2)}## & I think I noticed a ##2^{n}(n+2)!## in the denominators but I'm having trouble with the numerators. & ##a_{0}=a_{2}=0## since everything on the left-hand side has to add to zero. I could show you how I got that if you really want but these series solution problems are a lot of shovelling. (enough shovelling that I could have easily made a mistake :rolleyes:)
  4. Mar 11, 2015 #3
    Yea I got that and now I got a3 = a1/6 - a0/24 and a4 = a0/24-a1/24 and that doesn't make sense since the a0 is basically the same in both. ugh
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