1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Power series solution to degree 2 ODE
Loading...