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Series solution to diffeq, stuck on matching the indices

  1. Dec 6, 2013 #1
    1. The problem statement, all variables and given/known data

    find the series solution to y''+x^2*y'+y=0

    2. Relevant equations
    y=summation from n=0 to infinity Cn*x^n


    3. The attempt at a solution

    y=sum from 0 to inf Cnxn
    x^2*y'=sum from 1 to inf nC n xn+1 = sum from 2 to inf (n-1) C n-1 xn = sum from 1 to inf (n-1) C n-1 xn
    y''=(n+2)(n+1)Cn+2 xn from 0 to infinity

    basically, I cant get all three series to match at starting index n=0 and get them to have the terms x^n
     
    Last edited: Dec 6, 2013
  2. jcsd
  3. Dec 6, 2013 #2
    It's a little hard to read without Latex, but your y'' doesn't seem right. The y' alone will have terms of the form ##nC_nx^{n-1}## starting at n = 1 (since the constant term vanishes) , so the y'' terms should be ##n(n-1)C_nx^{n-2}## starting at n = 2.

    You do not have to fool around with the indices on ##x^2y'##. Keep things as simple as you can, and I think it will match up.
     
  4. Dec 6, 2013 #3
    @brmath my y'' is what you have w/ the indices shifted from 2 to 0. How do you manage not shift indices? Don't you need to combine everything into a single summation term and factor x^n, setting the rest equal to zero? That's what I been doing on every problem so far for finding series to diffeqs.
     
  5. Dec 7, 2013 #4

    vela

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    $$y = \sum_{n=0}^\infty c_n x^n$$
    $$x^2 y' = \sum_{n=1}^\infty n c_n x^{n+1} = \sum_{n=2}^\infty (n-1)c_{n-1} x^n = \sum_{n=1}^\infty (n-1)c_{n-1}x^n$$ Looks fine, though that last summation seems unnecessary.

    $$y'' = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n$$ That's fine.

    You don't have to get them to all start at n=0. You have
    $$y'' + x^2 y' + y = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n + \sum_{n=2}^\infty (n-1)c_{n-1} x^n + \sum_{n=0}^\infty c_n x^n$$ Just separate out the first two terms of the first and last series, e.g.
    $$y = c_0 + c_1x + \sum_{n=2}^\infty c_nx^n$$ and then you'll have three summations that all start from n=2 so you can combine them.
     
  6. Dec 7, 2013 #5
    what do I do with the extra C0 and C1X?
     
  7. Dec 7, 2013 #6

    HallsofIvy

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    Vela told you that- separate the n= 0 and n= 1 terms from the first and third sums.
    $$y'' + x^2 y' + y = \sum_{n=0}^\infty (n+2)(n+1)c_{n+2} x^n + \sum_{n=2}^\infty (n-1)c_{n-1} x^n + \sum_{n=0}^\infty c_n x^n= 0$$

    In order that a power series be 0 for all x, we must have each coefficient 0.

    If n= 0, [itex](n+2)(n+1)c_{n+2}= 2c_2[/itex] and [itex]c_n= c_0[/itex] so we must have 2c_2+ c_0= 0.

    If n= 1, [itex](n+2)(n+1)c_{n+2}= 6c_3[/itex] and [itex]c_n= c_1[/itex] so we must have [itex]6c_3+ c_1= 0[/itex].

    For [itex]n\ge 2[/itex] we have all three sums so
    [itex](n+2)(n+1)c_{n+2}+ (n- 1)c_{n-1}+ c_n= 0[/itex]

    In order to solve for [itex]c_2[/itex], you will need to know [itex]c_0[/itex] and, in order to solve for [itex]c_3[/itex], you will need to know [itex]c_1[/itex]. Those will be the two "constants" for the general solution.
     
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