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!

Homework Help: ODE power series method

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data
    using the fact that [tex]\frac{sint}{t}[/tex] = [tex]\sum^{}_{n=0}[/tex][tex]\frac{ (-1)[tex]^{n}[/tex] t[tex]^{2n}[/tex]}{(2n+1)!}[/tex]

    i wanted to write:

    using that sint/t = ∑ (-1)ⁿ·t²ⁿ/(2n+1)!

    find a power series solution of the equation tx'' +sint x = 0
    2. Relevant equations
    sorry i pushed the wrong button,

    3. The attempt at a solution
    I'm a beginner at LaTeX, a disaster at LaTeX

    i get something like ∑n(n-1)a_nt^(n-2) +sint∑a_n = 0
    Last edited: Nov 4, 2009
  2. jcsd
  3. Nov 5, 2009 #2

    You only need to use the tex-brackets at the beginning and at the end; don't clutter the formula with them. Also, you should use \sin (or\mathrm{sin}) instead of sin, makes the font right.

    [tex]\frac{\sin t}{t} = \sum_{n=0}\frac{ (-1)^{n} t^{2n}}{(2n+1)!}[/tex]

    Okay, so how do find a power series solution? You write the DE as a sum over powers of t and then require that the coefficient is zero for all n. That gives you a recursion relation between the coefficients a_n.

    You're missing a t and a t^n there. Then write out the series expansion for sint/t and you can find a formal solution of the form

    [tex] \sum_{n=0}^{\infty} A_n t^n = 0 [/tex] where now A_n is something quite complicated, but it contains the recursion relation for the coefficients a_n. I am not sure if you can find an answer in terms of elementary functions; the form of the equation suggests to me that you can't.
  4. Nov 5, 2009 #3
    Well this is how far i get:
    Using that

    [tex]\frac{\sin t}{t} = \sum_{n=0}^{\infty}\frac{(-1)^{n} t^{2n} }{(2n+1)!}[/tex]

    for t≠0 i rewrite tx''+sint x =0 as

    [tex](\aleph); x'' + \frac{\sin t}{t}x = 0 [/tex]

    [tex] Ansats: x(t) = \sum_{n=0}^{\infty}a_{n}t^{n} [/tex]


    [tex] x''(t) = \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2}[/tex]

    pluggin this and the ansats for x(t) into [tex](\aleph) [/tex]

    i get

    [tex] \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n} }{(2n+1)!}\sum_{n=0}^{\infty}a_{n}t^{n} [/tex]

    here is where i'm not 100% sure.

    i multiply the last two series getting

    [tex] \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n+1} }{(2n+1)!}\sum_{n=0}^{\infty}a_{n} [/tex]

    which is equivalent to

    [tex]\sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sin t\sum_{n=0}^{\infty}a_{n}[/tex]

    now what's next?
  5. Nov 5, 2009 #4
    You cannot use the same indices with the two sums. Use for example m and n. What you should have then is something proportional to [tex] t^{2m+n} [/tex]. Now you need to rescale the sums so that you will have something of the form [tex] \sum A_n t^n [/tex] where then A_n will be zero for all n. So for example for the first sum you'd write
    [tex] \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} t^{n} [/tex].
  6. Nov 5, 2009 #5
    \sum_{n=2}^{\infty}n(n-1)a_{n}t^{n-2} + \sum_{m=0}^{\infty}\frac{(-1)^{m}t^{2m} }{(2m+1)!}\sum_{n=0}^{\infty}a_{n}t^{n}


    [tex] \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}t^{n} + \sum_{^{m=0}_{n=0}}^{\infty}\frac{(-1)^{m}t^{2m+n} }{(2m+1)!}a_{n} [/tex]

    i know how to shift series and add them but now what?

  7. Nov 5, 2009 #6
    You shift the term on the right as well. Just notice when you do it: then you only sum over a finite number of m:s for each n.
  8. Nov 5, 2009 #7
    what's the substitution in the second series?

    give me a hint
  9. Nov 5, 2009 #8
    Since you want to have t^n there, you have to do n+2m -> n. Then a_n -> a_(n-2m), and the summation over m will be cut off by the fact that a_(-1) = 0. That's all there is to it.
  10. Nov 5, 2009 #9
    yes that is exactly what i tried but got confused.

    i used 2m+n = k then m = 1/2·(k-n) and (2m+1)! --> (k+1-n)! and [tex]a_{n} \rightarrow a_{k-2m}[/tex]
    Last edited: Nov 5, 2009
  11. Nov 5, 2009 #10
    You don't want to replace any m:s, only n:s. So the only things you need to touch are an and tn+2m. You should end up with

    [tex] \sum_{n=0}^{\infty} \left[ (n+2)(n+1) a_{n+2} + \sum_{m=0}^{n/2} a_{n-2m} \frac{(-1)^m}{(2m+1)!}\right] t^n, [/tex]

    where the sum over m goes to m=n/2 for even values of n and to m=(n-1)/2 for odd values.
  12. Nov 5, 2009 #11
    Danke Clamtrox
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook