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Frobenius Method for ODE

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the Frobenius method, for an expansion about x=0, to find ONE solution of

    xy''+y'+(1/4)y=0

    2. Relevant equations



    3. The attempt at a solution

    starting with an assumption of y1=[tex]\sum[/tex]anxn+r
    and plugging it into the ODE, i found

    y=[tex]\frac{-1}{4}[/tex][tex]\sum[/tex]a0xn/(n!)2

    i think this can be equated to:
    y=(-1/4)a0e2x/x, but when i plug it into the ODE it doesn't equal 0, and thus isn't a solution of the ODE, can anyone see where i've gone wrong?
     
  2. jcsd
  3. Sep 24, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi zass! Welcome to PF! :smile:
    Difficult to say without seeing your full calculations, but did you take into account that the equations for n = 0 and 1 are usually different from the equations for general n?
     
  4. Sep 25, 2009 #3
    I think i accounted for it
    Here's my working out, maybe u can see if i've made a mistake:
    Using the initial assumption i've found u' and u'', then subbed them into the ode:
    [tex]\sum[/tex][tex]^{inf}_{n=0}[/tex](n+r)(n+r-1)anxn+r-2 + [tex]\sum[/tex][tex]^{inf}_{n=0}[/tex](n+r)anxn+r-2 + (1/4)[tex]\sum[/tex][tex]^{inf}_{n=0}[/tex]anxn+r-1 = 0

    Simplifying and collecting terms and sum:
    =[tex]\sum[/tex][tex]^{inf}_{n=1}[/tex]((n+r)(n+r-1)an + (n+r)an + (1/4)an-1)xn+r-2 + (0+r)(0+r-1)a0x0+r-2 + (0+r)a0x0+r-2 = 0

    Equating coefficients i found
    r2 = 0

    And thus the series now becomes
    n(n-1)an + nan + (1/4)an-1 = 0

    Which gives the recurrence formula:
    an = [tex]\frac{-an-1}{4n2}[/tex] , n=1,2,3,....inf

    then taking the first few terms i found
    an = a0/(n!)2
    and then just subbed back into the initial assumption made. I can't see anything wrong with what i've done :confused:
     
  5. Sep 25, 2009 #4

    tiny-tim

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    Hi zass! :smile:

    (have a sigma: ∑ and an infinity: ∞ and if you're using tex, ∞ is \infty, and sub and sup are _{} and ^{} :wink:)

    You've left out (-1/4)n

    does that make it ok?:smile:
     
  6. Sep 26, 2009 #5
    ah of course!! for some reason i thought that'd just be a constant.
    finally got a solution that fits :biggrin:
     
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