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Homework Help: ODEs- Series Solutions Near a Regular Singular Point

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data
    6x2(x+1)2y''+0.5x(x+2)y'+y=0

    ii) Find all values of r for which there is a series solution of form
    xr[tex]\sum[/tex](anxn,n=0,inf)
    a0 [tex]\neq[/tex]0

    Find all values of r for which there is a series solution of form
    inf
    xr[tex]\sum[/tex]an(x-2)n
    n=0
    a0 [tex]\neq[/tex]0

    Do not try to solve the problem, just find the values of r without solving for any constants.

    2. Relevant equations


    3. The attempt at a solution

    What I tried to do was finding the indicial equation. For the first one I got 2 indicial equations, so I got 4 answers. Does it make sense ?

    r=1/2
    r=1/3
    r=0
    r=23/24
    Also, if I try to reach the Eular equation, I get only the first two, but not the last two. Have I made any mistakes ?

    And, I didn't know how to do the second one. Actually, I can do it but it would take me ages, so I believe there is a way to find r's without trying to reach the indicial equation.

    Maybe I can try and reach the Eular equation?

    Thanks,
    Roni.
     
    Last edited: May 23, 2010
  2. jcsd
  3. May 23, 2010 #2

    vela

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    Your second equation isn't an indicial equation; it just tells you how to calculate a1 in terms of a0. Only your first equation, the one involving only a0, dictates what the allowed values of r are.
    Think about when you need to use the Frobenius method as opposed to the regular old series solution.
     
    Last edited: May 23, 2010
  4. May 23, 2010 #3
    The thing is that the second equation is also for a0 and this is confusing me.
    ((12(r-1)*r+0.5*r)a0)x1+r=0


    We use Frobenius method near singular regular points...
    but I don't think I understand how to use the second one.
     
  5. May 23, 2010 #4

    vela

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    Your missing another term that contributes to the coefficient of xr+1.
     
  6. May 23, 2010 #5

    vela

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    Ignore my earlier question. :)

    Are you sure you have the correct form for the second series? I'm guessing they wanted you to expand about x=2, so the series should be

    [tex](x-2)^r \sum_{n=0}^\infty a_n (x-2)^n[/tex]

    whereas you only have xr out front.
     
  7. May 23, 2010 #6
    I went over it like 5 times, and I don't think I'm missing anything when I group a0

    There are two terms but for a1....


    And, how would you do the second one ?

    Thanks.
     
  8. May 23, 2010 #7

    vela

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    You don't group by an. You group by powers of x.
     
  9. May 23, 2010 #8

    I have only xr, it might be a typo, though...
    Say we have (x-2)r, should I go the regular Frobenius method ? Is it going to be like very very long for a question on an exam ? is there a shortcut ?

    Thanks.
     
  10. May 23, 2010 #9
    hmmm... wow... thanks a lot.. didnt know this ...
    appreciate it.
     
  11. May 23, 2010 #10

    vela

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    I'd try the substitution u=x-2, and then go with the regular Frobenius method with the new differential equation.
     
  12. May 23, 2010 #11
    substitution... hmm, very smart :\

    I'm going to try it...

    Thanks.
     
  13. Jun 10, 2010 #12
    because x=2 is reguler point, this eq has solution power serier.
    And we can apply Frobinous theorem
     
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