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Frobenius Method

  1. Apr 13, 2009 #1
    I have been looking at the Frobenius method for solving ODEs of the form. I have a few questions on it.


    (x^2)y'' + xby' + cy = 0

    If b and c are functions of x, does one use the Frobenius method, where as if they are constants, it is an Euler Cauchy equation and you use y = x^r ??

    Thats the first Q. anyway.

    Thanks folks!!!:smile:
     
  2. jcsd
  3. Apr 14, 2009 #2
    y=xr only works for the Euler-Cauchy equation, i.e. b and c are constants.

    If b and c are functions of x, you cannot use the try function y=xr.
     
  4. Apr 14, 2009 #3
    Thanks for cledaring that up.

    I'm having trouble applying the method. My textbook, (which I won't name but it's approach and exlanation in this section is absolutely terrible) isn't helping me much.

    I have been trying to solve, for example,

    xy'' + 5y' + xy = 0

    So I get

    SUM(n + r)(N + r -1)(a_n)x^(n + r -2) + SUM(5)(n + r)(a_n)x^(n + R -2) + SUM(a_n)x^(n + r +1) = 0

    where SUM is the sum to infinity from n = 0.

    and the general solution is of form y=(x^r)SUM(a_n)(x^n)


    I don't know what to do know. The book's next steps are done without explanation really.
    Can someone help me??
     
  5. Apr 14, 2009 #4
    x=0 is a regular singular point. There is at least one solution of the form y=(x^r)SUM(a_n)(x^n)
    where r satisfy the indicial equation r2+4r=0.
    r1=0 , r2=-4 and r1-r2 is an integer in this case.

    The Frobenius method only guarantee for r=0 (the larger root) but not for r=-4 (but there is no harm for trying)

    If you know anything about Bessel equation, I would suggest you solve the equation using the substitution z=x2y.
     
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