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Math Help - Finding Frobenius solutions of a diff equation

  1. #1
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    Finding Frobenius solutions of a diff equation

    Hi,
    I am having a little trouble finding the two linearly independent Frobenius solutions of the following differential equation:
    2xy''+(2x+1)y'+2y=0

    I have got as far as finding out the recurrence relations:
    note ak is a with subscript k and a(k-1) is a with subscript (k-1)
    CASE 1:
    ak=-(2a(k-1))/(2k-1)

    In the solutions I have it says the above equation and then hence:
    ak=(-2/(2k-1))(-2/(2k-3))...(-2/1)a0
    where a0 is a with subscript 0

    and also
    CASE 2:
    ak=(ak-1)/k

    The solutions again have the above line and hence:
    ak=(((-1)^k)/k!)a0


    However I really do not know how in the solutions they have got from each ak equation to the hence part.


    Any help appreciated.
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  2. #2
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    Quote Originally Posted by BlueEagle View Post
    I have got as far as finding out the recurrence relations:
    note ak is a with subscript k and a(k-1) is a with subscript (k-1)
    CASE 1:
    ak=-(2a(k-1))/(2k-1)
    .
    Let us assume that you got the recurrence relations correct.

    \boxed{ a_n = - \frac{2a_{n-1}}{2n-1} \ \  n=1,2,...}.

    Let a_0 be anything.

    Thus, a_1 = - \frac{2a_0}{1} = -2a_0.

    Thus, a_2 = - \frac{2a_1}{3} = \frac{2^2 a_0}{1\cdot 3}.

    Thus, a_3 = - \frac{2a_2}{5} = - \frac{2^3 a_0}{1\cdot 3\cdot 5}.

    In general,
    a_n = (-1)^n \cdot \frac{2^n a_0}{1\cdot 3\cdot ... \cdot (2n-1)}.

    Try doing the other one.
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  3. #3
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    Ha ha yeah of course! Thanks for the help. I get the other one now.

    a1=-a0/1=-a0
    a2=-a1/2=a0/2
    a3=-a2/3=(a1/2)/3=a1/6=-a0/6

    hence ak=(a0(-1)^k)/k!
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