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Frobenius method for a differential equations

  1. Feb 12, 2013 #1
    1. The problem statement, all variables and given/known data

    The function satisfies the differential equation f''(x) = xf(x) and has boundary conditions

    f(0) = 1 and f'(0) = 1

    Use Frobenius method to solve for f(x) with a taylor expansion of f(x) up to the quartic term a4x4

    2. Relevant equations

    f(x) = a0 + a1x + a2x2 + a3x3 + a4x4

    f'(x) = a1 + 2a2x + 3a3x2 + 4a4x3

    f'(x) = 2a2 + 6a3x + 12a4x2

    3. The attempt at a solution

    What I have done so far is set the differential equation to zero and applied the series to f''(x) and xf(x) which gave an series of a0x + a1x2 + a2(x3 +2) + a3(x4 + 6x) + a4(x5 + 12x2) = 0. I then set up equations for each power of x which is equal to zero. From the x powers I get a2 = 0, a3 = -a0/6, and a4 = -a4/12. Applying the boundary conditions shows that a0 = 1 and a1 = 1. When I use the values in the equation, the fractions within each power of x cancels each other out and you are left with no series and just a trivial statement of 0 = 0. What am I doing wrong in trying to find a series that describes the function produced by the differential equation?
     
  2. jcsd
  3. Feb 12, 2013 #2

    LCKurtz

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    Hard to say where your error is without seeing more steps. I get results like yours but without the minus signs. I get ##a_0 = 1,\, a_1 = 1,\, a_2 = 0,\, a_3= 1/6,\, a_4=1/12## so either you or I have signs wrong. But either way, that won't give you a zero series when you put them in ##f(x) = a_0+a_1x +a_2x^2+a_3x^3+a_4x^4##.
     
  4. Feb 12, 2013 #3
    Well, for example when I do the powers for x, I get the equations

    x0: 2a2 = 0 → a2 = 0
    x1: a0 + 6a3 =0 → a3 = -a0/6 = -1/6
    x0: a1 + 12a4 = 0 → a4 = -a1/12 = -1/12

    So when you get the values for the ai constants, you are suppose to apply them to the series for f(x) and not the differential equation? I think my problem is that I applied it to the differential equation and got a series such as:

    2a2 + x(a0 + 6a3) +x2(a1+12a4 ) +...

    where each fraction in each x power cancels out, whereas they wont if I just apply them to the series for f(x).

    For the series f(x), I get 1 + x - (1/6)x3 - (1/12)x4
     
    Last edited: Feb 12, 2013
  5. Feb 12, 2013 #4

    LCKurtz

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    I still think you have the minus signs wrong. Remember your equation is ##f''(x) - xf(x)=0##, with a minus sign there.
     
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