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Easy ODE to solve using frobenius method help?

  1. Jun 29, 2009 #1
    I can't seem to figure this one out:

    y'' + (1/x)*y' -c*y = -c*y0

    where c is just a constant.

    if someone could please go through the steps of the frobenius method, i would really appreciate it!

    thanks.
     
  2. jcsd
  3. Jun 29, 2009 #2

    HallsofIvy

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    Step 1: Since the constant solution, y= cy0 for all x, is clearly a solution you can solve the homogeneous equation y''+ (1/x)y'- cy= 0 and add cy0.

    Step 2: Multiply that equation by x to get xy''+ y'- cxy= 0.

    Step 3: Write
    [tex]y= \sum_{n=0}^\infty a_nx^{n+c}[/itex]
    differentiate term by term to find y' and y'' and substitute into that equation.

    Step 4: Look just at the terms having lowest power of x. That should be a quadratic equation in c times a0. Assume that [itex]a_0\ne 0[/itex] so that the coefficient must be 0 and solve that "indicial equation" for two values of c.

    Step 5: Put each of those values of c into your equation in turn to find two independent solutions to the homogeneous equation. Setting each coefficient of x to a power to 0, you will get a recursive equation for an.
     
  4. Jun 29, 2009 #3
    i thought frobenius method was to look at the asymptotic solutions? halls what explained was simply how find a series solution?
     
  5. Jun 29, 2009 #4

    HallsofIvy

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  6. Jun 30, 2009 #5
    thank you for such a clear response!

    these are my steps:
    [tex]
    y = \sum_{n=0}^{\infty} a_{n}x^{n + c}
    [/tex]
    [tex]
    \frac{dy}{dx} = \sum_{n=0}^\infty (n + c)a_{n}x^{n + c -1}
    [/tex]
    [tex]
    \frac{d^{2}y}{dx^{2}} = \sum_{n=0}^{\infty} (n + c)(n + c -1)a_{n}x^{n + c - 2}
    [/tex]

    now i substitute them into the original equation multiplied by x:

    [tex]
    \sum_{n=0}^{\infty} (n + c)(n + c -1)a_{n}x^{n + c -1} + (n + c)a_{n}x^{n + c -1} - {\alpha}a_{n}x^{n + c + 1} = 0
    [/tex]

    the first two terms have the same power of x, so the equation can be simplified:

    [tex]
    \sum_{n=0}^{\infty} (n + c)^{2}a_{n}x^{n + c - 1} - {\alpha}a_{n}x^{n + c + 1} = 0
    [/tex]

    does this mean that the only solution is c = 0? have i done the right things so far?

    thanks for your help!!
     
  7. Jun 30, 2009 #6

    HallsofIvy

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