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Non-homogeneous Difference Equations

  1. Oct 8, 2008 #1
    Hi everyone, this is my first post but I've been reading through the forum for a quite sometime now. I've recently began trying to do some difference equations, for the most I've been doing good, but just a few things have puzzled me.
    I have two questions relating to the particular solutions of non-homogeneous equation.
    First one is, if the equation is x(k+1) - x(k) = 8log2(k) and I wanted to get the particular solution:
    f(x) = 8long2(k) but what is particular solution? I know for example if f(x) = 8e^k it would be c*e^k.

    Secondly if the equation is x(3k) = 2x(k) for 1,3,9,27,..., with x(1) = 1, how do I get this into the standard form? Or is it already in the standard form.
    My first step for this one was:
    let k = 2^m and x(k) = y(m) hence,
    3k = 2^(m+2) and x(3k) = y(m+2)
    But have no idea of this is the right approach.

    Sorry I know they are stupid questions, but they've really stumped me. Any help would be great. Thanks.
     
    Last edited: Oct 8, 2008
  2. jcsd
  3. Oct 12, 2008 #2
    Any ideas anyone?
     
  4. Oct 12, 2008 #3

    arildno

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    Well, let us exponentiate, and we get:
    [tex]\frac{y_{k+1}}{y_{k}}=k^{8}, y_{k}=2^{x_{k}}[/tex]

    Furthermore, introduce [tex]z_{k}=\sqrt[8]{y_{k}}[/tex], whereby we get the difference equation:
    [tex]z_{k+1}-kz_{k}=0[/tex]
    The solution of this is well-known:
    [tex]z_{k}=B(k-1)![/tex]
    Thus, we get:
    [tex]y_{k}=C((k-1)!)^{8}[/tex]
    And the particular solution for x becomes:
    [tex]x_{k}=D+8\log_{2}((k-1)!)[/tex]

    Also note that this includes the homogenous solution, i.e a constant, so this is, as it happens, the general solution.
     
    Last edited: Oct 12, 2008
  5. Oct 19, 2008 #4
    Ah I see, I'm still a bit confused but that helped a great deal. Thank you arildno, I appreciate it.
     
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