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Thread: Difference Equations

  1. #1
    Junior Member
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    Difference Equations

    The question states -

    For each of rhe following difference equations find:
    (i) the general solutions
    (ii) the solution corresponding to the given initial conditions

    (c) $\displaystyle x_{n+2} - 6x_{n+1} + 8x_{n} = 3.5^n, x_{0} = 3, x_{1} = 6 $

    Thanks for the help.. im not too sure what to do.. so any help wouldbe appreciated
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  2. #2
    Senior Member Peritus's Avatar
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    first we solve the homogeneous difference equation:

    $\displaystyle
    x_{n + 2} - 6x_{n + 1} + 8x_n = 0

    $

    we assume the solution is of the following form:

    $\displaystyle
    r^n
    $

    we now plug in this expression into the DE:

    $\displaystyle \begin{gathered}
    r^{n + 2} - 6r^{n + 1} + 8r^n = 0 \hfill \\
    \hfill \\
    \Leftrightarrow r^n \left( {r^2 - 6r + 8} \right) = 0 \hfill \\
    \end{gathered} $

    and solve for r:

    r = 4, 2 thus:

    $\displaystyle
    x_h = C_1 2^n + C_2 4^n
    $

    now to find the particular solution we use the method of undetermined coefficients namely we assume that the solution has the form of the inhomogeneous term:

    $\displaystyle
    x_p = D3.5^n

    $

    we proceed to substituting the particular solution in the DE:

    $\displaystyle
    \begin{gathered}
    D3.5^{n + 2} - 6D3.5^{n + 1} + 8D3.5^n = 3.5^n \hfill \\
    \hfill \\
    \Leftrightarrow D3.5^n \left( {3.5^2 - 6 \cdot 3.5 + 8} \right) = 3.5^n \hfill \\
    \hfill \\
    \Leftrightarrow D = \frac{1}
    {{3.5^2 - 6 \cdot 3.5 + 8}} \hfill \\
    \end{gathered}

    $

    so the general solution is:

    $\displaystyle
    x = x_h + x_p = C_1 2^n + C_2 4^n + D3.5^n
    $

    now all you have to do is apply the initial conditions to find the constants...
    Last edited by Peritus; Jan 25th 2008 at 01:10 PM.
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  3. #3
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    i was able to do that bit by simply factorising the orginial equation, whilst treating it as a quadratic, thanks for showing me a new method though. How do i then find the constants?
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  4. #4
    Senior Member Peritus's Avatar
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    well I've already told you: apply the initial conditions:

    you're told that: $\displaystyle
    x_0 = 3
    $

    which means that:
    $\displaystyle
    x[n = 0] = 3

    $

    thus:$\displaystyle
    x_0 = C_1 2^0 + C_2 4^0 + D3.5^0 = C_1 + C_2 + D = 3

    $

    after applying the second initial condition you'll get another equation ....
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  5. #5
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    I wrote down the wrong initial conditions, but i got the answer eventually!.. Thanks for the help
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