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Thread: More Difference Equations

  1. #1
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    More Difference Equations

    I thought id post this in a new thread, i got the answer to the last one, but am kinda unsure of what to do in this -

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


    (A) $\displaystyle
    x_{n} - 3x_{n-1} = 2.3^n + n, x_{0} = 1/4
    $

    I am able to get Yn, but im confused with the rest of the question

    Also, with this question i am unsure of what to do with the '4', should i represent it as C?


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

    (B) $\displaystyle
    x_{n+2} - 2x_{n+1} + x_{n} = 4, x_{0} = 3, x_{1} = 6
    $

    Thanks everybody!
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle
    x_n - 3x_{n - 1} = 2.3^n + n

    $

    solve the homogeneous equation, using the method I've already shown you.

    Then assume a particular solution of the following form:

    $\displaystyle
    x_p = C_1 2.3^n + C_2 n
    $

    and apply the initial conditions...

    (B) $\displaystyle x_{n + 2} - 2x_{n + 1} + x_n = 4$

    we know that the difference equation should hold for any positive n, in particular for n=n+1:

    $\displaystyle x_{n + 3} - 2x_{n + 2} + x_{n + 1} = 4
    $

    subtract the two equations:

    $\displaystyle
    x_{n + 3} - 3x_{n + 2} + 3x_{n + 1} - x_n = 0

    $

    and this is a homogeneous equation you've already know how to solve...
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  3. #3
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    For the second question, i resulted in $\displaystyle (x-1)^3 $ is that right?
    And then i used $\displaystyle A(1)^n + Bn(1)^n + Cn^2(1)^n $.. does that seem right?
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  4. #4
    Senior Member Peritus's Avatar
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    Right
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  5. #5
    Senior Member Peritus's Avatar
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    Please read the forum rules:

    15)Create a new thread for each new problem. Do not add new problems to existing threads especially when they are not your own thread.
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  6. #6
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    This is how i did the following question, but i seem to get stuck -

    $\displaystyle x_{n} = 3x_{n-1} + 2.3^n + n, x_{0} = 1/4 $

    $\displaystyle y_{n} = A3^n $

    $\displaystyle z_{n} = Bn3^n + Cn $

    $\displaystyle Bn3^n + Cn - 3(B(n-1)3^{n-1} + C(n-1)) = 2.3^n + n $

    $\displaystyle B3^n - 2Cn + 3C = 2.3^n + n $

    $\displaystyle B=2 $

    How do i obtain C, is that right so far?
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