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Math Help - Difference equation

  1. #1
    Senior Member Twig's Avatar
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    Difference equation

    Hi!

    Problem: (1) \; \; y_{n+1}-y_{n}=e^{n} \; , y_{0}=0

    Solution: Characteristic equation:  r-1=0 .

     r = 1

    y_{n}^{h} = C_{1}\cdot 1^{n}

    Particular solution:  y_{n}^{p}=An\cdot e^{n} <-- Is this correct ?

    When I insert this into (1) I get, e^{n}\left[A(n+1)\cdot e - An\right] = e^{n}

    Thx
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  2. #2
    Super Member Random Variable's Avatar
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    Wouldn't the the form of the particular solution be just  Ae^{n} ?

     Ae^{n+1} - Ae^{n} = e^{n}

     A = \frac{e^{n}}{e^{n+1}-e^{n}} = \frac{1}{e-1}

    so  y^{p}_{n} = \frac{1}{e-1}e^{n}
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  3. #3
    Senior Member Twig's Avatar
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    Hi!

    The book says, if right side is  (polynomial)\cdot k`{n} , then you should try y_{n}^{p}=(polynomial)\cdot k^{n} \cdot n^{m}

    where m is the number of solutions to the charateristic solution that is equal to  k .

    And I think, we always have 1^{n} on the right side to to speak.

    They do something like that in a previous example.



    Edit: I tried your suggestion, it worked. I guess I misinterpreted the book. Here we only have e^{n} , so we donīt mulitply by any n .
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