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Math Help - Some more diffeq help

  1. #1
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    Some more diffeq help

    I'm reading this example in my book and I don't get how the author gets this solution.

    X1(n) = 2X1(n-1) ; x-sub-1
    X2(n) = 2X2(n-1) ; x-sub-2

    The solutions to these DEs, according to the book, are

    X1(n) = (2^n) * X1(0)
    X2(n) = (3^n) * X2(0)

    I can't figure out how he makes this leap.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Codename46 View Post
    I'm reading this example in my book and I don't get how the author gets this solution.

    X1(n) = 2X1(n-1) ; x-sub-1
    X2(n) = 3X2(n-1) ; x-sub-2 probably a three here

    The solutions to these DEs, according to the book, are

    X1(n) = (2^n) * X1(0) (**)
    X2(n) = (3^n) * X2(0) (**)

    I can't figure out how he makes this leap.

    Thanks in advance.
    These are both differrence equations. Both have the same form

    (1)\;\;\;x_n = r x_{n-1},\;\;\;x_0=a

    Here we seek a solution of the form

    x_n = k \rho ^ n

    Subsituting into (1) gives

    k \rho ^{n+1} = r k \rho ^ n\;\;\; \Rightarrow\;\;\;\rho = r

    Thus, x_n = k r^n. Now x_0 = a so

    x_0 = k r^0 = k = a. Thus,

    x_n = a r^n. For your solutions (**) a = x_1(0),\;\;x_2(0)
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