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Math Help - Solve the initial-value proble. X[n+2]- x[n+1]-2x[n] = 0, for x[1]= 0, x[2]= 1?

  1. #1
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    Solve the initial-value proble. X[n+2]- x[n+1]-2x[n] = 0, for x[1]= 0, x[2]= 1?

    can you help with that, step by step
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  2. #2
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    Hello, saihat!

    Solve the initial value problem: . X(n\!+\!2) -X(n\!+\!1)-2X(n)\:=\:0,\;\;X(1) = 0,\;X(2) = 1

    We conjecture that: . X(n) \,=\,x^n

    The equation becomes: . x^{n+2} -x^{n+1} - 2x^n\:=\:0

    Divide by x^n\!:\;\;x^2 - x - 2 \:=\:0 \quad\Rightarrow\quad (x + 1)(x-2) \:=\:0 \quad\Rightarrow\quad x \:=\:-1,2

    So we have two solutions: . \begin{array}{ccc}f(n) &=& (\text{-}1)^n \\ f(n) &=& 2^n \end{array}

    Form a linear combination of the two solutions: . f(n) \:=\:A(\text{-}1)^n + B(2^n)


    We know the first two values of the sequence:

    . . \begin{array}{ccccccc}f(1) = 0\!: & \text{-}A + 2B &=& 0 & {\color{blue}[1]} \\ f(2) = 1\!: & A + 4B &=& 1 & {\color{blue}[2]} \end{array}

    Add [1] and [2]: . 6B \:=\:1 \quad\Rightarrow\quad B \,=\,\tfrac{1}{6}

    Substitute into [2]: . A + 4\left(\tfrac{1}{6}\right) \:=\:1 \quad\Rightarrow\quad A \,=\,\tfrac{1}{3}


    Hence: . f(n) \;=\;\frac{1}{3}(\text{-}1)^n + \frac{1}{6}(2^n) \;=\;\frac{(\text{-}1)^n}{3} + \frac{2^n}{6} \;=\;\frac{(\text{-}1)^n}{3} + \frac{2^{n-1}}{3}

    Therefore: . f(n) \;=\;\frac{2^{n-1} + (\text{-}1)^n}{3}

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