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Math Help - 2nd-order difference equation

  1. #1
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    2nd-order difference equation

    Solve the difference equation:
    a_{n+2} - a_{n+1} - 2 a_n = 3(-1)^n

    a_0 = 3 , a_1 = -1

    any help on this is appreciated, thanks!
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  2. #2
    Junior Member slovakiamaths's Avatar
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    <br />
a_{n+2} - a_{n+1} - 2 a_n = 3(-1)^n
    Since <br />
a_{0} = 3 , <br />
a_{1} = -1
    put n=0,we have
    a_{2}-a_{1}-2 a_{0} = 3(-1)^0
    a_{2}-(-1)-2(3) = 3(1)
    therefore a_2=3+6-1=8
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  3. #3
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    Since ,
    put n=0,we have


    therefore =3+6-1=8
    Thanks for your input slovak,
    I understand what you've written, but maybe I should have specified that by "a solution" I'm talking about function such that f(n) = a_n

    I believe you start by finding a characteristic polynomial for the homogeneous equation (or something like that)
    which for this would be x^2-x-2=0, factoring to get (x+1)(x-2)

    So the homogeneous is A(-1)^n+B(2)^n?

    Then your supposed to find a particular solution of the form C(-1)^n any help from here I'm stuck
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  4. #4
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    Hello, jef3189!

    Solve the difference equation:

    . . a_{n+2} \:=\: a_{n+1} + 2 a_n + 3(\text{-}1)^n \qquad a_0 = 3 ,\;\; a_1 = \text{-}1

    The first few terms are: . \begin{Bmatrix}a_0 &=& 3 \\ a_1 &=& \text{-}1 \\ a_2 &=& 8 \\ & \vdots \end{Bmatrix}


    \begin{array}{ccccc}\text{We are given:} & a_{n+2} &=& a_{n+1} + 2a_n + 3(\text{-}1)^n \\ \\[-3mm]<br />
\text{The next term:} & a_{n+3} &=& a_{n+2} + 2a_{n+1} - 3(\text{-}1)^n \end{array}


    Add: . a_{n+3} + a_{n+2} \;=\; a_{n+2} + a_{n+1} + 2a_{n+1} + 2a_n \quad\Rightarrow\quad a_{n+3} - 3a_{n+1} - 2a_n \:=\:0


    Let: X^n \,=\,a_n

    Substitute: . X^{n+3} - 3X^{n+1} - 2X^n \:=\:0

    Divide by X^n\!:\;\;X^3 - 3X - 2 \:=\:0

    Factor: . (X+1)^2(X-2) \:=\:0 \quad\Rightarrow\quad X \:=\:\text{-}1,\:\text{-}1,\:2


    The function is of the form: . f(n) \;=\;A(\text{-}1)^n + Bn(\text{-}1)^n + C(2^n)


    Use the first three terms of the sequence:

    . . \begin{array}{cccccccccc}<br />
f(0) \:=\:3 & A + C &=& 3 \\<br />
f(1) \:=\:\text{-}1 & -A - B + 2C &=& \text{-}1 \\<br />
f(2) \:=\:8 & A + 2B + 4C &=& 8 \end{array}

    Solve the system: . A =2,\;B=1,\;C=1


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


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  5. #5
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    Let  a_n = (-1)^n b_n

    then we have

     b_{n+2} + b_{n+1} - 2b_n = 3

    Take 1st difference ,


    \delta (b_{n+2} + b_{n+1} - 2b_n) = \delta(3)
    (b_{n+3} - b_{n+2}) + ( b_{n+2} -  b_{n+1}) - 2( b_{n+1}- b_{n}) = 0
    b_{n+3} - 3  b_{n+1} + 2  b_{n} = 0
    Now , calculate the characteristic roots :
    t^3 - 3t + 2 = 0
    (t-1)^2 (t+2) = 0
    t = 1 , t= - 2

    Be careful that t=1 is the double root , we have

    b_n = A + Bn + C(-2)^n

    a_n = A(-1)^n + B(-1)^n n + C(2)^n

    First obtain a_2 = 10 , then find A,B,C by solving the system of linear equtions .
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