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Math Help - difference equations

  1. #1
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    Smile difference equations

    hi!

    ive got three difference equations but could do with some help becuase i didnt think that they would be too long but they are my homework for this week and am slightly confused why i would get such a short homework!

    a) Xn+2 - 5Xn+1 + 6Xn =0


    b) Xn+2 - 5Xn+1 + 6Xn = 4

    c) Xn+2 - 6Xn+1 + 9Xn =0

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  2. #2
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    Quote Originally Posted by charlotte_usa View Post
    a) Xn+2 - 5Xn+1 + 6X[SIZE=1][SIZE=2][SIZE=1]n [SIZE=2]=0
    x_{n+2}-5x_{n+1}+6x_n=0

    The charachteristic equation,
    k^2 - 5k + 6 = (x-2)(x-3)=0

    So,
    x_n = C_1 2^n + C_2 3^n
    But we cannot find,
    C_1,C_2
    Without initial conditions.
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  3. #3
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    Xn+2 - 5Xn+1 + 6Xn = 4
    Here we need a particular solution,
    Look for a solution of the form x_n = A
    Then,
    x_n=A \ \ x_{n+1} = A \ \ x_{n+2} = A
    Thus,
    A-5A+6A = 4 implies A=2

    Thus, the general solution is,
    x_n =  C_1 2^n + C_2 3^n + 2
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  4. #4
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    Xn+2 - 6Xn+1 + 9Xn =0
    The charachteristic equation is,
    k^2 - 6k + 9 = (k-3)(k-3)=0

    Since there is only one solution for k=3

    It means the general solution is,
    x_n = C_13^n + C_2n3^n
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    x_{n+2}-5x_{n+1}+6x_n=0

    The charachteristic equation,
    k^2 - 5k + 6 = (x-2)(x-3)=0

    So,
    x_n = C_1 2^n + C_2 3^n
    But we cannot find,
    C_1,C_2
    Without initial conditions.
    but to get the complementary solution

     Yn= A2^n + B3^n

    right? and since Dn = 0 then there is no particular solution?
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    Here we need a particular solution,
    Look for a solution of the form x_n = A
    Then,
    x_n=A \ \ x_{n+1} = A \ \ x_{n+2} = A
    Thus,
    A-5A+6A = 4 implies A=2

    Thus, the general solution is,
    x_n =  C_1 2^n + C_2 3^n + 2
    here the c1 = 2 and c2 = 3 so ive been taught that for a particular solution, since c1 doesnt equal c2 then  Zn = DnC1^n<br />

    so then to try Zn = Cn

    so from Xn+2 - 5Xn+1 + 6Xn = 4
    it will become C(n+2) - 5(c(n+1)) + 6(Cn) = 4

    and from setting n=-1 i get c-6c=4 so c doesnt equal 2 :S

    where have i gone wrong?
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  7. #7
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    Quote Originally Posted by charlotte_usa View Post
    but to get the complementary solution

     Yn= A2^n + B3^n

    right? and since Dn = 0 then there is no particular solution?
    Yes that is the full solution because it is a homogenous difference equation. Thus, there is no particular solution.


    here the c1 = 2 and c2 = 3 so ive been taught that for a particular solution, since c1 doesnt equal c2 then

    so then to try Zn = Cn

    so from Xn+2 - 5Xn+1 + 6Xn = 4
    it will become C(n+2) - 5(c(n+1)) + 6(Cn) = 4

    and from setting n=-1 i get c-6c=4 so c doesnt equal 2 :S

    where have i gone wrong?
    I have no idea what you did.
    Why C_1=2 and C_2=3?
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  8. #8
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    Hello, charlotte_usa!

    I believe the second problem requires special handling . . .


    b)\;X_{n+2} - 5X_{n+1} + 6X_n\;=\; 4
    \begin{array}{ccccc}\text{We have:} & X_{n+2} - 5X_{n+1} + 6X_n  & = & 4 & (1)\\<br />
\text{and also:} & X_{n+3} - 5X_{n+2} + 6X_{n+1} & = & 4 & (2)\end{array}

    Subtract (1) from (2): . X_{n+3} - 6X_{n+2} + 11X_{n+1} - 6X_n \;=\;0

    The characteristic equation is: . m^3 - 6m^2 + 11m - 6 \:=\:0

    . . which factors: . (m-1)(m-2)(m-3) \;=\;0

    . . and has roots: . m \:=\:1,\:2,\:3


    Can you finish it now?

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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    Yes that is the full solution because it is a homogenous difference equation. Thus, there is no particular solution.



    I have no idea what you did.
    Why C_1=2 and C_2=3?
    The charachteristic equation,


    i mean that say r=2 and t=3 (the solutions to the characheristic equation) and since r doesnt equal t, then ive been taught to try Zn = Cn for the particular solution. then i put cn into the original equation

    so from Xn+2 - 5Xn+1 + 6Xn = 4
    it will become C(n+2) - 5(c(n+1)) + 6(Cn) = 4

    and from setting n=0 then 2c-5c=4 so c=-3/4 and Zn (the particular solution) is 3/4n

    so the general solution is  yn = A2^n + B3^n +3/4n
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  10. #10
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    is this the right answer?
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  11. #11
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    hmm im not to sure...
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