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Math Help - difference equation

  1. #1
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    difference equation

    Am I on the right track with this problem
    I am trying to find the particular solution
    a_{n+2} - 2a _{n+1} -63a_{n} = 64n, n>= 0 with a_{0} = 0  a_{1} = 7

    (x-9)(x+7) = 0
    x=9, x=-7

    u_{n} = A*9^n + B * -7^n for arbitrary constants A and B

    RHS
    64n =
    V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D)

    substitute a_{n} = V_{n}
    a_{n+2} - 2a_{n+1} - 63an
    = (C(n+2)+D) - 2(C(n+1)+D -63 (Cn + D) = 64n
    then
    (-64 + -C) = 0
    -64D = 0

    implying C = -64
    D = 0

    thus Vn = n+ -64

    Therefore the general solution is
    A9^n + B(-7)^n + n - 64 for arbitrary constants A and B

    a_{0} =0 =  A * 9^0 + B*-7^0 + 0 -64
    a_{1} = 7 =  A * 9^1 + B*-7^1 + 1 -64 <-- I will work on these if I am on the right track

    how am I doing?
    (no good I think)
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  2. #2
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    Quote Originally Posted by dunsta View Post
    Am I on the right track with this problem
    I am trying to find the particular solution
    a_{n+2} - 2a _{n+1} -63a_{n} = 64n, n>= 0 with a_{0} = 0 a_{1} = 7

    (x-9)(x+7) = 0
    x=9, x=-7

    u_{n} = A*9^n + B * -7^n for arbitrary constants A and B

    RHS
    64n =
    V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D)

    substitute a_{n} = V_{n}
    a_{n+2} - 2a_{n+1} - 63an
    = (C(n+2)+D) - 2(C(n+1)+D) -63 (Cn + D) = 64n Mr F says: The red bracket will make a difference.

    then
    (-64 + -C) = 0
    -64D = 0

    Mr F says: The above two equations make no sense. Please show clearly where they come from.

    implying C = -64
    D = 0

    thus Vn = n+ -64

    Therefore the general solution is
    A9^n + B(-7)^n + n - 64 for arbitrary constants A and B

    a_{0} =0 = A * 9^0 + B*-7^0 + 0 -64
    a_{1} = 7 = A * 9^1 + B*-7^1 + 1 -64 <-- I will work on these if I am on the right track

    how am I doing?
    (no good I think)
    Your particular solution is wrong. I get a_n = -n as a particular solution.
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  3. #3
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    = (C(n+2)+D) - 2(C(n+1)+D) -63 (Cn + D) = 64n Mr F says: The red bracket will make a difference.

    LHS
    =(Cn + 2C + D) - 2(Cn + C + D) - 63Cn - 63CD
    = Cn + 2C + D -2Cn -2C -2D -63Cn -63CD
    = Cn - 2Cn -63Cn +2C -2C +D -2D -63D
    = -64Cn -64D

    RHS = 64n

    -64Cn -64D = 64n
    -64Cn -64n -64D = 0
    n(-64C -64) -64D = 0

    (after this I split the equation into two, as below, and dropped the n (I am just going off the most similar example in my notes)


    then
    (-64 + -C) = 0
    -64D = 0

    Mr F says: The above two equations make no sense. Please show clearly where they come from.
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  4. #4
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    Quote Originally Posted by dunsta View Post
    = (C(n+2)+D) - 2(C(n+1)+D) -63 (Cn + D) = 64n Mr F says: The red bracket will make a difference.

    LHS
    =(Cn + 2C + D) - 2(Cn + C + D) - 63Cn - 63CD Mr F says: ?
    = Cn + 2C + D -2Cn -2C -2D -63Cn -63CD Mr F says: ?
    = Cn - 2Cn -63Cn +2C -2C +D -2D -63D
    = -64Cn -64D

    RHS = 64n

    -64Cn -64D = 64n
    -64Cn -64n -64D = 0
    n(-64C -64) -64D = 0

    (after this I split the equation into two, as below, and dropped the n (I am just going off the most similar example in my notes)

    then
    (-64 + -C) = 0 Mr F says: Shouldn't this be -64C - 64 = 0?
    -64D = 0 Mr F says: I got -63D, but I can't be bothered re-checking. D = 0 is the correct result in both cases.

    [snip]
    Did you check the particular solution you found (in both of your posts) to see that it worked? If you had done so, you would have realised it was wrong.
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  5. #5
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    Can anybody point me towards a similar problem with a solution. So far in Discrete maths I have had the best results from looking at a problem with a soution and then working out the hows and whys, I have read my notes on this topic over and over but they are long winded explanations in words, with no examples.
    So if any one can point me to a similar problem/exercise I would be appreciative. At the moment I seem to be lost with this problem.



    a_{n+2} - 2a _{n+1} -63a_{n} = 64n, n>= 0 with a_{0} = 0  a_{1} = 7

    (x-9)(x+7) = 0
    x=9, x=-7 <---is this part correct?

    u_{n} = A*9^n + B * -7^n for arbitrary constants A and B

    RHS
    64n =
    V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D) <-- is this correct?? Mr Fantastic is this the particular solution you refer to that should be a_{n} = -n


    A9^n + B(-7)^n + n - 64 for arbitrary constants A and B

    a_{0} =0 =  A * 9^0 + B*-7^0 + 0 -64
    a_{1} = 7 =  A * 9^1 + B*-7^1 + 1 -64 <-- I think my goal is to try and find A and B for these expressions
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  6. #6
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    Quote Originally Posted by dunsta View Post
    Can anybody point me towards a similar problem with a solution. So far in Discrete maths I have had the best results from looking at a problem with a soution and then working out the hows and whys, I have read my notes on this topic over and over but they are long winded explanations in words, with no examples.
    So if any one can point me to a similar problem/exercise I would be appreciative. At the moment I seem to be lost with this problem.



    a_{n+2} - 2a _{n+1} -63a_{n} = 64n, n>= 0 with a_{0} = 0 a_{1} = 7

    (x-9)(x+7) = 0
    x=9, x=-7 <---is this part correct?

    u_{n} = A*9^n + B * -7^n for arbitrary constants A and B Mr F says: To check if it's correct, substitute it into the homogenous difference equation. Does it work? (Do you understand the thinking behind what you have done above?)

    RHS
    64n =
    V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D) <-- is this correct?? Mr Fantastic is this the particular solution you refer to that should be a_{n} = -n Mr F says: Yes.


    A9^n + B(-7)^n + n - 64 for arbitrary constants A and B

    a_{0} =0 = A * 9^0 + B*-7^0 + 0 -64
    a_{1} = 7 = A * 9^1 + B*-7^1 + 1 -64 <-- I think my goal is to try and find A and B for these expressions
    ..
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