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Math Help - Difference Equation - Non Homogeneous need help

  1. #1
    Member zzizi's Avatar
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    Difference Equation - Non Homogeneous need help

    Hi


    Can someone help me with this and provide a step by step response?

    Suppose I have the following difference equation:

    u_{n}= -u_{n-1}+6u_{n-2} +7 with
    u_{0} =1, u_{1} =2

    I have solved the characteristic eqn to be \lambda = 2,-3

    But how do I go about solving the particular solution?


    Many thanks in advance!
    Last edited by zzizi; December 22nd 2012 at 02:54 PM.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by zzizi View Post
    Hi


    Can someone help me with this and provide a step by step response?

    Suppose I have the following difference equation:

    u_{n}= -u_{n-1}+6u_{n-2} +7 with
    u_{0} =1, u_{1} =2

    I have solved the characteristic eqn to be \lambda = 2,-3

    But how do I go about solving the particular solution?


    Many thanks in advance!
    Hi zzizi!

    Wiki explains it better than I can:
    Recurrence relation - Wikipedia, the free encyclopedia


    The equation in the above example was [[homogeneous differential equation|homogeneous]], in that there was no constant term. If one starts with the non-homogeneous recurrence
    b_{n}=Ab_{n-1}+Bb_{n-2}+K

    with constant term ''K'', this can be converted into homogeneous form as follows: The [[steady state]] is found by setting b_n = b_{n-1} =b_{n-2} = b^* to obtain
    b^{*} = \frac{K}{1-A-B}

    Then the non-homogeneous recurrence can be rewritten in homogeneous form as
    [b_n - b^{*}]=A[b_{n-1}-b^{*}]+B[b_{n-2}-b^{*}]

    which can be solved as above.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    I would use the technique of symbolic differencing to obtain a homogeneous linear recurrence:

    u_{n}=-u_{n-1}+6u_{n-2}+7

    u_{n+1}=-u_{n}+6u_{n-1}+7

    Subtracting the former from the latter, we obtain:

    u_{n+1}=7u_{n-1}-6u_{n-2}

    The characteristic roots are r=-3,\,1,\,2 hence the closed form is:

    u_n=k_1(-3)^n+k_2+k_32^n

    We may use initial conditions to determine the parameters k_i:

    u_0=k_1+k_2+k_3=1

    u_1=-3k_1+k_2+2k_3=2

    3_2=9k_1+k_2+4k_3=11

    Solving this system, we find:

    k_1=\frac{7}{20},\,k_2=-\frac{7}{4},\,k_3=\frac{12}{5} and so we have:

    u_n=\frac{7(-3)^n+3\cdot2^{n+4}-35}{20}
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  4. #4
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by ILikeSerena View Post
    Hi zzizi!

    Wiki explains it better than I can:
    Recurrence relation - Wikipedia, the free encyclopedia


    The equation in the above example was [[homogeneous differential equation|homogeneous]], in that there was no constant term. If one starts with the non-homogeneous recurrence
    b_{n}=Ab_{n-1}+Bb_{n-2}+K

    Thank you very much, for your reply.
    So I was incorrect to think it was non-homogeneous?
    I wasn't aware that it could be solved this way, so I will have to perhaps look into it further.
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  5. #5
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by MarkFL2 View Post
    I would use the technique of symbolic differencing to obtain a homogeneous linear recurrence:

    u_{n}=-u_{n-1}+6u_{n-2}+7

    u_{n+1}=-u_{n}+6u_{n-1}+7

    Subtracting the former from the latter, we obtain:

    u_{n+1}=7u_{n-1}-6u_{n-2}

    The characteristic roots are r=-3,\,1,\,2 hence the closed form is:

    u_n=k_1(-3)^n+k_2+k_32^n

    We may use initial conditions to determine the parameters k_i:

    u_0=k_1+k_2+k_3=1

    u_1=-3k_1+k_2+2k_3=2

    3_2=9k_1+k_2+4k_3=11

    Solving this system, we find:

    k_1=\frac{7}{20},\,k_2=-\frac{7}{4},\,k_3=\frac{12}{5} and so we have:

    u_n=\frac{7(-3)^n+3\cdot2^{n+4}-35}{20}
    Thank you very much for this solution.

    How would I go about checking it? I tried to find u(3) from the original recursion and the closed form but they didn't correlate.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    From the inhomogeneous recurrence you gave:

    u_2=-2+6+7=11

    u_3=-11+12+7=8

    Using the closed form I gave:

    u_3=\frac{7(-3)^3+3(2)^7-35}{20}=\frac{160}{20}=8
    Thanks from zzizi
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  7. #7
    Super Member ILikeSerena's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by zzizi View Post
    Thank you very much, for your reply.
    So I was incorrect to think it was non-homogeneous?
    I wasn't aware that it could be solved this way, so I will have to perhaps look into it further.
    You were right. It is non-homogeneous.

    In your case you have
    Quote Originally Posted by zzizi View Post
    u_{n}= -u_{n-1}+6u_{n-2} +7 with
    u_{0} =1, u_{1} =2

    I have solved the characteristic eqn to be \lambda = 2,-3
    If you compare that to
    b_{n}=Ab_{n-1}+Bb_{n-2}+K

    you'll see that you have:
    b_n = u_n
    A = -1
    B = 6
    K = 7
    b^{*} = \frac{K}{1-A-B} = \frac{7}{1-(-1)-6} = - \frac 7 4


    As a result, your solution takes the form:
    u_n = C (-3)^n + D 2^n + b^{*}

    If you fill in your boundary conditions, you get 2 equations with 2 unknowns (C and D), which can be solved with substitution.
    Last edited by ILikeSerena; December 23rd 2012 at 05:47 AM.
    Thanks from zzizi
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  8. #8
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by MarkFL2 View Post
    From the inhomogeneous recurrence you gave:

    u_2=-2+6+7=11

    u_3=-11+12+7=8

    Using the closed form I gave:

    u_3=\frac{7(-3)^3+3(2)^7-35}{20}=\frac{160}{20}=8

    Umm ... I guess I miscounted

    Thanks MarkFL2 I just saw your response on yahoo answers!!
    Last edited by zzizi; December 23rd 2012 at 06:05 AM.
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  9. #9
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by ILikeSerena View Post
    You were right. It is non-homogeneous.

    In your case you have


    If you compare that to
    b_{n}=Ab_{n-1}+Bb_{n-2}+K

    you'll see that you have:
    b_n = u_n
    A = -1
    B = 6
    K = 7
    b^{*} = \frac{K}{1-A-B} = \frac{7}{1-(-1)-6} = - \frac 7 4


    As a result, your solution takes the form:
    u_n = C (-3)^n + D 2^n + b^{*}

    If you fill in your boundary conditions, you get 2 equations with 2 unknowns (C and D), which can be solved with substitution.
    Thank you very much for your explanation! Much appreciated.
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  10. #10
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    I have a query;

    Is there another way, by substituting values to find the particular solution?
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  11. #11
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by ILikeSerena View Post
    You were right. It is non-homogeneous.

    In your case you have


    If you compare that to
    b_{n}=Ab_{n-1}+Bb_{n-2}+K

    you'll see that you have:
    b_n = u_n
    A = -1
    B = 6
    K = 7
    b^{*} = \frac{K}{1-A-B} = \frac{7}{1-(-1)-6} = - \frac 7 4


    As a result, your solution takes the form:
    u_n = C (-3)^n + D 2^n + b^{*}

    If you fill in your boundary conditions, you get 2 equations with 2 unknowns (C and D), which can be solved with substitution.

    What if I end up with zero for the b* value? I used this method to solve my difference equation but I got zero for the constant. If this happens would the whole solution be complete or do I try another method?
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  12. #12
    Super Member ILikeSerena's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by zzizi View Post
    What if I end up with zero for the b* value? I used this method to solve my difference equation but I got zero for the constant. If this happens would the whole solution be complete or do I try another method?
    The only way that b* can be zero is if the constant K is zero.
    In that case the difference equation is a homogeneous difference equation instead of an in-homogeneous one.

    But this is not applicable to your current problem statement.....
    Am I misunderstanding you?
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  13. #13
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by ILikeSerena View Post
    The only way that b* can be zero is if the constant K is zero.
    In that case the difference equation is a homogeneous difference equation instead of an in-homogeneous one.

    But this is not applicable to your current problem statement.....
    Am I misunderstanding you?
    You are quite right. I wanted to understand the concept because I have another problem for my homework that I am working which is like this:

    u_{n}= -3u_{n-1}+4u_{n-2}+9

    When I applied the formula I got this:

    b^{*} = \frac{K}{1-A-B} = \frac{9}{1-(-3)-4} =  0


    SO would this be considered a homogeneous D.Eqn?
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  14. #14
    Super Member ILikeSerena's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Quote Originally Posted by zzizi View Post
    You are quite right. I wanted to understand the concept because I have another problem for my homework that I am working which is like this:

    u_{n}= -3u_{n-1}+4u_{n-2}+9

    When I applied the formula I got this:

    b^{*} = \frac{K}{1-A-B} = \frac{9}{1-(-3)-4} =  0

    I just wondered what I should do in this situation
    I'm afraid you miscalculated b*.

    b^* = \frac 9 0 = \infty

    Either way, it means this method does not work.

    Next method in line is the symbolic differentiation method MarkFL2 described.
    That one works for this problem.
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  15. #15
    Member zzizi's Avatar
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    Re: Difference Equation - Non Homogeneous need help

    Thank you.
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