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Differential equation with repeated roots

  1. Oct 25, 2013 #1
    Hi,
    I'm somewhat new here, only posted a few times, and would like some help from you guys here if possible
    I'm stuck with a problem on the topic mentioned.

    x'=Ax

    A is a 2*2 matrix
    A =
    [-5 1]
    [-1 -3]

    Now I managed to find the eigenvalues which is -4, repeated twice (multiplicity 2)
    And the eigenvector as well which is

    [1]
    [1]

    Now I'm really confused as to what to do next, I know the equation is e-4t
    But I am having trouble deriving the homogenous and particular solutions to the problem.
    Would appreciate any input
     
    Last edited: Oct 25, 2013
  2. jcsd
  3. Oct 25, 2013 #2

    tiny-tim

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    hi hops1! :smile:
    (don't you mean
    [-5 1]
    [-1 -3] ? :confused:)

    for a repeated root, the solutions are usually eAt and teAt :wink:
     
  4. Oct 25, 2013 #3
    You are correct thanks, but what I'm actually having trouble with is that the second vector should come out to be [t;1]
    I don't understand how that comes about.
    The first colution I understand is C1[1;1]e-4t, but for the second solution I had the feeling that I should use teAt, but to get the second vector, that's what's got me confused
     
  5. Oct 25, 2013 #4

    tiny-tim

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    hi hops1! :smile:

    try an easier one …

    [1 2]
    [0 1] :wink:
     
  6. Oct 25, 2013 #5
    Ok thanks, I tried the one you posted.
    I hope I did it right
    x'=Ax

    I split it into a system of equations
    x1' = x1 + 2x2
    x2' = x2

    x2 could be found easily,
    since x2' - x2 = 0
    Thus follows x2 = C2et

    Then to find x1 -> x1' - x1 = 2x2
    I first find the homogenous solution thus x1' - x1 = 0
    Which yields the same solution as x2, so x1 = C1et

    Now to find the particular solution
    x1' - x1 = 2x2
    Tried Atet

    Aet + Atet - Atet = 2C2et
    Solving this yields A = 2C2

    Thus
    x1 = C1et + 2C2tet
    x2 = C2et

    x = C1 [1;0] et + C2[2t;1] et
     
    Last edited: Oct 25, 2013
  7. Oct 25, 2013 #6

    tiny-tim

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    i'm really not following your reasoning :confused:

    and if we put C1 = 0, C2 = 1, into your solution, it gives x = [(t/2);1] et, which doesn't work :redface:
     
  8. Oct 25, 2013 #7
    :frown: I'm worse of than I thought.

    Well I think et is correct.
    I'm guessing I messed up in the particular solution for x1 ??


    I think I found the error in my calculation, will try to run through it again
     
    Last edited: Oct 25, 2013
  9. Oct 25, 2013 #8
    I think I got it now, I made an error in the particular solution

    So it should be
    C1 [1;0]et + C2 [2t;1] et
     
  10. Oct 25, 2013 #9
    Now I went back to my initial problem.
    I'm stuck when trying to get the particular solution, since if you rearrange the matrix
    x1' + 5x1 = x2
    x2' + 3x1 = -x1

    So let's say I take the top equation.

    x1' + 5x1 = x2

    I choose a particular solution to test say Cte-4t
    I end up with
    Ce-4t-4Cte-4t + 5Cte-4t = x2
    But I have yet to solve x2 , and doing the same for x2 would give
    Ce-4t-4Cte-4t + 3Cte-4t = x1
    And except for combining the Cte-4t in each individual equation I don't see where to go next from there. :confused:
     
  11. Oct 26, 2013 #10

    tiny-tim

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    (just got up :zzz:)

    i don't understand where you think a particular solution comes into it :confused:

    x' - Ax = 0 is a homogeneous equation (it has no t)

    x' - Ax = B(t) is not, and its solutions are all the solutions to the homogeneous equation plus one particular solution to the whole equation

    when the RHS is 0, no particular solution is needed

    (because every homogeneous solution is a particular solution)
     
  12. Oct 26, 2013 #11
    I understand that the Homogenous equation doesn't require the t.
    I have already gotten the homogenous solution. That's where I get the e-4t from.
    So found the homogenous solution already using the eigenvalues of the system

    But the equation given as
    x1' + 5x1 = x2
    requires the solution to be a sum of the homogenous solution and the particular solution, since it's not set equal to 0.
     
    Last edited: Oct 26, 2013
  13. Oct 26, 2013 #12

    tiny-tim

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    ohh, i see what you mean now …

    you're treating x2 as if it's a known function of t, instead of as a variable

    as you say, the problem with that is it's not a known function (unless of course you make a guess as to what x2 is)
    technically, no, the homogeneous solution to x1' + 5x1 = x2 is x1 = Ce-5t, isn't it? :redface:

    can you adapt the solution you already found in the easier case? ([1 2][0 1])
     
  14. Oct 26, 2013 #13
    Actually I wanted to ask about that as well.
    Why does the eigenvalue give a different solution to the equation than if I solve the individual homogenous equation.
    That's really been confusing me

    That's why I'm having trouble adapting the other solution as well.
     
  15. Oct 26, 2013 #14

    tiny-tim

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    sorry, got to rush out

    absolutely no time :redface:
     
  16. Oct 26, 2013 #15
    No prob,
    I'll see how far I can get in the mean time.
    Hope some more people can give a little input as well
     
  17. Oct 26, 2013 #16

    tiny-tim

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    (i'm back o:))

    ok, the reason why your solution doesn't work (though it did work for the simpler case)

    is that the "plain" eigenvector is neither x1 nor x2 (in the simpler case, it was)

    first you need to find that eigenvector, ie the one that corresponds to e-4t (with no extra t) …

    just solve for Ax = -4x :smile:
     
  18. Oct 26, 2013 #17
    Hi, thanks for taking your time with me on this.
    I understand now why it wouldn't work since, the two equations are dependent on each other pretty much.
    So the eigenvector that corresponds to the given eigenvalue is
    [1]
    [1]

    I actually managed to find some notes online regarding this.
    Stating that I should solve
    (A-rI)w=v
    Where r is my eigenvalue and v is my eigenvector and solve the equations to find the w vector which would then be a generalized eigenvector for the maxtrix A.
    So this would give me a particular solution with "t" multiplied the original eigenvector plus a term without "t" multiplied by the vector gotten from solving the above equation.
    Using the above method I got this answer after going through the calculations

    x= C1Ve^(-4t) + C2Vte^(-4t) + C2We^(-4t)
    Where V = [1;1] and W = [-1;0]

    So I think I managed to get the answer using this method.
    Our lecturer had told us about the other method I was trying to do, but when explaining gave a simple example much similar to what you had given me, but then gave the other one in a mock exam we had and said we were supposed to use the method for the simpler version. Anyway the solution he gave turned out to be incorrect as well after closer inspection.
     
  19. Oct 26, 2013 #18

    tiny-tim

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    yes, Ve-4t, where V = [1 1], is a solution :smile:

    yes, or you could write it

    x = C1Ve-4t + C2Ue-4t
    Where V = [1;1] and U = [t-1;t] :wink:
     
  20. Oct 26, 2013 #19
    Thanks a million man.
     
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