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Solving a System of Differential Equations

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Y' =
    [ -14 0 12
    24 -2 24
    16 0 14 ] Y
    Subject to the condition [itex] y_1 (0) = -6, y_2 (0) = 11, y_3 (0) = 8, find: y_1(t), y_2(t), y_3(t) [/itex]
    2. Relevant equations


    3. The attempt at a solution
    This is more just to see if my answers are correct than anything, so I begin by finding the eigenvalues of the matrix (checked with an internet calculator so I know this is correct).
    Eigenvalues end up being: -2, -2, and 2.
    I solve for the eigenvectors, which end up being :
    [itex] v_1 = [-1, 0 ,1], v_2 = [0, 1, 0], v_3 = [-3, 6, 4] [/itex]
    Therefore I have the general solution: [itex] Y = C_1 v_1 e^{-2t} + C_2 v_2 t e^{-2t} + C_3 v_3 e^{2t} [/itex]

    Using the initial values given, putting it into the matrix and solving it, I end up with:
    [itex] C_1 = -3,
    C_2 = 13/2, C_3 = 4
    [/itex]

    However, the question wants me to find [itex] y_1 (t), y_2 (t), and y_3 (t) [/itex]
    It also does not allow me to enter a vector as an answer, so does anyone know how my answer is supposed to be formatted? Or am I going about it wrong? Thank you for any assistance you may provide.
     
  2. jcsd
  3. Nov 8, 2014 #2

    Mark44

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    You have this:
    ##Y = C_1 v_1 e^{-2t} + C_2 v_2 t e^{-2t} + C_3 v_3 e^{2t} ##
    With the three vectors and the constants now included, it looks like this:
    $$Y = -3 \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} e^{-2t} + 13/2 \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix} e^{-2t} + 4 \begin{bmatrix} -3 \\ 6 \\ 4\end{bmatrix} e^{2t} $$

    You can get your three solutions out of the above.

    Edit: I assumed without checking that the work shown was correct.
     
    Last edited: Nov 9, 2014
  4. Nov 8, 2014 #3

    Simon Bridge

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    How are y1, y2 and y3 related to Y? (Mark beat me to it!)
     
  5. Nov 9, 2014 #4

    ehild

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    Are you sure that those are the eigenvalues?
     
  6. Nov 9, 2014 #5
    Hmm, I don't know, I've been reading the internet and the textbook but I can't find any clues on what it means.

    Also - I'm sure those are the eigenvalues, pretty sure at least, I did them by hand and then checked with wolfram.
     
  7. Nov 9, 2014 #6

    ehild

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    What is your characteristic equation?
    It is also possible that you did not copy the matrix correctly. Check, please.
     
    Last edited: Nov 9, 2014
  8. Nov 9, 2014 #7
    Oh yes, I'm sorry, for some reason I missed a negative sign on the 12 in the first row.

    I kind of get the relationship too, Y = [y1, y2, y3], where y2 = y1' and y3 = y2' if I'm not mistaken, I just don't see how this could be used to get a solution.
     
    Last edited: Nov 9, 2014
  9. Nov 9, 2014 #8

    Mark44

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    There's a technique that can be used to convert a second- or higher-order differential equation into a system of linear differential equations, and possibly that's what happened here. If y1 = y, then from the above, y2 = y' and y3 = y'' and y3' = y'''. So an equation in y, y', y'', and y''' can be rewritten as a system of linear equations in y1, y2, and y3.
     
  10. Nov 9, 2014 #9
    Hmm, I just tried that but it seems that the variable 'y' cannot be used in my answer.
     
  11. Nov 9, 2014 #10

    Mark44

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    y is a vector. I'm pretty sure you need to specify y1, y2, and y3 separately.
     
  12. Nov 9, 2014 #11

    ehild

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    I tried to convert the system of equations into a third-order differential equation, but to my surprise, it fell apart to three independent second-order equations: y1"=4y1, y2"=4y2, y3"=4y3. :)
     
  13. Nov 9, 2014 #12

    ehild

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    I got something different. The general solution does not involve a term with t e-2t. (Mark, you omitted the factor t in the second term. Was it a typo?)
     
  14. Nov 10, 2014 #13

    Mark44

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    Yes. I didn't notice the factor of t in the second term of the solution the OP showed in post 1.
     
  15. Nov 10, 2014 #14

    ehild

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    @Kevin: y1, y2, y3 are the components of the vector Y, and the matrix equation can be expanded into a system of linear differential equation for y1, y2, y3:
    y1'=14y1 - 12y3
    y2'=24y1 -2y2 +24y3
    y3'=16y1+ 14y3.

    If you have repeated roots λ of the characteristic equation, take one solution as v1eλt, and the other solution in the form (v2+tv3)eλt.
     
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