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Systems of DEs

  1. Jul 26, 2005 #1
    I have this system of DEs, but after getting through all the simplifications I can't factor the auxiliary equation:

    x1' = -2x1 + x2 + x3
    x2' = x1 - x2 + 3x3
    x3' = -x2 - 3x3

    Then I eliminated all variables except for x2, using Differentiation operator.
    my AE from x2 is this:

    r3 + 6r2 + 12r + 12 = 0

    And I'm stuck!

    thank you in advance.
     
  2. jcsd
  3. Jul 26, 2005 #2

    George Jones

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    Staff Emeritus
    Science Advisor
    Gold Member

    For what it's worth, this factors into

    [tex]
    \left( r + a \right) \left(r^2 + \left( 6 - a \right) r + \frac{12}{a} \right) = 0,
    [/tex]

    with

    [tex]
    a = 2 + 2^{\frac{2}{3}}
    [/tex]

    Regards,
    George
     
  4. Jul 26, 2005 #3
    I get a different equation for x2.

    I get: -(D+2)(D^2 + 4D + 5)x2 = 0
    or (D^3 + 6D^2 + 13D + 10)x2 = 0
    the aux has roots: -2, -2 +i, -2 - i.

    We can check that the eigenvalues of
    [tex] \left( \begin{array}{ccc}
    -2 & 1 & 1 \\
    1 & -1 & 3 \\
    0 & -1 & -3
    \end{array} \right) [/tex]
    are -2, -2+i, -2-i

    so the solutions should be linear combinations
    of exp(-2t), exp(-2t)*sin(t), exp(-2t)*cos(t).
     
  5. Jul 27, 2005 #4
    How did you get that equation?
    I tried eliminating x1 first, after that I have 2 equations:

    x2 + (D + 3)x3 = 0
    (D + 1)(D + 2)x2 - x2 - x3 - 3(D + 2)x3 = 0

    so from here

    x2 + (D + 3)x3 = 0
    [(D + 1)(D + 2) - 1]x2 - [3(D + 2) + 1]x3 = 0

    is that what you had?
    because I can't get your answer.
    thanks.
     
    Last edited: Jul 27, 2005
  6. Jul 27, 2005 #5
    starting here
    x2 + (D + 3)x3 = 0
    [(D + 1)(D + 2) - 1]x2 - [3(D + 2) + 1]x3 = 0

    therefore
    [3(D+2)+1]x2 + (D+3)[3(D+2)+1]x3 = 0
    (D+3)[(D + 1)(D + 2) - 1]x2 - (D+3)[3(D + 2) + 1]x3 = 0

    add those.

    and some mindless algebra gives:
    [3(D+2)+1 + (D+3)[(D+1)(D+2) -1]] x2 = 0;
    [3D + 6+1 + (D+3)[D^2 + 3D + 1]]x2 = 0
    [3D +7 + D^3 + 3D^2 + D + 3D^2 + 9D + 3]x2 = 0
    [D^3 + 6D^2 + 13D + 10]x2 = 0
     
    Last edited: Jul 27, 2005
  7. Jul 27, 2005 #6
    argh.... Ok, thanks much! I guess I just expected a nice and simple solution...
     
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