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Homework Help: Rewrite equation as two first order ODE's Help Needed

  1. Apr 12, 2010 #1
    1. rewrite the following equation as two first order ODE's

    x'' + 2(x^(2) - 2).x' + x = 0

    2. Relevant equations



    This is what I have so far:

    x'' + 2(x^(2) - 2) . x' + x = 0
    x'' = - 2(x^(2) - 2) . x' - x

    x' = y
    y' = -2(x(^2)-2). y - x

    Are these correct?
     
  2. jcsd
  3. Apr 12, 2010 #2

    Mark44

    Staff: Mentor

    Yes, they're fine.
     
  4. Apr 12, 2010 #3
    Thanks :)

    Just a follow on, how would I go about finding the Jacobian matrix for the two first order ODE's?

    Im a little confused with the second one:

    y' = -2(x^(2)-2) . y - x
     
  5. Apr 12, 2010 #4

    Mark44

    Staff: Mentor

    You have
    x' = y
    y' = -x -2y(x^2 - 2)

    Jacobian matrix will have four entries in it. Do you know what they are, in general?
     
  6. Apr 12, 2010 #5
    Yes I know how they are derived, Im just unsure of the bottom 2 entries of the matrix.
    So far I have half the matrix:

    (0 1)/(X Y) Where X and Y are yet to be found.

    How would I derive the x and y parts of y'?
     
  7. Apr 12, 2010 #6

    Mark44

    Staff: Mentor

    I didn't ask how they are derived, just what they are, in general (not specific to this problem). If you explain how you got the 0 and 1 in the top row, it will go a long way to you understanding what goes in the bottom row.
     
    Last edited: Apr 13, 2010
  8. Apr 13, 2010 #7
    ok so I have now found the Jacobian matrix, all I need now is to discover the fixed points of the two frsit order ODE's at the top of page. How would I go about doing that?
     
  9. Apr 13, 2010 #8

    Mark44

    Staff: Mentor

    Here's a link to a wikipedia article that might be helpful, not sure - http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

    See the section titled "In dynamical systems."

    However, I'm not sure what the Jacobian will do for you. As you posted the problem in the other thread, it says to analyze the behavior using the eigenvalues.

    You really should get in the habit of posting the complete problem in one thread, rather than in dribs and drabs in multiple threads.
     
  10. Apr 13, 2010 #9
    Yes I do wish to analyze the behaviour using eigenvalues, with me having the jacobian matrix I know simply need the fixed points. I used Derive and it gave me the fixed points of (0,0) however Im nt sure if that is correct.
     
  11. Apr 13, 2010 #10

    Mark44

    Staff: Mentor

    The fixed point(s) are the values of x and y for which x' and y' = 0. Solve the equations x' = y and y' = -2y(x2 - 2) - x simultaneously.
     
  12. Apr 13, 2010 #11
    Ok I know im being slow here but could you start me off on the simulataneous equation, each time I attempt it I only get x and y to equal 0. Thanks
     
  13. Apr 13, 2010 #12

    Mark44

    Staff: Mentor

    That's what I get, too, so the only fixed point seems to be at (0, 0).
     
  14. Apr 13, 2010 #13
    Oh ok that has lifted a huge weight from my shoulders, thanks for all your help:)
     
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