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Dx/dt = y; dy/dt = x?

  1. Sep 28, 2009 #1
    x'' = y; y'' = -x?

    I'm working on vector fields and I've got these two equations; no teacher in my highschool knows what to do from here.

    initial conditions:
    t = 0
    x, y = 0, 1 (the initial position)
    dx/dt = 0
    dy/dt = 0

    [itex]\frac{d^2x}{dt^2} = y[/itex]
    [itex]\frac{d^2y}{dt^2} = -x[/itex]

    How do I solve this?

    Thanks a bundle :)
    -Unit
     
    Last edited: Sep 28, 2009
  2. jcsd
  3. Sep 28, 2009 #2

    LCKurtz

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    Your title gives one system and your post gives a different one. I assume your system is

    x'' = y
    y'' = x
    plus your initial conditions. Differentiate the first equation twice:

    x'''' = y'' = x, so you have:

    x''''(t) - x(t) = 0.
    This is a constant coefficient equation with characteristic equation:
    r4 - 1 = 0.

    Do you know how to write the solution for x(t) from that? If so, then you can use

    y = x''

    to get y. Plug in your initial conditions along the way to keep it simple.
     
  4. Sep 28, 2009 #3
    Yes, sorry for the discrepancy. Your assumption was right, I was indeed talking about x'' = y and y'' = -x. I fixed the thread title!

    And omgsh that makes perfect sense! I was thinking I had to make a substitution of sorts, something crazy with polar coordinates or something. The fact that it's a fourth derivative equaling the first derivative makes it look like a sine/cosine function!

    I'll work on the answer now, using r^4 - 1 = 0, thank you :)
     
  5. Sep 28, 2009 #4
    Waiiiiiit. x''''(t) = -x(t)
    x''''(t) + x(t) = 0

    That is what I'm solving. Does this make any difference?
     
  6. Sep 28, 2009 #5

    LCKurtz

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    Yes, a little. Characteristic equation now r4 + 1 = 0.

    If you don't know how to proceed search constant coefficient differential equations on Google. Good luck.
     
  7. Oct 3, 2009 #6
    Re: x'' = y; y'' = -x?


    I would rather convert the problem to the following IVP and solve

    [tex]\frac{dZ}{dt} = AZ[/tex]

    subject to Z(0)=(0 0 1 0)t

    where
    [tex] A = \left(\begin{array}{cccc}0&1&0&0\\
    0&0&1&0\\
    0&0&0&1\\
    -1&0&0&0\end{array}\right)
    [/tex] [tex] Z = \left(\begin{array}{c}x\\ \dot{x}\\ y \\ \dot{y} \end{array}\right)
    [/tex]
     
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