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Coupled Differential Equations

  1. May 26, 2010 #1
    Hi all,

    I want to solve equations of the form:
    [tex]\dot x + x + y = sin(\omega t)[/tex]
    [tex]\dot y = \dot x - y[/tex]

    This is not a standard type of form for Runge-Kutta or linear systems of equations because
    [tex]\dot y = f(\dot x, y, t)[/tex]
    instead of
    [tex]\dot y = f(x, y, t)[/tex].
    Any hints or links to place for help would be appreciated! Thanks!
     
  2. jcsd
  3. May 26, 2010 #2

    tiny-tim

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    Hi mkrems! :wink:

    Can't you get it into f(x, y, t) form by substituting for x' from the first equation? :smile:
     
  4. May 27, 2010 #3
    Use your first equation to isolate y, namely,

    [tex]y = \sin{\omega t} - x^{\prime} - x[/tex]

    Now, differentiate this to get y prime,

    [tex]y^{\prime} = \omega \cos{\omega t} - x^{\prime \prime} - x^{\prime}[/tex]

    and substitute these into your second equation to get...

    [tex]\omega \cos{\omega t} - x^{\prime \prime} - x^{\prime} =
    x^{\prime} - \sin{\omega t} + x^{\prime} + x[/tex]

    which may be rearranged to give you a (soluble) second-order equation in x only.

    [tex]x^{\prime \prime} + 3 x^{\prime} + x = \omega \cos{\omega t} + \sin{\omega t}[/tex]
     
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