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Converting 2nd Order ODE to 1st Order ODE

  1. Dec 6, 2004 #1
    I'm stuck as to where to start with this question:

    The position function x(t) in a certain nonlinear system is described by the second order ODE:

    < equation.gif >

    Transform this ODE into a pair of first order ODEs for x1=x and x2=dx/dt. (Note that x2 represents the velocity in this system.)

    I've thought about calculating the homogeneous equation and then the particular integral, but (a) how is this done with a sin(dx/dt) on the RHS and (b) how does this yield a pair of solutions? Or is there another way to go about it?

    Attached Files:

  2. jcsd
  3. Dec 6, 2004 #2


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    Show us what happens when you substitute x2=dx/dt. Isn't that a 1st order equation? What's the other one? Hint: it is given in this very post.
  4. Dec 6, 2004 #3


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    Krab,u should have given him the result and let him strive to find the solution for the messy equation he gets:
    [tex]\frac{dx_{1}}{dt}=x_{2} [/tex]
    [tex] \frac{dx_{2}}{dt}=-\int x_{2}dt- \alpha\sin x_{2} [/tex].
    That should fully answer your problem.If u want to solve the second equation,try numerical methods.It is a nonlinear first order integro-differential equation with constant coefficients.Impossible to solve analytically.For almost all cases.
    Last edited: Dec 6, 2004
  5. Dec 6, 2004 #4
    Thanks, I guess I was expecting the problem to be harder than that :rolleyes: . Yeah, the second part of the question involves evaluation by the Runge-Kutta method. :smile:
    Last edited: Dec 6, 2004
  6. Dec 6, 2004 #5


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    These are the first order equations:
    [tex]\frac{dx_1}{dt}=x_2 [/tex]
    [tex] \frac{dx_2}{dt}=-x_1- \alpha\sin x_2 [/tex]
    In this form, they can be used directly in a canned Runge-Kutta routine.
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