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Converting 2nd Order ODE to 1st Order ODE

  • #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?
 

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  • #2
krab
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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.
 
  • #3
dextercioby
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krab said:
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.
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.
 
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  • #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:
  • #5
krab
Science Advisor
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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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