Converting 2nd Order ODE to 1st Order ODE

In summary, the task is to transform a second order ODE into a pair of first order ODEs for the variables x1 and x2. This can be done by substituting x2=dx/dt and using the given equations. The resulting equations can then be solved using numerical methods such as the Runge-Kutta method.
  • #1
johnwalton84
16
0
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
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
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:
 
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  • #5
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.
 

1. How do you convert a 2nd order ODE to a 1st order ODE?

To convert a 2nd order ODE to a 1st order ODE, you need to introduce a new variable. Let y' = z, then the 2nd order ODE can be rewritten as a system of two 1st order ODEs: y' = z and z' = f(x, y, z).

2. Why do we need to convert a 2nd order ODE to a 1st order ODE?

Converting a 2nd order ODE to a 1st order ODE allows us to solve the ODE using numerical methods such as Euler's method or Runge-Kutta methods. These methods can only be applied to 1st order ODEs.

3. What are the advantages of converting a 2nd order ODE to a 1st order ODE?

Converting a 2nd order ODE to a 1st order ODE allows us to use numerical methods to solve the ODE, making it easier to obtain numerical solutions. It also allows us to solve more complicated ODEs that cannot be solved analytically.

4. Can any 2nd order ODE be converted to a 1st order ODE?

Yes, any 2nd order ODE can be converted to a 1st order ODE by introducing a new variable and rewriting the ODE as a system of two 1st order ODEs. However, the resulting 1st order ODE may not always be easier to solve than the original 2nd order ODE.

5. Are there any limitations to converting a 2nd order ODE to a 1st order ODE?

There are no inherent limitations to converting a 2nd order ODE to a 1st order ODE. However, the resulting 1st order ODE may be more difficult to solve or may require more computational resources. It is important to consider the trade-offs and choose the best approach for solving the specific ODE at hand.

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