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Homework Help: Convert 2nd order ODE to system of 1st order

  1. Apr 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Convert the following second-order differential equation into a system of first-order equations and solve y(1) and y'(1) with 4th-order Runge-kutta for h=0.5.
    ##y''(t)+sin(y(t))=0,\ y(0)=1,\ y'(0)=0##

    2. Relevant equations
    The Runge-kutta method might be applicable, but I know how to do that part no problem.

    3. The attempt at a solution
    The issue I'm having is converting it into a proper system of equations.

    I let:
    ##x_1=y,\ x_2=y'##
    ##x_1'=x_2,\\ x_2'=y''=-sin(y)=-sin(x_1)##
    ##x_1(0)=1,\ x_2(0)=0##

    This yields:

    I need to put this in the form:
    so that I can solve it using runge-kutta. Any ideas?
  2. jcsd
  3. Apr 30, 2015 #2


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    Homework Helper

    Correct so far.

    The ODE isn't linear with constant coefficients, so it can't necessarily be put into that form. But it isn't necessary to do so to use Runge-Kutta (or any other ODE solution algorithm). As long as you have a first-order vector ODE of the form [tex]
    \dot{\mathbf{x}} = \mathbf{f}(\mathbf{x})[/tex] and an initial condition on [itex]\mathbf{x}[/itex] you can proceed.
  4. Apr 30, 2015 #3
    Ok, so I wrote up the following code,
    Code (Text):

    #System of Equations - Problem 5
    #Hayden Smotherman
    #Math 428
    #30 April 2015

    x1=1; #y(0)
    x2=0; #y'(0)
    x= [x1,x2];
    step=0.5; #Step size
    boundary = 1;

    for i=1:n
    which gives me x= 1.000000000000000 -0.719893726895089. This feels wrong to me, as x(1)=y(t). Any idea what' s up?
    Last edited by a moderator: May 1, 2015
  5. May 1, 2015 #4


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    Homework Helper

    Take a closer look at the assignments to k1, etc. Do you want k1 = [x2,-sin(x1)] or k1 = [x(2),-sin(x(1))]?
  6. May 1, 2015 #5
    Haha, x(2) or it's not going to iterate. Thanks, that was a silly mistake.
  7. May 1, 2015 #6

    D H

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    Staff Emeritus
    Science Advisor

    A couple of things. First notice how I edited your post. I didnt change the content, just the appearance of your code. It's now marked as code. You can click on the "+" symbol in the edit menubar, or just do it manually by using code tags.

    That's not what I wanted to write about. You created a bug in your code. A number of things contributed to this:
    • You used very similar names for things: x1, x2, and x. That's what got you in trouble.
    • It looks like you used cut-and-paste and then modified each line.
    • You didn't use Matlab to anything close to its full power.
    • You didn't use the debugger.
    It's been a long time since I've used Matlab, and I don't have it now. So this is untested code.

    Code (Text):

    # First, let's get rid of the potential for the x1 versus x(1) bug.
    y0=1; #y(0)
    ydot0=0; #y'(0)
    x = [y0, ydot0];

    step=0.5; #Step size
    boundary = 1;

    # Define the derivative function.
    # Someone wants you to integrate a different function?
    # Just change the derivative function.
    f = @(x) [x(2), -sin(x(1))];

    for i=1:n
      k2=f(x + (step/2)*k1);
      k3=f(x + (step/2)*k2);
      k4=f(x + step*k2);
    You could get even more elegant and use the Butcher tableau instead those hard-coded statements inside the loop. Now you can swap out the integration technique for a different Runge-Kutta technique as well as swapping out the function to be integrated.
  8. May 1, 2015 #7
    Yeah, I copy and pasted some old code I wrote for a non-vector-valued Runge-Kutta. Thanks for the tips. What exactly is the debugger? I'm running Octave on Linux; is that still a thing?
  9. May 2, 2015 #8

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

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