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Linear ODE Systems in Numerical Methods.

  1. Dec 9, 2004 #1
    I feel so embarrased asking this question, but this is the place to get answers.

    I have a 2nd order ODE with a forcing function that needs to be manipulated and put into a matrix for a numerical method solution, ie Matlab. My question is: Is the matrix composed of a particular solution in the top row and a homogenous solution in the bottom row? Does this satisfy the requirement for two equations? Maybe I should say equation, not solution.

    My work:

    m d^2x/dt^2 + c dx/dt + kx = 0

    d^2x/dt^2 = dx/dt

    Substituting y2 for d^2t/dx^2 and y1 for dx/dt, and realizing that y2 is the derivative of y1, I end up with, in matrix form:

    (I am using periods to hold the spacing)


    Thank you


    On edit, I realized I forgot the signs in the first equation.
    On second edit, I changed the lower equation to simplify what I was after.
    Last edited: Dec 10, 2004
  2. jcsd
  3. Dec 11, 2004 #2
    Any body care to comment?

  4. Jan 29, 2005 #3
    The way you describe the problem is a little confusing. Let me try to paraphrase.
    Your original problem is:
    [tex] m \frac{d^2 x}{d t^2} + c \frac{dx}{dt} + kx = 0 [/tex]
    [tex] m \frac{d^2 x}{d t^2} - \frac{dx}{dt} = 0 [/tex]

    And you would like to transfer it to matrix form, right ?
    You might do something as follows:

    [tex] y_{1}=x [/tex]
    [tex] y_{2}= \frac{dx}{dt} [/tex]
    And let
    [tex] X = \left(\begin{array}{c} y_{1} \\ y_{2} \end{array}\right) [/tex]
    [tex] Y = \left(\begin{array}{c}\frac{d y_{1}}{dt} \\ \frac{dy_{2}}{dt} \end{array}\right) [/tex]

    And you might turn this problem into

    [tex] \left(\begin{array}{cc} c & m \\ -1 & m \end{array}\right) Y=
    \left(\begin{array}{cc} -k & 0 \\ 0 & 00 \end{array}\right)X

    Is this what you are trying to ask ?

    Probably you can just start from here and check the other websites
    to solve this question in martrix form. You might as well just take a look


    It might not solve your question directly. But it is with a lot of links
    on differential equation. You could start from there to collect some information...
  5. Jan 29, 2005 #4


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    Homework Helper

    I think the original question is a bit confussing. This is how I'd present converting a high-order ODE to a system of first-order ODEs:

    \frac {d^2y} {dx^2} + \ln{y} = yx

    To convert this to a system of ODEs,


    z[x]=\frac{dy} {dx}


    \frac{dy} {dx} = z

    \frac {dz} {dx} =yx-\ln{y}

    You can use the same method for higher-order ODEs, just assigning different variables to each higher-order derivative. I used Mathematica to solve this system with initial conditions:



    I attached a plot of the solution which you can find in "additional options".


    Attached Files:

  6. Jan 29, 2005 #5


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    Homework Helper

    Oh yea, I thought the talking dictionary on line was cool, but LaTeX rocks!

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