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A 2nd Order ODE

  1. Jul 31, 2006 #1
    Dear All,

    I have a Problem about a 2nd order ode. I dont know how it can be solved with Matlab. If someone know about it then please let me know. I need to get the values of x & y. All other values are known.

    The equation is:

    [ M + mf mf
    mf mf ][ ¨x
    ¨y ]+
    [ C 0
    0 cf ][ x˙
    y˙ ]+[ K 0
    0 kf][ x
    y ] = [ Fe(t)
    0 ]

    Thanks Alot
  2. jcsd
  3. Jul 31, 2006 #2
    you are going to have to make the equation more clear. What are all the 0s? Try to put it up in tex.
  4. Jul 31, 2006 #3


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    Homework Helper

    I think this is how it's supposed to look:
    M + m_f & m_f \\
    m_f & m_f
    \ddot{x} \\
    \right) +
    C & 0 \\
    0 & c_f
    \right) +
    K & 0 \\
    0 & k_f
    \right) =

    I don't know how to use MATLAB to solve it, though.
    Last edited: Jul 31, 2006
  5. Aug 1, 2006 #4


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    There should be an [tex]\dot{x}[/tex] after the damping terms (c's) and an [tex]x[/tex] after the stiffness terms (k's)...

    For the simulation, first write it in first-order form.

    It's quite simple to solve this forced msk system as an IVP in Matlab, check the help files on odes...

  6. Aug 1, 2006 #5
    yes. This is an equation of motion for a Tuned Liquid Column Damper with (xdot & ydot) after damping terms and (x & y) after the stiffness matrix.I dont know how i can handle the matrics if i change it to first order. If you know something then please explain a little more about the problem. How to handle the matrics to get a first order system.

    The zeros 0s are 0.There is no entry where there is zero.
  7. Aug 2, 2006 #6


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    [tex]u=\dot{x}[/tex] and [tex]v=\dot{y}[/tex]


    [tex]\dot{u}=\ddot{x}[/tex] and [tex]\dot{v}=\ddot{y}[/tex]

    ie. you now have 4 first-order equations.
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