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Math Help - Help on solving second order equation with constant

  1. #1
    fuz
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    Smile Help on solving second order equation with constant

    Hi,

    I am hoping someone can give me a nudge in the right direction on understanding and solving the following equations:

    Au''(t)+Bu'(t)-Cu(t)+Dp(t)+F=G(t)

    -Du'(t) + Hp'(t) = -I(t)

    where u(t) and p(t) are functions of time. I have taken diff eq classes several years ago, and am refreshing my memory but would like an expert to point out anything unusual. The F is a constant and not associated with any time function.
    Thanks!
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  2. #2
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    Well, one interesting feature is that, assuming u and p are the unknown functions for which you wish to solve, and G and I are known functions, then one of your equations can be integrated directly to produce

    \displaystyle-Du(t)+Hp(t)=-\int I(t)\,dt:=i(t), from which we may solve for

    Dp(t)=\dfrac{Di(t)+D^{2}u}{H}.

    Hence, the first ODE becomes

    A\ddot{u}+B\dot{u}-Cu+\dfrac{Di(t)+D^{2}u}{H}=G(t)-F, or

    \displaystyle A\ddot{u}+B\dot{u}+\left(\frac{D^{2}}{H}-C\right)u=G(t)-F-\dfrac{Di(t)}{H}.

    Now you have a regular ol' inhomogeneous second-order ODE with constant coefficients. Use your favorite method here.
    Last edited by Ackbeet; March 14th 2011 at 05:53 PM. Reason: Correct tense.
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  3. #3
    fuz
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    Thank you more than I can express! I am going to try and get this into MATLAB.
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  4. #4
    A Plied Mathematician
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    You're welcome! Let me know if you have any further difficulties.
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  5. #5
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    \displaystyle-Du(t)+Hp(t)=-\int I(t)\,dt:=i(t),

    Okay, now that the OP has got it I can ask one of my own. I had gotten to the point of integrating the second equation to find p(t) as a function of u(t). But is there still a solution if I(t) is not integrable?

    -Dan
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  6. #6
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    Hmm. Well, you might be able to work with an eigenvalue approach, where the integrability of I(t) is never assumed. If you let x_{1}(t)=u(t), x_{2}(t)=\dot{u}(t), and x_{3}(t)=p(t), then you get the following system (it just works better with three coordinates than four):

    \displaystyle\frac{d}{dt}\begin{bmatrix}x_{1}\\x_{  2}\\x_{3}\end{bmatrix}=\begin{bmatrix}0 &1 &0\\C/A &-B/A &-D/A\\0 &D/H &0\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\  end{bmatrix}+\begin{bmatrix}0\\(G(t)-F)/A\\-I(t)/H\end{bmatrix}.

    You could use the usual eigenvalue procedure to find the homogeneous solution, and then use undetermined coefficients to find a particular solution, perhaps.
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