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Math Help - Reducing Systems of higher order equations

  1. #1
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    Reducing Systems of higher order equations

    I missed the lecture and was wondering if somebody could clue me in on the method. I as far as I can tell (and to prove that I am trying over here ) We introduce
    x1=x(t) x2=x'(t) x3=x''(t)
    and because my 2 equations contain dy/dx and d2y/dx2 terms I assume I have to introduce y1=y(t) y2=y'(t) y3=y''(t). But I dont understand where I go from here.

    Silent
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  2. #2
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    Quote Originally Posted by SilenceInShadows View Post
    I missed the lecture and was wondering if somebody could clue me in on the method. I as far as I can tell (and to prove that I am trying over here ) We introduce
    x1=x(t) x2=x'(t) x3=x''(t)
    and because my 2 equations contain dy/dx and d2y/dx2 terms I assume I have to introduce y1=y(t) y2=y'(t) y3=y''(t). But I dont understand where I go from here.

    Silent
    First you are going to have to write out the question clearly. You say first
    " We introduce x1=x(t) x2=x'(t) x3=x''(t) "
    which has x as dependent variable and t as independent variable, and then
    "because my 2 equations contain dy/dx and d2y/dx2 terms I assume I have to introduce y1=y(t) y2=y'(t) y3=y''(t)" where your equation seems to have y as dependent variable and x as independent variable but then you have switched to "t" as independent variable in the second line.

    What is the specific problem you are trying to solve?
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  3. #3
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    sorry, that was quite clumsy you are quite right though. I should have said d2x/dt2 and d2y/dt2.

    The problem is:

    5x''(t)+ 3(t)^0.5y'(t)-7x-8y=sinht
    tan(t) y''(t)+7x'(t)+9ln(t)x=cosht

    But this is just an example I picked off a work sheet, im more directly after the method.

    Silent
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  4. #4
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    You have:

    5x''+3t^{1/2}y'-7x+8y=\sinh(t)

    \tan(t)y''+7x'+9\ln(t)x=\cosh(t)

    I can then let x'=u and y'=v and arrive at the following first-order system of four equations:

    x'=u

    5u'+3t^{1/2}v-7x+8y=\sinh(t)

    y'=v

    \tan(t)v'+7x'+9\ln(t)x=\cosh(t)
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  5. #5
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    I was expecting to arrive at a matrix M(t) and an inhomogenous term H(t)
    can this be drawn from your equations?
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  6. #6
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    Yes: \textbf{X}^{'}=\textbf{M}(t)\textbf{X}+\textbf{H}(  t)

    Line them all up in the form of x, y, u, v ok. So the first row in the matrix corresponding to the first equation would be (0,0,1,0) right? The second row would be (0,0,0,1). You can do the rest.
    Last edited by shawsend; November 23rd 2008 at 06:40 AM. Reason: corrected second row if in form x y u v
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  7. #7
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    Thanks your a star

    Silent
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  8. #8
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    \left[\begin{array}{c} x \\ y \\ u \\ v\end{array}\right]^{'} =\left[\begin{array}{cccc}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 7/5 & -8/5 & 0 & -3/5 t^{1/2} \\-\frac{9\ln(t)}{\tan(t)} & 0 & -\frac{7}{\tan(t)} & 0\end{array}\right] \left[\begin{array}{c} x\\  y \\ u \\ v\end{array}\right] =\left[\begin{array}{c}0 \\ 0 \\ \frac{\sinh(t)}{5} \\ \frac{\cosh(t)}{\tan(t)}\end{array}\right]

    Just curious if I could get all the latex in there (may need to be checked though)
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