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Math Help - using matrix-differential eqns-help exam quite soon

  1. #1
    n22
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    using matrix-differential eqns-help exam quite soon

    Hello,
    I am really not sure about this one .Please help.thanks
    find the particular solution to the system of differential eqns.
    dx/dt=x+3z
    dy/dt=y-5z
    dz/dt=2z
    given the initial condition x(o)=y(0)=z(0)=1
    once I chuck the coefficients of the differential eqns into the matrix im not sure what to do?
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  2. #2
    MHF Contributor

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    Re: using matrix-differential eqns-help exam quite soon

    I presume you have written that as a matrix differential equation as you title this "using matrix differential equations":
    \frac{d\begin{bmatrix}x \\ y \\ z \end{bmatrix}}{dt}=  \begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -5\\ 0 & 0 & 2\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}

    Surely you have some instruction in this?

    Start by finding the eigenvalues and eigenvectors of the matrix. Then B^{-1}AB= D where A is the matrix, B is a matrix having the eigenvectors of A as its columns, and D is a diagonal matrix having the eigenvalues of A on its diagonal. That is dX/dt= AX, which, after multiplying by B^{-1} gives d(A^{-1}X)/dt= B^{-1}AX= B^{-1}ABB^{-1}X. Let Y= A^{-1}X and the equation becomes dY/dt= DY where D is that diagonal matrix. Because D is diagonal, that is easy to solve and, once you have Y, X= AY.

    HOWEVER, in this particular problem, there is no point in doing that or even writing this in terms of a matrix. The third equation is dz/dt= 2z which is easily solvable: z= Ce^t and z(0)= C= 1. Then dy/dt= y- 5z= y- 5 which has general solution y= Ce^t+ 5 and y(0)= C+ 5= 1 so C= -4. Finally, dx/dt= x+ 3z= x+ 3 has general solution z= Ce^t- 3 and z(0)= C- 3= 1 so C= 4.

    (Okay, not all matrices are diagonalizable, and this particular one is not.)
    Last edited by HallsofIvy; June 12th 2013 at 05:04 PM.
    Thanks from n22
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