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Math Help - Solving a coupled pair of differential equations using matrices

  1. #1
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    Solving a coupled pair of differential equations using matrices

    I am asked to find the eigenvalues and eigenvectors of the matrix: A = \left(\begin{array}{cc}5&3\\1&7\end{array}\right)

    I find them as \lambda_1 = 8 , \lambda_2 = 4 and  v_1 = (1,1) ,  v_2 = (-3,1)

    The equations I need to solve are:  \frac{dx}{dt} = 5x +3 y and \frac{dy}{dt} = x + 7y. These equations are coupled (whatever that means...)

    I turn these into a pair of simultaneous equations sort of thing to form the matrix equation  X' = AX , where X' is \left(\begin{array}{cc}\frac{dx}{dt}\\\frac{dy}{dt  }\end{array}\right) and X = \left(\begin{array}{cc}x&\\y&\end{array}\right)

    As far as I know, I have diagonalise A which will give me  A = P^{-1}DP. I know  P = \left(\begin{array}{cc}1&-3\\1&1\end{array}\right) and  D = \left(\begin{array}{cc}8&0\\0&4\end{array}\right) using the eigenvalues.

    So I go along...

     X' = P^{-1}DPX

     PX' = DPX

    (PX)' = D(PX)

    Subbing matrices in...

    \left(\begin{array}{cc}1&-3\\1&1\end{array}\right) * \left(\begin{array}{cc}\frac{dx}{dt}\\\frac{dy}{dt  }\end{array}\right) = \left(\begin{array}{cc}8&0\\0&4\end{array}\right)*  \left(\begin{array}{cc}1&-3\\1&1\end{array}\right)*\left(\begin{array}{cc}x&  \\y&\end{array}\right)

    Can someone check if what I've done is correct so far? And if so, am I allowed to expend all the differentials and all the matrices then treat them as simultaneous equations. Also, I'm not too sure how to get rid of the dt when there is no  t variable in the equations...
    Last edited by jgv115; May 11th 2013 at 02:08 AM.
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  2. #2
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    Re: Solving a coupled pair of differential equations using matrices

    (The two equations are "coupled" means that each equation involve both of the functions.)

    Yes, what you have done so far is correct.

    The point is that letting Z= PX, you have
    \frac{dZ}{dt}= DZ
    \begin{pmatrix}\frac{dz_1}{dt} \\ \frac{dz_2}{dt}\end{pmatrix}= \begin{pmatrix}8 & 0 \\ 0 & 4 \end{pmatrix}\begin{pmatrix{z_1 \\ z_2 \end{pmatrix}
    \begin{pmatrix}\frac{dz_1}{dt} \\ \frac{dz_2}{dt}\end{pmatrix}= \begin{pmatrix}8z_1 \\ 4z_2\end{pmatrix}
    which is equivalent to the two "uncoupled" equations
    \frac{dz_1}{dt}= 8z_1 and \frac{dz_2}{dt}= 4z_2

    Those obviously have solutions z_1= C_1e^{8t} and z_2= C_2e^{4t} so that
    Z(t)= \begin{pmatrix}C_1e^{8t} \\ C_2e^{4t}\end{pmatrix}

    Since Z= PX, X= P^{-1}Z:
    \begin{pmatrix}x(t) \\ y(t)\end{pmatrix}= \begin{pmatrix}\frac{1}{4} & \frac{3}{4} \\ \frac{-1}{4} & \frac{1}{4} \end{pmatrix}\begin{pmatrix}C_1e^{8t} \\ C_2e^{4t}\end{pmatrix}
    \begin{pmatrix}x(t) \\ y(t) \end{pmatrix}= \begin{pmatrix}\frac{1}{4}C_1e^{8t}+ \frac{3}{4}C_2e^{4t} \\ -\frac{1}{4}C_1e^{8t}+ \frac{1}{4}C_2e^{4t}\end{pmatrix}
    Last edited by HallsofIvy; May 11th 2013 at 04:29 AM.
    Thanks from jgv115
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  3. #3
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    Re: Solving a coupled pair of differential equations using matrices

    I hope I don't break the rules by posting on this but I believe this question is valid.

    I know that it can be proven that the general solution to the D.E can be written as  x(t) = C_{1}e^{\lambda_{1}*t}v_{1} + C_{2}e^{\lambda_{2}*t}v_2. I just learnt this but I'm not too sure where it came from.

    So that makes x(t) = C_{1}e^{8x} - 3C_{2}e^{4x} and y(x) = C_{1}e^{8x} + C_{2}e^{4x} which is different to the solutions given by diagonalising matrix A. Am I misunderstanding something here?
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