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Thread: first order system

  1. #1
    Junior Member
    May 2010

    first order system

    dy/dt = (0 1)
    (-3 -4 )y y(0) = (-2)

    where y is a vector-valued function

    I showed that this is equal to y'' + 4y + 3y = 0 y(0) = 2 y'(0) = -4
    where y is a scalar function of t

    Then the question asked me to find the analytic solution which i found was

    y = exp(-3x) + exp(-x)

    The next part of the question says hence, or otherwise, find the solution for the first order system which is the first equation with the matrix I said before but I'm not sure what this is actually asking me to do if can anyone explain what it is asking?
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  2. #2
    MHF Contributor

    Apr 2005
    Most internet readers do not preserve spaces at the beginning of lines so matrices are very hard to read.

    It is much better to use LaTex:
    \frac{dY(t)}{dt}= \begin{pmatrix} 0 & 1 \\ - 3 & -4\end{pmatrix}Y(t)
    Y(0)= \begin{pmatrix}-2 \\ -4\end{pmatrix}

    Double click on that to see the code.

    Also I used the capital Y because it would be very bad notation to use the same symbol for both the vector function and the scalar function it is supposed to be "equal" to.

    If we set Y= \begin{pmatrix}y(t) \\ x(t)\end{pmatrix} then that matrix equation becomes
    \frac{d}{dt}\begin{pmatrix}y(t) \\ x(t)\end{pmatrix}= \begin{pmatrix} 0 & 1 \\ - 3 & -4\end{pmatrix}\begin{pmatrix}y \\ x \end{pmatrix} = \begin{pmatrix} x \\-3y- 4x\end{pmatrix}

    Which then reduces to the two scalar equations \frac{dy}{dt}= x and \frac{dx}{dt}= -3y- 4x with the intial conditions x(0)= -2, y(0)= -4.

    Differentiate the first equation twice to get \frac{d^2y}{dt^2}= \frac{dx}{dt} and from the second equation that is equal to -3y- 4x: \frac{d^2y}{dt^2}= -3y- 4x. From the first equation x= \frac{dy}{dt} so we can replace it by that: \frac{d^2y}{dt^2}= -3y- 4\frac{dy}{dt} or \frac{d^2y}{dt^2}+ 3y+ 4= 0. Since x(0)= -2 and x= y', y(0)= -4, y'(0)= -2.

    To find the solution to the intial problem means, of course, to find the vector function Y(t)= \begin{pmatrix}y(t) \\ x(t)\end{pmatrix}= \begin{pmatrix}y(t)\\ y'(t)\end{pmatrix} (since x= y').

    Since you have found that y(t)= e^{-3t}+ e^{-t} (NOT " e^{-3x}+ e^{-x}" because y is a function of t, not x!), x(t)= y'= -3e^{-3t}- e^{-t} and
    Y(t)= \begin{pmatrix}y(t)\\ x(t)\end{pmatrix}= \begin{pmatrix}e^{-3t}+ e^{-t} \\ -3e^{-3t}- e^{-t}\end{pmatrix}.
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