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Simultaneous Differential Equations watch

    • Thread Starter

    Note that

    \frac{\mathrm{d}x}{dt} = 2x + 3y

    \frac{\mathrm{d}y}{dt} = x + 4y has the general solution

    \begin{pmatrix} x  \\y \end{pmatrix} = Ae^t\begin{pmatrix} -3  \\1 \end{pmatrix} + Be^{5t}\begin{pmatrix} 1  \\1 \end{pmatrix}

    Given x(0) = -2/3 and y(0) = 1/3, solve

    \frac{\mathrm{d}x}{dt} = 2x + 3y + 2e^{2t}

    \frac{\mathrm{d}y}{dt} = x + 4y + 3e^{2t}

    So I thought we just need to add a term to our previous solution such that it differentiates to gives 2e^(2t) for x and 3e^(2t) so we don't really need to worry about the other stuff. I did it in a rather hack way:

    \begin{pmatrix} x  \\y \end{pmatrix} = Ae^t\begin{pmatrix} -3  \\1 \end{pmatrix} + Be^{5t}\begin{pmatrix} 1  \\1 \end{pmatrix} + Ce^{2t}\begin{pmatrix} p  \\q \end{pmatrix}

    Differentiate and set equal to the dx/dt and dy/dt expressions above and you get

    C = -2/(3q) and

    pC + 2qC + 3 = 0

    Which, upon substitution, leads to

    2p = 5q

    So if we define p = 5, we get q = 2 and hence C = -1/3. Now if we use the initial conditions to find A and B we get A = 0 and B = 1. In short:

    x = e^{5t} - \frac{5}{3}e^{2t}

    y = e^{5t} - \frac{2}{3}e^{2t}

    (you can verify by differentiation that this does satisfy the above equations)

    Now my questions

    1) Is there any elegant way of getting values for C, p or q? My first instinct was to guess p = 3 and q = 2 but that was wrong. Was C = -2/(3q) predictable?

    2) Why are there an infinite number of solutions? I choose p = 5, but p = 10 would be perfectly fine also surely? What initial condition is missing?

    This C in your particular integral is not necessary.
    you should be able to work out why?
    This is why use have a degree of freedom in choices of C,p,q
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Updated: November 18, 2008

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