Mathematical methods for economic theory

Martin J. Osborne
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8.7 Exercises on systems of first-order linear differential equations

  1. Find the general solution of the system of equations
    x'(t)  = 4y(t)
    y'(t)  = −x(t) + 4y(t).

    Solution

    Isolating y(t) in the first equation we have y(t) = x'(t)/4, so that y'(t) = x"(t)/4. Substituting these expressions into the second equation we get
    x"(t)/4 = −x(t) + x'(t),
    or
    x"(t) − 4x'(t) + 4x(t) = 0.
    The characteristic equation is (r − 2)2, which has the repeated root r = 2. Thus the general solution of this equation is
    x(t) = (A + Bt)e2t.
    Given y(t) = x'(t)/4, we thus have
    y(t) = (1/2)(A + Bt)e2t + (1/4)Be2t = [(1/4)B + (1/2)A + (1/2)Bt]e2t.

  2. Find the general solution of the following system of equations and the particular solution with x(0) = 1 and y(0) = 0.
    x'(t)  = x(t) − 5y(t)
    y'(t)  = 2x(t) − 5y(t).

    Solution

    The two equations generate the second-order equation
    x"(t) + 4x'(t) + 5x(t) = 0.
    The characteristic equation has complex roots. The solution of the second-order equation is
    x(t) = e−2t(A cos t + B sin t).
    Calculating x' and substituting into the first equation we obtain
    y(t) = (1/5)e−2t((3A − B)cos t + (A + 3B)sin t).
    For the given initial conditions we have A = 1 and B = 3.

  3. Find the solution of the system
    x'(t)  = −by(t)
    y'(t)  = bx(t)
    given the initial conditions x(0) = C1 and y(0) = C2.

    Solution

    We have x"(t) = −by'(t) = −b2x(t), or
    x"(t) + b2x(t) = 0.
    The characteristic equation is r2 + b2 = 0, which has complex roots. Thus, given the general form of the solution to such an equation, the solution of our equation is
    x(t) = C1cos bt − C2sin bt.
    Using y(t) = −x'(t)/b, we have y(t) = C1sin bt + C2cos bt. (Remember that the derivative of sin bt is bcos bt, and the derivative of cos bt is −bsin bt.)