Mathematical methods for economic theory

Martin J. Osborne
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8.6 Exercises on second-order differential equations

  1. Find the general solutions of the following differential equations and the solutions of the associated initial value problems for the given initial conditions. In each case, determine whether the equilibrium is stable.
    1. x"(t) + 3x'(t) − 4x(t) = 12. Initial conditions: x(0) = 4, x'(0) = 2.
    2. x"(t) − 2x'(t) + x(t) = 3. Initial conditions: x(0) = 4, x'(0) = 2.
    3. x"(t) + 4x'(t) + 8x(t) = 2. Initial conditions: x(0) = 9/4, x'(0) = 4.

    Solution

    1. The roots of the characteristic equation are 1 and −4. Thus the general solution of the homogeneous equation is
      Aet + Be−4t.
      One solution of the original equation is x(t) = −3. Thus the general solution of the original equation is
      Aet + Be−4t − 3.
      For the given initial conditions we have A = 6 and B = 1. The equilibrium is not stable.
    2. x(t) = A1et + A2tet + 3; A1 = 1 and A2 = 1. The equilibrium is not stable.
    3. x(t) = e−2t(A1cos 2t + A2sin 2t) + 1/4; A1 = 2 and A2 = 4. The equilibrium is stable.

  2. Find a solution of the differential equation
    x"(t) + 2x'(t) + x(t) = t.

    Solution

    x(t) = t − 2.

  3. Find the general solution of the differential equation x"(t) − 4x'(t) + 4x(t) = 5 and the solution of the associated initial value problem for the initial conditions x(0) = 4 and x'(0) = 6.

    Solution

    The general solution is x(t) = A1e2t + A2te2t + 5/4. For the initial condition x(0) = 4 we need A1 + 5/4 = 4, or A1 = 11/4, and for the condition x'(0) = 6 we need A2 + 2A1 = 6, so that A2 = 6 − 22/4 = 1/2.

  4. Find the general solution of the differential equation
    x"(t) − x'(t) + 2x(t) = 2t3.

    Solution

    The characteristic equation of the homogeneous equation is r2 − r + 2 = 0, so the general solution of the homogeneous equation is
    Aet/2cos(√7t/2) + Bet/2sin(√7t/2).
    Now find a solution of the original equation. Try f(t) = C + Dt + Et2 + Ft3. We have f'(t) = D + 2Et + 3Ft2 and f"(t) = 2E + 6Ft, so for f to be a solution we need
    2F   =  2
    −3F + 2E   =  0
    6F − 2E + 2D   =  0
    2E − D + 2C   =  0.
    These equations yield F = 1, E = 3/2, D = −3/2, and C = −9/4. Thus the general solution of the original differential equation is
    Aet/2cos(√7t/2) + Bet/2sin(√7t/2) − 9/4 − 3t/2 + (3/2)t2 + t3.

  5. Find the general solution of the differential equation
    x"(t) − 4x'(t) + 4x(t) = −4t + 4t2.

    Solution

    The characteristic equation of the homogeneous equation is r2 − 4r + 4 = 0, for which 2 is a repeated root. Thus the general solution of the homogeneous equation is
    (A + Bt)e2t.
    Now find a solution of the original equation. Try f(t) = C + Dt + Et2. We have f'(t) = D + 2Et and f"(t) = 2E, so for f to be a solution we need
    4E   =  4
    −8E + 4D   =  −4
    2E − 4D + 4C   =  0.
    These equations yield C = 1/2 and D = E = 1. Thus the general solution of the original differential equation is
    (A + Bt)e2t + 1/2 + t + t2.

  6. Find the general solution of the differential equation
    x"(t) + x(t) = 1 + t2.

    Solution

    The characteristic equation of the homogeneous equation is r2 + 1 = 0, which has complex roots. Thus the general solution of the homogeneous equation is
    x(t) = Acos t + Bsin t.
    To find a solution of the original equation, try x(t) = a + bt + ct2. We have x"(t) = 2c, so that for x(t) to solve the original equation we need a = −1, b = 0, and c = 1. We conclude that the general solution of the original equation is
    x(t) = Acos t + Bsin t − 1 + t2.