Mathematical methods for economic theory

Martin J. Osborne
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8.4 Exercises on linear first-order differential equations

  1. Find the general solution of x'(t) + (1/2)x(t) = 1/4. Determine the equilibrium state and examine its stability.

    Solution

    x(t) = Ce−t/2 + 1/2. Equilibrium: x* = 1/2; stable.

  2. Find the general solutions of the following differential equations, and in each case solve the associated initial value problem for x(0) = 1.
    1. x'(t) − 3x(t) = 5.
    2. 3x'(t) + 2x(t) + 16 = 0.
    3. x'(t) + 2x(t) = t2.

    Solution

    1. x(t) = Ce3t − 5/3; C = 8/3.
    2. x(t) = Ce−2t/3 − 8; C = 9.
    3. x = Ce−2t + (1/2)t2 − (1/2)t + 1/4 (integrate by parts twice); C = 3/4.

  3. Solve the following differential equations.
    1. tx'(t) + 2x(t) + t = 0 for t ≠ 0.
    2. x'(t) − x(t)/t = t for t > 0.
    3. x'(t) − tx(t)/(t2 − 1) = t for t > 1.
    4. x'(t) − 2x(t)/t + 2a2/t2 = 0 for t > 0.

    Solution

    1. Writing the equation in the standard form, we have
      x'(t) + (2/t)x(t) = −1.
      Now, ∫t(2/s)ds = 2 ln t, so from the general solution of an equation of the form x'(t) + a(t)x(t) = b(t) we have
      x(t) = e-2 ln t[C − ∫te2 ln udu]
      or
      x(t) = (1/t2)[C − ∫tu2du]
      or
      x(t) = (1/t2)[C − t3/3] = C/t2 − t/3.
    2. We have ∫(−1/t)dt = −ln t, so the integrating factor is 1/t. Hence the solution is x(t) = t(C + t).
    3. x(t) = C(t2−1)1/2 + t2 − 1.
    4. x(t) = Ct2 + 2a2/3t.

  4. In the theory of auctions, an initial value problem of the form
    x'(t)G(t) + x(t)G'(t) = tG'(t) with x(t0) = t0,
    where G is a known function, arises. (The variable t is interpreted as a player's valuation of the object for sale in the auction.) Solve this equation, expressing the solution in terms of the function G (but not its derivative). (If you start by converting the equation into the standard form and finding the integrating factor, reflect afterwards on the fact that you did not need to do so.)

    Solution

    The left-hand side of the differential equation is the derivative of x(t)G(t). Thus integrating both sides leads to
    x(t)G(t) = ∫tsG'(s)ds.
    Applying integration by parts to the right-hand side produces
    tG(t) − ∫tG(s)ds.
    Thus the general solution of the differential equation is
    x(t) = t − (∫tG(s)ds)/G(t).
    The condition x(t0) = t0 implies that the solution of the initial value problem is
    x(t) = t − (∫t
    t0     G(s)ds)/G(t).

  5. Solve the equation
    x'(t) − x(t)/(t2 − 1) = 1,
    restricting to t > 1. (This question is more difficult than the others.)

    Solution

    Let a(t) = −1/(t2 − 1) and b(t) = 1 for all t. Using the fact that
    1/(s2 − 1) = (1/2)[1/(s − 1) − 1/(s + 1)],
    we have
    ∫t a(s)ds = −∫t (1/(s2−1))ds = −(1/2)ln((t−1)/(t + 1))
    and hence
    e−∫ta(s)ds = ((t−1)/(t + 1))1/2.

    Also

    ∫te−∫ua(s)dsb(u)du  = ∫te−(1/2)ln((u−1)/(u+1))du
      = ∫t((u+1)/(u−1))1/2du
      = ∫t[(u+1)(u+1)/((u−1)(u+1))]1/2du
      = ∫t(u+1)/(u2−1)1/2du
      = ∫tu/(u2−1)1/2du + ∫t1/(u2−1)1/2du
      = (u2−1)1/2 + ln[u + (u2−1)1/2].
    (To calculate the second integral, use the substitution x = sec θ, so that dx = sec θ tan θ dθ and the integrand is sec θ, with integral ln|sec θ + tan θ| + C.)

    Thus the solution of the equation is

    x(t) = ((t−1)/(t+1))1/2[C + (t2 − 1)1/2 + ln(t + (t2 − 1)1/2)].