Mathematical methods for economic theory

Martin J. Osborne
Previous section Previous page Contents Text Exercises Search Next page Next section

8.3 Exercises on separable first-order differential equations

  1. Solve the following differential equations.
    1. (x(t))2x'(t) = t + 1.

      Solution

      x(t) = ((3/2)t2 + 3t + 3C)1/3. (Separable.)

    2. x'(t) = t3 − t.

      Solution

      x(t) = t4/4 − t2/2 + C. (Direct integration.)

    3. x'(t) = tet − t.

      Solution

      x(t) = tet − et − (1/2)t2 + C. (Direct integration.)

    4. ex(t)x'(t) = t + 1.

      Solution

      x(t) = ln((1/2)t2 + t + C). (Separable.)

    5. x'(t) = e−x(t)/tx(t).

      Solution

      The equation is separable:
      ∫xexdx = ∫(1/t)dt,
      so (integrating by parts on the left) xex − ex = ln t + C. Thus the solution is defined by the condition
      (x(t) − 1)ex(t) = ln t + C.

    6. x'(t) = 4tx(t) + t.

      Solution

      The equation is separable:
      ∫(1/(4x + 1))dx = ∫t dt,
      so that
      (1/4)ln|4x + 1| = (1/2)t2 + C,
      or
      x(t) = Cexp(2t2) − 1/4.

  2. Solve the following initial value problems.
    1. tx'(t) = x(t)(1 − t), (t, x) = (1, 1/e).

      Solution

      x(t) = Cte−t; C = 1.

    2. (1 + t3)x'(t) = t2x(t), (t, x) = (0, 2).

      Solution

      x(t) = C(1 + t3)1/3; C = 2.

    3. x(t)x'(t) = t, (t, x) = (√2, 1).

      Solution

      x(t) = √(t2 + C); C = −1.

    4. e2tx'(t) − (x(t))2 − 2x(t) − 1 = 0, (t, x) = (0, 0).

      Solution

      x(t) = (2 − C − e−2t)/(C + e−2t); C = 1.

    5. x'(t) = 4tx(t) + t, (t, x) = (0, 0).

      Solution

      x(t) = Ce2t2 − 1/4; C = 1/4.