4 Ijesha Close, Ilupeju, Lagos
+2347 086 296 002

Differential Equations

  • The type of differential we solve are linear, second order, constant coefficient differential equations.
  • There are two types:
    • Homogeneous
    • Non-homogeneous
Homogeneous Equations
  • This type of differential equations will be of the form:
    • ad²y/dx² + bdy/dx + cy = 0
  • First step to solving them is to find the complementary function by forming an auxiliary equation:
    • aλ² + bλ + c = 0
  • The complementary function formed depends on the discriminant of the auxiliary function
  • Solve the quadratic equation and use the following cases to work out the complementary function
  • The complementary function is also the general solution and we need to substitute values,  find A and B to get the particular solution
Non-homogeneous Equations
  • This type of differential equations will be of the form:
    • ad²y/dx² + bdy/dx + cy = f(x)
  • We find a complementary function in the same way as above, by making f(x) = 0
  • After finding a complementary function, we have to add a particular integral
  • The particular integral we add is dependent upon f(x) and the trial solution used:
  • Once we have the trial solution, we equate it to y and differentiate it to find y’ and differentiate again to find y” in terms of P, Q , and R
  • Then substitute these into the initial differential equation and solve simultaneously for P, Q, and R
  • The trial solution with substituted values of P, Q, and R becomes the particular integral
  • Addition of complementary function and particular integral gives use the particular solution
Method Summary

Examples
  1. The value of the assets of a large commercial organization at time t, measured in years, is $(108y + 109). The variables y and t are related by the differential equation:
    d²y/dt² + 5dy/dt + 6y = 15 cos 3t – 3 sin 3t
    Find y in terms of t, given that y = 3 and dy/dt = 2 when t = 0

    • Form the auxiliary equation by assuming f(x) = 0:
      • d²y/dt² + 5dy/dt + 6y = 0
      • ∴ λ² + 5λ + 6 = 0
    • Solve the quadratic equation:
      • (λ + 3)(λ + 2) = 0
      • ∴ λ = 3 or 2
    • The answers are two real solutions, so the complementary function is the following:
      • y = Ae-3t + Be-2t
    • Now we have to find the particular integral using a trial solution:
      • f(x) = 15 cost 3t – 3 sin 3t
    • Thus, our trial solution and its derivatives are:
      • y = P cos 3t + Q sin 3t
      • y’ = -3P sin 3t + 3Q cos 3t
      • y” = -9P cos 3t – 9Q sin 3t
    • Substitute into the original equation and simplify to get:
      • (-3P + 15Q) cos 3t + (-3Q + 15P) sin 3t
    • Take the coefficient from both sides to form the simultaneous equations:
      • -3P + 15Q = 15
      • -3Q + 15P = -3
    • Solve these to get P and Q
      • P = 0 ad Q = 1
    • Thus, the particular integral is y = sin 3t
    • Our general solution is the addition of the complementary function and the articular integral:
      • y = Ae-3t + Be-2t + sin 3t
    • First step to finding the particular solution is to differentiate the general solution:
      • dy/dt = -3Ae-3t – 2Be-2t + cos 3t
    • Substitute the given information into the equations above to form the simultaneous equations:
      • From y: 3 = A + B
      • From dy/dt: 3A + 2B = -1
    • Solve the simultaneous equations to find A and B
      • A = -7 and B = 10
    • Thus, our particular solution is:
      • y = -7e-3t + 10e-2t + sin 3t
Minimum 4 characters