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Differential Equations

  • Form a differential equation from the information given:
    • If something is proportional, add a constant of proportionality, k
    • If rate is decreasing, add a negative sign
  • Separate variables, bring dx and dt on opposite sides
  • Integrate both sides to form an equation
  • Use the conditions given to find c and/or k
Examples
  1. A biologist is investigating the spread of a weed in a particular region. At time t weeks, the area covered by the weed is A m². The biologist claims that the rate of increase of A is proportional to √(2A – 5)
    1. Write down a differential equation with the given info
      • dA/dt ∝ √(2A – 5)
      • dA/dt = k√(2A – 5)
    2. At the start of the investigation, the area covered by the weed was 7 m². 10 weeks later, the area covered was 27 m². Find the are covered 20 weeks after the start of the investigation.
      • Using dA/dt above
      • Separate variables: dA1/√(2A – 5) = kdt
      • Integrate both sides: kt + c = (2A – 5)½
      • When t = 0:
        • A = 7
        • ∴ c = 3
        • ∴ kt + 3 = (2A – 5)½
      • When t = 10:
        • 10k + 3 = (2(27) – 5)½
        • 10k = √(49) – 3
        • ∴ k = 0.4
      • Therefore the differential equation is 0.4t + 3 = (2A – 5)½
      • Now substitute 20 as t and then find A:
        • 0.4(20) + 3 = (2A – 5)½
        • 11 = (2A – 5)½
        • 121 = 2A – 5
        • ∴ A = 63 m²
  2. Liquid is flowing into a small tank which has a leak. Initially the tank empty, and t minutes later, the volume of the liquid in the tank is V cm³. The liquid is flowing into the tank at  a constant rate of 80 cm³ per minute.
    Because of the leak, liquid is being lost from the tank at a rate which, at any instant, is equal to kV cm³ per minute, where k is a positive constant

    1. Write down a differential equation describing this situation and solve it to show that V = 1/k (80 – 80e-kt)
      • Represent the given information as a derivative: dV/dt = 80 – kV
      • Proceed to solve the differential equation:
        • dt/dV = 1/80 – kV
        • dt = dv/80 – kV
        • ∫dt = ∫dV/80 – kV
        • t + c = – 1/k ln|80 – kV|
      • Use the given information; when t = 0, V = 0: ∴c = 1/k ln(80)
      • Substitute c into the equation:
        • t – 1/k ln(80) = – 1/k ln|80 – kV|
        • t = 1/k ln(80) – 1/k ln|80 – kV|
        • t = 1/k ln(80/80 – kV)
        • kt = ln(80/80 – kV)
        • ekt = 80/80 – kV
        • 80 – kV = 80/ekt
        • kV = 80 – 80e-kt
        • ∴ V = 1/k (80 – 80e-kt)
    2. V = 500 when t = 15, show k = 4 – 4e-15k/25. Find k using iterations, initially k = 0.1
      • 500 = 1/k (80 – 80e-15k)
      • k = (80 – 80e-kt)/500
      • k = 4 – 4e-15k/25
      • Doing the mishwaar iterations gives k = 0.14
    3. Work out the volume of the liquid at t = 20 and state what happens to the volume after a long time
      • Using k above: V = 1/0.14 (80 – 80e-15k)
      • Substitute t = 20 in V above: ∴ V = 537 cm³
      • The volume of the liquid in the tank after a long time approaches the maximum volume:
        • ∴ V = 1/0.14 (80)
        • ∴ V = 571 cm³
  3. A tank containing water is in the form of a cone with vertex C. The axis is vertical and the semi-vertical angle is 60°, as shown in the diagram below.

    At time t = 0, the tank is full and the depth of the water is H. At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional √h, where h is the depth of the water at time t. The tank becomes empty when t = 60.

    1. Show that h and t satisfy a differential equation of the form dh/dt = -Ah3/2, where A is a positive constant
      • Represent the information given as an equation:
        • V = 1/3 πr²h
        • r = tan 60 × h = h√3
        • ∴ V = 1/3 π(h√3)²h = πh³
        • dV/dt ∝ -√h = -kh½
      • Find the rate of change of h:
        • dh/dt = dV/dt ÷ dV/dh
        • dh/dt = –kh½/3πh² = – k/ h-3/2
    2. Solve the differential equation given in (a) and obtain an expression for t in terms of h and H
      • dt = dh/-Ah3/2
      • ∫Adt = ∫dh/-h-3/2
      • At + c = –2/5 h5/2
      • Use the information given to find the unknowns
        • when t = 0:
          • -A(0) + c = 2/5 (H)5/2
          • ∴ c = 2/5 H5/2
        • when t = 60:
          • -A(60) + c = 0
          • c = 60A
          • ∴ A = 1/150 H5/2
      • Thus the initial equation becomes:
        • – 1/150 H5/22/5 H5/2 = 2/5 h5/2
        • H5/2 (-t/150 + 2/5) = 2/5 h5/2
        • t/150 + 2/5 = 2h5/2/5H5/2
        • t/150 = 2/5 – 2h5/2/5H5/2
        • t = 150(2/5 – 2h5/2/5H5/2)
        • t = 60 – 60h5/25H-5/2
        • t = 60 (1 – (h/H)5/2)
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