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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
• 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)