#### 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__

__Examples__

- 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)
- Write down a differential equation with the given info
^{dA}/_{dt}∝ √(2A – 5)^{dA}/_{dt}= k√(2A – 5)

- 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: dA
^{1}/_{√(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

- 10k + 3 = (2(27) – 5)
- 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

- 0.4(20) + 3 = (2A – 5)

- Using

- Write down a differential equation with the given info
- 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- 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}) - e
^{kt}=^{80}/_{80 – kV} - 80 – kV =
^{80}/_{ekt} - kV = 80 – 80e
^{-kt} - ∴ V =
^{1}/_{k}(80 – 80e^{-kt})

- t –

- Represent the given information as a derivative:
- 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

- 500 =
- 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

- ∴ V =

- Using k above: V =

- Write down a differential equation describing this situation and solve it to show that V =
- 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.- Show that h and t satisfy a differential equation of the form
^{dh}/_{dt}= -Ah^{–3/2}, 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 = -kh^{½}

- V =
- Find the rate of change of h:
^{dh}/_{dt}=^{dV}/_{dt}÷^{dV}/_{dh}^{dh}/_{dt}= –^{kh½}/_{3πh²}= –^{k}/_{3π}h^{-3/2}

- Represent the information given as an equation:
- Solve the differential equation given in (a) and obtain an expression for t in terms of h and H
- dt =
^{dh}/_{-Ah–3/2} - ∫Adt = ∫
^{dh}/_{-h-3/2} - At + c = –
^{2}/_{5}h^{5/2} - Use the information given to find the unknowns
- when t = 0:
- -A(0) + c =
^{2}/_{5}(H)^{5/2} - ∴ c =
^{2}/_{5}H^{5/2}

- -A(0) + c =
- when t = 60:
- -A(60) + c = 0
- c = 60A
- ∴ A =
^{1}/_{150}H^{5/2}

- when t = 0:
- Thus the initial equation becomes:
- –
^{1}/_{150}H^{5/2}+^{2}/_{5}H^{5/2}=^{2}/_{5}h^{5/2} - H
^{5/2}(-^{t}/_{150}+^{2}/_{5}) =^{2}/_{5}h^{5/2} - –
^{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 – 60h
^{5/2}5H^{-5/2} - t = 60 (1 – (
^{h}/_{H})^{5/2})

- –

- dt =

- Show that h and t satisfy a differential equation of the form