4.10h Coupled systems: simultaneous first order DEs

8 A biologist is studying the effect of pesticides on crops. On a certain farm pesticide is regularly applied to a particular crop which grows in soil. Over time, pesticide is transferred between the crop and the soil at a rate which depends on the amount of pesticide in both the crop and the soil. The amount of pesticide in the crop after \(t\) days is \(x\) grams. The amount of pesticide in the soil after \(t\) days is \(y\) grams. Initially, when \(t = 0\), there is no pesticide in either the crop or the soil. At first it is assumed that no pesticide is lost from the system. The biologist further assumes that pesticide is added to the crop at a constant rate of \(k\) grams per day, where \(k > 6\). After collecting some initial data, the biologist suggests that for \(t \geqslant 0\), this situation can be modelled by the following pair of first order linear differential equations. \(\frac { d x } { d t } = - 2 x + 78 y + k\) \(\frac { d y } { d t } = 2 x - 78 y\)

1. 1. Show that \(\frac { d ^ { 2 } x } { d t ^ { 2 } } + 80 \frac { d x } { d t } = 78 \mathrm { k }\).
   2. Determine the particular solution for \(x\) in terms of \(k\) and \(t\). If more than 250 grams of pesticide is found in the crop, then it will fail food safety standards.
   3. The crop is tested 50 days after the pesticide is first added to it. Explain why, according to this model, the crop will fail food safety standards as a result of this test. Further data collection suggests that some pesticide decays in the soil and so is lost from the system. The model is refined in light of this data. The particular solution for \(x\) for this refined model is \(\mathrm { x } = \mathrm { k } \left( 20 - \mathrm { e } ^ { - 41 \mathrm { t } } \left( 20 \cosh ( \sqrt { 1677 } \mathrm { t } ) + \frac { 819 } { \sqrt { 1677 } } \sinh ( \sqrt { 1677 } \mathrm { t } ) \right) \right.\).
2. Given now that \(k < 12\), determine whether the crop will fail food safety standards in the long run according to this refined model. In the refined model, it is still assumed that pesticide is added to the crop at a constant rate.
3. Suggest a reason why it might be more realistic to model the addition of pesticide as not being at a constant rate.

17 Two similar species, X and Y , of a small mammal compete for food and habitat. A model of this competition assumes, in a particular area, the following.

• In the absence of the other species, each species would increase at a rate proportional to the number present with the same constant of proportionality in each case.
• The competition reduces the rate of increase of each species by an amount proportional to the number of the other species present.

So if the numbers of species X and Y present at time \(t\) years are \(x\) and \(y\) respectively, the model gives the differential equations \(\frac { d x } { d t } = k x - a y\) and \(\frac { d y } { d t } = k y - b x\),
where \(k , a\) and \(b\) are positive constants.

1. 1. Show that the general solution for \(x\) is \(x = A e ^ { ( k + n ) t } + B e ^ { ( k - n ) t }\), where \(n = \sqrt { a b }\) and \(A\) and \(B\) are arbitrary constants.
   2. Hence find the general solution for \(y\) in terms of \(A , B , k , n , a\) and \(t\). Observations suggest that suitable values for the model are \(k = 0.015 , a = 0.04\) and \(b = 0.01\). You should use these values in the rest of this question.
2. When \(t = 0\), the numbers present of species X and Y in this area are \(x _ { 0 }\) and \(y _ { 0 }\) respectively.
   1. Show that \(\mathrm { x } = \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } - 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { 0.035 \mathrm { t } } + \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } + 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { - 0.005 \mathrm { t } }\).
   2. Hence show that \(y = \frac { 1 } { 4 } \left( x _ { 0 } + 2 y _ { 0 } \right) e ^ { - 0.005 t } - \frac { 1 } { 4 } \left( x _ { 0 } - 2 y _ { 0 } \right) e ^ { 0.035 t }\).
3. Use initial values \(x _ { 0 } = 500\) and \(y _ { 0 } = 300\) with the results in part (b) to determine what the model predicts for each of the following questions.
   1. What numbers of each species will be present after 25 years?
   2. In this question you must show detailed reasoning. When will the numbers of the two species be equal?
   3. Does either species ever disappear from the area? Justify your answer.
4. Different initial values will apply in other areas where the two species compete, but previous studies indicate that one species or the other will eventually dominate in any given area.
   1. Identify a relationship between \(x _ { 0 }\) and \(y _ { 0 }\) where the model does not predict this outcome.
   2. Explain what the model predicts in the long term for this exceptional case.

14 Solve the simultaneous differential equations \(\frac { \mathrm { d } x } { \mathrm {~d} t } + 2 x = 4 y , \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } + 3 x = 5 y\),
given that when \(t = 0 , x = 0\) and \(y = 1\).

1. A scientist is studying the effect of introducing a population of white-clawed crayfish into a population of signal crayfish.
   At time \(t\) years, the number of white-clawed crayfish, \(w\), and the number of signal crayfish, \(s\), are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } w } { \mathrm {~d} t } = \frac { 5 } { 2 } ( w - s ) \\ & \frac { \mathrm { d } s } { \mathrm {~d} t } = \frac { 2 } { 5 } w - 90 \mathrm { e } ^ { - t } \end{aligned}$$

1. Show that $$2 \frac { \mathrm {~d} ^ { 2 } w } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} w } { \mathrm {~d} t } + 2 w = 450 \mathrm { e } ^ { - t }$$
2. Find a general solution for the number of white-clawed crayfish at time \(t\) years.
3. Find a general solution for the number of signal crayfish at time \(t\) years. The model predicts that, at time \(T\) years, the population of white-clawed crayfish will have died out. Given that \(w = 65\) and \(s = 85\) when \(t = 0\)
4. find the value of \(T\), giving your answer to 3 decimal places.
5. Suggest a limitation of the model.

1. Two compounds, \(X\) and \(Y\), are involved in a chemical reaction. The amounts in grams of these compounds, \(t\) minutes after the reaction starts, are \(x\) and \(y\) respectively and are modelled by the differential equations

$$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = - 5 x + 10 y - 30 \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = - 2 x + 3 y - 4 \end{aligned}$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 50$$
2. Find, according to the model, a general solution for the amount in grams of compound \(X\) present at time \(t\) minutes.
3. Find, according to the model, a general solution for the amount in grams of compound \(Y\) present at time \(t\) minutes. Given that \(x = 2\) and \(y = 5\) when \(t = 0\)
4. find
   1. the particular solution for \(x\),
   2. the particular solution for \(y\). A scientist thinks that the chemical reaction will have stopped after 8 minutes.
5. Explain whether this is supported by the model.

1. A scientist is studying the effect of introducing a population of type \(A\) bacteria into a population of type \(B\) bacteria.

At time \(t\) days, the number of type \(A\) bacteria, \(x\), and the number of type \(B\) bacteria, \(y\), are modelled by the differential equations $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = x + y \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 3 y - 2 x \end{aligned}$$

1. Show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 4 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 0$$
2. Determine a general solution for the number of type \(A\) bacteria at time \(t\) days.
3. Determine a general solution for the number of type \(B\) bacteria at time \(t\) days. The model predicts that, at time \(T\) hours, the number of bacteria in the two populations will be equal. Given that \(x = 100\) and \(y = 275\) when \(t = 0\)
4. determine the value of \(T\), giving your answer to 2 decimal places.
5. Suggest a limitation of the model.

1. A patient is treated by administering an antibiotic intravenously at a constant rate for some time.

Initially there is none of the antibiotic in the patient.
At time \(t\) minutes after treatment began

• the concentration of the antibiotic in the blood of the patient is \(x \mathrm { mg } / \mathrm { ml }\)
• the concentration of the antibiotic in the tissue of the patient is \(y \mathrm { mg } / \mathrm { ml }\)

The concentration of antibiotic in the patient is modelled by the equations $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.025 y - 0.045 x + 2 \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.032 x - 0.025 y \end{aligned}$$

1. Show that $$40000 \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} t ^ { 2 } } + 2800 \frac { \mathrm {~d} y } { \mathrm {~d} t } + 13 y = 2560$$
2. Determine, according to the model, a general solution for the concentration of the antibiotic in the patient's tissue at time \(t\) minutes after treatment began.
3. Hence determine a particular solution for the concentration of the antibiotic in the tissue at time \(t\) minutes after treatment began. To be effective for the patient the concentration of antibiotic in the tissue must eventually reach a level between \(185 \mathrm { mg } / \mathrm { ml }\) and \(200 \mathrm { mg } / \mathrm { ml }\).
4. Determine whether the rate of administration of the antibiotic is effective for the patient, giving a reason for your answer.

1. At the start of the year 2000, a survey began of the number of foxes and rabbits on an island.

At time \(t\) years after the survey began, the number of foxes, \(f\), and the number of rabbits, \(r\), on the island are modelled by the differential equations $$\begin{aligned} & \frac { \mathrm { d } f } { \mathrm {~d} t } = 0.2 f + 0.1 r \\ & \frac { \mathrm {~d} r } { \mathrm {~d} t } = - 0.2 f + 0.4 r \end{aligned}$$

1. Show that \(\frac { \mathrm { d } ^ { 2 } f } { \mathrm {~d} t ^ { 2 } } - 0.6 \frac { \mathrm {~d} f } { \mathrm {~d} t } + 0.1 f = 0\)
2. Find a general solution for the number of foxes on the island at time \(t\) years.
3. Hence find a general solution for the number of rabbits on the island at time \(t\) years. At the start of the year 2000 there were 6 foxes and 20 rabbits on the island.
4. 1. According to this model, in which year are the rabbits predicted to die out?
   2. According to this model, how many foxes will be on the island when the rabbits die out?
   3. Use your answers to parts (i) and (ii) to comment on the model.

10 In a predator-prey environment the population, at time \(t\) years, of predators is \(x\) and prey is \(y\). The populations of predators and prey are measured in hundreds. The populations are modelled by the following simultaneous differential equations. $$\frac { \mathrm { d } x } { \mathrm {~d} t } = y \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } = 2 y - 5 x$$

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 5 x\).
2. 1. Find the general solution for \(x\).
   2. Find the equivalent general solution for \(y\). Initially there are 100 predators and 300 prey.
3. Find the particular solutions for \(x\) and \(y\).
4. Determine whether the model predicts that the predators will die out before the prey.

14 On an isolated island some rabbits have been accidently introduced. In order to eliminate them, conservationists have introduced some birds of prey.
At time \(t\) years \(( t \geq 0 )\) there are \(x\) rabbits and \(y\) birds of prey.
At time \(t = 0\) there are 1755 rabbits and 30 birds of prey.
When \(t > 0\) it is assumed that:

• the rabbits will reproduce at a rate of \(a \%\) per year
• each bird of prey will kill, on average, \(b\) rabbits per year
• the death rate of the birds of prey is \(c\) birds per year
• the number of birds of prey will increase at a rate of \(d \%\) of the rabbit population per year.

This system is represented by the coupled differential equations: $$\begin{aligned} & \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.4 x - 13 y \\ & \frac { \mathrm {~d} y } { \mathrm {~d} t } = 0.01 x - 1.95 \end{aligned}$$ 14

1. State the value of \(a\), the value of \(b\), the value of \(c\) and the value of \(d\) [0pt] [2 marks]
   14
2. Solve the coupled differential equations to find both \(x\) and \(y\) in terms of \(t\)

5 In a predator-prey environment the population, at time \(t\) years, of predators is \(x\) and prey is \(y\). The populations of predators and prey are measured in hundreds. The populations are modelled by the following simultaneous differential equations. \(\frac { \mathrm { d } x } { \mathrm {~d} t } = y \quad \frac { \mathrm {~d} y } { \mathrm {~d} t } = 2 y - 5 x\)

1. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } - 5 x\).
2. 1. Find the general solution for \(x\).
   2. Find the equivalent general solution for \(y\). Initially there are 100 predators and 300 prey.
3. Find the particular solutions for \(x\) and \(y\).
4. Determine whether the model predicts that the predators will die out before the prey.

An isolated island is populated by rabbits and foxes. At time \(t\) the number of rabbits is \(x\) and the number of foxes is \(y\). It is assumed that: • The number of foxes increases at a rate proportional to the number of rabbits. When there are 200 rabbits the number of foxes is increasing at a rate of 20 foxes per unit period of time. • If there were no foxes present, the number of rabbits would increase by 120% in a unit period of time. • When both foxes and rabbits are present the foxes kill rabbits at a rate that is equal to 110% of the current number of foxes. • At time \(t = 0\), the number of foxes is 20 and the number of rabbits is 80.

1. 1. Construct a mathematical model for the number of rabbits. [9 marks]
   2. Use this model to show that the number of rabbits has doubled after approximately 0.7 units of time. [1 mark]
2. Suggest one way in which the model that you have used for the number of rabbits could be refined. [1 mark]

\includegraphics{figure_15} Two tanks, A and B, each have a capacity of 800 litres. At time \(t = 0\) both tanks are full of pure water. When \(t > 0\), water flows in the following ways: • Water with a salt concentration of \(\mu\) grams per litre flows into tank A at a constant rate • Water flows from tank A to tank B at a rate of 16 litres per minute • Water flows from tank B to tank A at a rate of \(r\) litres per minute • Water flows out of tank B through a waste pipe • The amount of water in each tank remains at 800 litres. At time \(t\) minutes (\(t \geq 0\)) there are \(x\) grams of salt in tank A and \(y\) grams of salt in tank B. This system is represented by the coupled differential equations \begin{align} \frac{dx}{dt} &= 36 - 0.02x + 0.005y \tag{1}
\frac{dy}{dt} &= 0.02x - 0.02y \tag{2} \end{align}

1. Find the value of \(r\). [2 marks]
2. Show that \(\mu = 3\) [3 marks]
3. Solve the coupled differential equations to find both \(x\) and \(y\) in terms of \(t\). [9 marks]

A children's play centre has two rooms, a room full of bouncy castles and a room full of ball pits. At any given instant, each child in the centre is playing either on the bouncy castles or in the ball pits. Each child can see one room from the other room and can decide to change freely between the two rooms. It is assumed that such changes happen instantaneously. The number of children playing on the bouncy castles at time \(t\) hours, is denoted by \(C\) and the corresponding number of children playing in the ball pits is \(P\). Because the number of children is large for most of the time, \(C\) and \(P\) are modelled as being continuous. When there is a different number of children in each room, some children will move from the room with more children to the room with fewer children. A researcher therefore decides to model \(C\) and \(P\) with the following coupled differential equations. $$\frac{dP}{dt} = \alpha(P-C) + \gamma t$$ $$\frac{dC}{dt} = \alpha(C-P)$$

1. Explain why \(\alpha\) must be negative. [1]

After examining data, the researcher chooses \(\alpha = -2\) and \(\gamma = 32\).

1. Show that \(P\) satisfies the second order differential equation \(\frac{d^2P}{dt^2} + 4\frac{dP}{dt} = 64t + 32\). [2]
2. 1. Find the complementary function for the differential equation from part (b). [1]
   2. Explain why a particular integral of the form \(P = at + b\) will not work in this situation. [1]
   3. Using a particular integral of the form \(P = at^2 + bt\), find the general solution of the differential equation from part (b). [3]

At a certain time there are 55 children playing in the ball pits and 24 children per hour are arriving at the ball pits.

1. Use the model, starting from this time, to estimate the number of children in the ball pits 30 minutes later. [4]
2. Explain why the model becomes unreliable as \(t\) gets very large. [1]

During an industrial process substance \(X\) is converted into substance \(Z\). Some of the substance \(X\) goes through an intermediate phase, and is converted to substance \(Y\), before being converted to substance \(Z\). The situation is modelled by $$\frac{dy}{dt} = 0.3x + 0.2y \text{ and } \frac{dz}{dt} = 0.2y + 0.1x$$ where \(x\), \(y\) and \(z\) are the amounts in kg of \(X\), \(Y\) and \(Z\) at time \(t\) hours after the process starts. Initially there is 10 kg of substance \(X\) and nothing of substances \(Y\) and \(Z\). The amount of substance \(X\) decreases exponentially. The initial rate of decrease is 4 kg per hour.

1. Show that \(x = Ae^{-0.4t}\), stating the value of \(A\). [3]
2. 1. Show that \(\frac{dx}{dt} + \frac{dy}{dt} + \frac{dz}{dt} = 0\). [2]
   2. Comment on this result in the context of the industrial process. [2]
3. Express \(y\) in terms of \(t\). [5]
4. Determine the maximum amount of substance \(Y\) present during the process. [3]
5. How long does it take to produce 9 kg of substance \(Z\)? [2]

Two species of insects, \(X\) and \(Y\), co-exist on an island. The populations of the species at time \(t\) years are \(x\) and \(y\) respectively, where \(x\) and \(y\) are measured in millions. The situation can be modelled by the differential equations $$\frac{\mathrm{d}x}{\mathrm{d}t} = 3x + 10y,$$ $$\frac{\mathrm{d}y}{\mathrm{d}t} = x + 6y.$$

1. 1. Show that \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} - 9\frac{\mathrm{d}x}{\mathrm{d}t} + 8x = 0\).
   2. Find the general solution for \(x\) in terms of \(t\). [7]
2. Find the corresponding general solution for \(y\). [4]
3. When \(t = 0\), \(\frac{\mathrm{d}x}{\mathrm{d}t} = 5\) and the population of species \(Y\) is 4 times the population of species \(X\). Find the particular solution for \(x\) in terms of \(t\). [6]

The following simultaneous equations are to be solved. $$\frac{\mathrm{d}x}{\mathrm{d}t} = 4x + 2y + 6e^{3t}$$ $$\frac{\mathrm{d}y}{\mathrm{d}t} = 6x + 8y + 15e^{3t}$$

1. Show that \(\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} - 12\frac{\mathrm{d}x}{\mathrm{d}t} + 20x = 0\). [5]
2. Given that \(\frac{\mathrm{d}x}{\mathrm{d}t} = 9\) and \(\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} = 10\) when \(t = 0\), find the particular solution for \(x\) in terms of \(t\). [7]

During an industrial process substance \(X\) is converted into substance \(Z\). Some of the substance \(X\) goes through an intermediate phase, and is converted to substance \(Y\), before being converted to substance \(Z\). The situation is modelled by $$\frac{dy}{dt} = 0.3x - 0.2y \quad \text{and} \quad \frac{dz}{dt} = 0.2y + 0.1x$$ where \(x\), \(y\) and \(z\) are the amounts in kg of \(X\), \(Y\) and \(Z\) at time \(t\) hours after the process starts. Initially there is 10 kg of substance \(X\) and nothing of substance \(Y\) and \(Z\). The amount of substance \(X\) decreases exponentially. The initial rate of decrease is 4 kg per hour.

1. Show that \(x = Ae^{-0.4t}\), stating the value of \(A\). [3]
2. Show that \(\frac{dx}{dt} + \frac{dy}{dt} + \frac{dz}{dt} = 0\). Comment on this result in the context of the industrial process. [4]
3. Express \(y\) in terms of \(t\). [5]

The quantity of grass on an island at time \(t\) years is \(x\), in appropriate units. At time \(t = 0\) some rabbits are introduced to the island. The population of rabbits on the island at time \(t\) years is \(y\), in units of \(100\)s of rabbits. An ecologist who is studying the island suggests that the following pair of simultaneous first order differential equations can be used to model the population of rabbits and quantity of grass for \(t \geq 0\). $$\frac{dx}{dt} = 3x - 2y,$$ $$\frac{dy}{dt} = y + 5x$$

1. 1. Show that \(\frac{d^2x}{dt^2} = a\frac{dx}{dt} + bx\) where \(a\) and \(b\) are constants which should be found. [2]
   2. Find the general solution for \(x\) in real form. [3]
2. Find the corresponding general solution for \(y\). [3]

At time \(t = 0\) the quantity of grass on the island was \(4\) units. The number of rabbits introduced at this time was \(500\).

1. Find the particular solutions for \(x\) and \(y\). [5]
2. The ecologist finds that the model predicts that there will be no grass at time \(T\), when there are still rabbits on the island. Find the value of \(T\). [1]
3. State one way in which the model is not appropriate for modelling the quantity of grass and the population of rabbits for \(0 \leq t \leq T\). [1]