1.08k Separable differential equations: dy/dx = f(x)g(y)

An entomologist is studying the population of insects in a colony. Initially there are 300 insects in the colony and in a model, the entomologist assumes that the population, \(P\), at time \(t\) weeks satisfies the differential equation $$\frac{dP}{dt} = kP,$$ where \(k\) is a constant.

1. Find an expression for \(P\) in terms of \(k\) and \(t\). [5]

Given that after one week there are 360 insects in the colony,

1. find the value of \(k\) to 3 significant figures. [2]

Given also that after two and three weeks there are 440 and 600 insects respectively,

1. comment on suitability of the model. [2]

An alternative model assumes that $$\frac{dP}{dt} = P(0.4 - 0.25\cos 0.5t).$$

1. Using the initial data, \(P = 300\) when \(t = 0\), solve this differential equation. [4]
2. Compare the suitability of the two models. [3]

The gradient at any point \((x, y)\) on a curve is proportional to \(\sqrt{y}\). Given that the curve passes through the point with coordinates \((0, 4)\),

1. show that the equation of the curve can be written in the form $$2\sqrt{y} = kx + 4,$$ where \(k\) is a positive constant. [5]

Given also that the curve passes through the point with coordinates \((2, 9)\),

1. find the equation of the curve in the form \(y = \text{f}(x)\). [4]

\includegraphics{figure_2} Figure 2 shows a vertical cross-section of a vase. The inside of the vase is in the shape of a right-circular cone with the angle between the sides in the cross-section being \(60°\). When the depth of water in the vase is \(h\) cm, the volume of water in the vase is \(V\) cm\(^3\).

1. Show that \(V = \frac{1}{9}\pi h^3\). [3]

The vase is initially empty and water is poured in at a constant rate of 120 cm\(^3\) s\(^{-1}\).

1. Find, to 2 decimal places, the rate at which \(h\) is increasing
   1. when \(h = 6\),
   2. after water has been poured in for 8 seconds. [7]

The number of people, \(n\), in a queue at a Post Office \(t\) minutes after it opens is modelled by the differential equation $$\frac{dn}{dt} = e^{0.5t} - 5, \quad t \geq 0.$$

1. Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue. [3]
2. Given that there are 20 people in the queue when the Post Office opens, solve the differential equation. [4]
3. Explain why this model would not be appropriate for large values of \(t\). [1]

At time \(t = 0\), a tank of height 2 metres is completely filled with water. Water then leaks from a hole in the side of the tank such that the depth of water in the tank, \(y\) metres, after \(t\) hours satisfies the differential equation $$\frac{dy}{dt} = -ke^{-0.2t},$$ where \(k\) is a positive constant.

1. Find an expression for \(y\) in terms of \(k\) and \(t\). [4]

Given that two hours after being filled the depth of water in the tank is 1.6 metres,

1. find the value of \(k\) to 4 significant figures. [2]

Given also that the hole in the tank is \(h\) cm above the base of the tank,

1. show that \(h = 79\) to 2 significant figures. [3]

The rate of increase in the number of bacteria in a culture, \(N\), at time \(t\) hours is proportional to \(N\).

1. Write down a differential equation connecting \(N\) and \(t\). [1]

Given that initially there are \(N_0\) bacteria present in a culture,

1. Show that \(N = N_0 e^{kt}\), where \(k\) is a positive constant. [6]

Given also that the number of bacteria present doubles every six hours,

1. find the value of \(k\), [3]
2. Find how long it takes for the number of bacteria to increase by a factor of ten, giving your answer to the nearest minute. [2]

An entomologist is studying the population of insects in a colony. Initially there are 300 insects in the colony and in a model, the entomologist assumes that the population, \(P\), at time \(t\) weeks satisfies the differential equation $$\frac{dP}{dt} = kP,$$ where \(k\) is a constant.

1. Find an expression for \(P\) in terms of \(k\) and \(t\). [5]

Given that after one week there are 360 insects in the colony,

1. find the value of \(k\) to 3 significant figures. [2]

Given also that after two and three weeks there are 440 and 600 insects respectively,

1. comment on suitability of the modelling assumption. [2]

An alternative model assumes that $$\frac{dP}{dt} = P(0.4 - 0.25 \cos 0.5t).$$

1. Using the initial data, \(P = 300\) when \(t = 0\), solve this differential equation. [3]
2. Compare the suitability of the two models. [2]

In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dM}{dt} = \frac{M}{t(1+t^2)}.$$

1. Find \(\int \frac{t}{1+t^2} dt\). [3]
2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1+t^2)} = \frac{A}{t} + \frac{Bt+C}{1+t^2}.$$ [5]
3. Use integration, together with your results in parts (i) and (ii), to show that $$M = \frac{Kt}{\sqrt{1+t^2}},$$ where \(K\) is a constant. [6]
4. When \(t = 1\), \(M = 25\). Calculate \(K\). What is the mass of the chemical in the long term? [4]

The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h,$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]

The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).

1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}.$$ [9]
2. What does this solution indicate about the long-term height of the tree? [1]
3. After a year, the tree has grown to a height of 2 m. Which model fits this information better? [3]

\includegraphics{figure_3} Fig. 9 shows a hemispherical bowl, of radius 10 cm, filled with water to a depth of \(x\) cm. It can be shown that the volume of water, \(V\) cm\(^3\), is given by $$V = \pi(10x^2 - \frac{1}{3}x^3).$$ Water is poured into a leaking hemispherical bowl of radius 10 cm. Initially, the bowl is empty. After \(t\) seconds, the volume of water is changing at a rate, in cm\(^3\) s\(^{-1}\), given by the equation $$\frac{dV}{dt} = k(20 - x),$$ where \(k\) is a constant.

1. Find \(\frac{dV}{dx}\), and hence show that \(\pi x \frac{dx}{dt} = k\). [4]
2. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where \(T = \frac{50\pi}{k}\). [5]

Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(kx\) cm\(^3\) s\(^{-1}\).

1. Show that, \(t\) seconds later, \(\pi(20 - x) \frac{dx}{dt} = -k\). [3]
2. Solve this differential equation. Hence show that the bowl empties in 37 seconds. [6]

A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after \(t\) seconds it is \(v\) m s\(^{-1}\). Its terminal (long-term) velocity is 5 m s\(^{-1}\). A model of the particle's motion is proposed. In this model, \(v = 5(1 - e^{-2t})\).

1. Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model. [3]
2. Verify that \(v\) satisfies the differential equation \(\frac{dv}{dt} = 10 - 2v\). [3]

In a second model, \(v\) satisfies the differential equation $$\frac{dv}{dt} = 10 - 0.4v^2.$$ As before, when \(t = 0\), \(v = 0\).

1. Show that this differential equation may be written as $$\frac{10}{(5-v)(5+v)} \frac{dv}{dt} = 4.$$ Using partial fractions, solve this differential equation to show that $$t = \frac{1}{4} \ln\left(\frac{5+v}{5-v}\right).$$ [8] This can be re-arranged to give \(v = \frac{5(1-e^{-4t})}{1+e^{-4t}}\). [You are not required to show this result.]
2. Verify that this model also gives a terminal velocity of 5 m s\(^{-1}\). Calculate the velocity after 0.5 seconds as given by this model. [3]

The velocity of the particle after 0.5 seconds is measured as 3 m s\(^{-1}\).

1. Which of the two models fits the data better? [1]

Fig. 7 illustrates the growth of a population with time. The proportion of the ultimate (long term) population is denoted by \(x\), and the time in years by \(t\). When \(t = 0\), \(x = 0.5\), and as \(t\) increases, \(x\) approaches 1. \includegraphics{figure_7} One model for this situation is given by the differential equation $$\frac{dx}{dt} = x(1-x).$$

1. Verify that \(x = \frac{1}{1+e^{-t}}\) satisfies this differential equation, including the initial condition. [6]
2. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. [3]

An alternative model for this situation is given by the differential equation $$\frac{dx}{dt} = x^2(1-x),$$ with \(x = 0.5\) when \(t = 0\) as before.

1. Find constants \(A\), \(B\) and \(C\) such that \(\frac{1}{x^2(1-x)} = \frac{A}{x^2} + \frac{B}{x} + \frac{C}{1-x}\). [4]
2. Hence show that \(t = 2 + \ln\left(\frac{x}{1-x}\right) - \frac{1}{x}\). [5]
3. Find how long it will take, according to this model, for the population to reach three-quarters of its ultimate value. [2]

Scientists can estimate the time elapsed since an animal died by measuring its body temperature.

1. Assuming the temperature goes down at a constant rate of 1.5 degrees Fahrenheit per hour, estimate how long it will take for the temperature to drop
   1. from 98°F to 89°F,
   2. from 98°F to 80°F. [2]

In practice, rate of temperature loss is not likely to be constant. A better model is provided by Newton's law of cooling, which states that the temperature \(\theta\) in degrees Fahrenheit \(t\) hours after death is given by the differential equation $$\frac{d\theta}{dt} = -k(\theta - \theta_0),$$ where \(\theta_0\)°F is the air temperature and \(k\) is a constant.

1. Show by integration that the solution of this equation is \(\theta = \theta_0 + Ae^{-kt}\), where \(A\) is a constant. [5]

The value of \(\theta_0\) is 50, and the initial value of \(\theta\) is 98. The initial rate of temperature loss is 1.5°F per hour.

1. Find \(A\), and show that \(k = 0.03125\). [4]
2. Use this model to calculate how long it will take for the temperature to drop
   1. from 98°F to 89°F,
   2. from 98°F to 80°F. [5]
3. Comment on the results obtained in parts (i) and (iv). [1]

Some years ago an island was populated by red squirrels and there were no grey squirrels. Then grey squirrels were introduced. The population \(x\), in thousands, of red squirrels is modelled by the equation $$x = \frac{a}{1 + kt},$$ where \(t\) is the time in years, and \(a\) and \(k\) are constants. When \(t = 0\), \(x = 2.5\).

1. Show that \(\frac{dx}{dt} = -\frac{kx^2}{a}\). [3]
2. Given that the initial population of 2.5 thousand red squirrels reduces to 1.6 thousand after one year, calculate \(a\) and \(k\). [3]
3. What is the long-term population of red squirrels predicted by this model? [1]

The population \(y\), in thousands, of grey squirrels is modelled by the differential equation $$\frac{dy}{dt} = 2y - y^2.$$ When \(t = 0\), \(y = 1\).

1. Express \(\frac{1}{2y - y^2}\) in partial fractions. [4]
2. Hence show by integration that \(\ln\left(\frac{y}{2-y}\right) = 2t\). Show that \(y = \frac{2}{1 + e^{-2t}}\). [7]
3. What is the long-term population of grey squirrels predicted by this model? [1]

A curve has equation $$x^2 + 4y^2 = k^2,$$ where \(k\) is a positive constant.

1. Verify that $$x = k\cos\theta, \quad y = \frac{k}{2}\sin\theta,$$ are parametric equations for the curve. [3]
2. Hence or otherwise show that \(\frac{dy}{dx} = -\frac{x}{4y}\). [3]
3. Fig. 8 illustrates the curve for a particular value of \(k\). Write down this value of \(k\). [1]

\includegraphics{figure_8}

1. Copy Fig. 8 and on the same axes sketch the curves for \(k = 1\), \(k = 3\) and \(k = 4\). [3]

On a map, the curves represent the contours of a mountain. A stream flows down the mountain. Its path on the map is always at right angles to the contour it is crossing.

1. Explain why the path of the stream is modelled by the differential equation $$\frac{dy}{dx} = \frac{4y}{x}.$$ [2]
2. Solve this differential equation. Given that the path of the stream passes through the point (2, 1), show that its equation is \(y = \frac{x^4}{16}\). [6]

In a chemical process, the mass \(M\) grams of a chemical at time \(t\) minutes is modelled by the differential equation $$\frac{dM}{dt} = -\frac{M}{2(1 + \frac{t}{2})}$$

1. Find \(\int \frac{1}{1 + \frac{t}{2}} dt\) [3]
2. Find constants \(A\), \(B\) and \(C\) such that $$\frac{1}{t(1 + \frac{t}{2})} = \frac{A}{t} + \frac{Bt + C}{1 + \frac{t}{2}}$$ [5]
3. Use integration, together with your results in parts (i) and (ii), to show that $$M \sim \frac{K}{.1 + \frac{t}{2}}$$ where \(K\) is a constant. [6]
4. When \(t = 1\), \(M = 25\). Calculate \(K\) What is the mass of the chemical in the long term? [4]

The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).

1. Write down the value of \(h\) for which \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]

The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).

1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}$$ [9]
2. What does this solution indicate about the long-term height of the tree? [1]
3. After a year, the tree has grown to a height of 2m. Which model fits this information better? [3]

A stone, of mass \(m\), falls vertically downwards under gravity through still water. At time \(t\), the stone has speed \(v\) and it experiences a resistance force of magnitude \(\lambda mv\), where \(\lambda\) is a constant.

1. Show that $$\frac{\text{d}v}{\text{d}t} = g - \lambda v$$ [2 marks]
2. The initial speed of the stone is \(u\). Find an expression for \(v\) at time \(t\). [6 marks]

The acceleration \(a\) ms\(^{-2}\) of a particle \(P\) moving in a straight line away from a fixed point \(O\) is given by \(a = \frac{k}{1+t}\), where \(t\) is the time that has elapsed since \(P\) left \(O\), and \(k\) is a constant.

1. By solving a suitable differential equation, find an expression for the velocity \(v\) ms\(^{-1}\) of \(P\) in terms of \(t\), \(k\) and another constant \(c\). [3 marks]

Given that \(v = 0\) when \(t = 0\) and that \(v = 4\) when \(t = 2\),

1. show that \(v \ln 3 = 4 \ln (1 + t)\). [3 marks]
2. Calculate the time when \(P\) has a speed of 8 ms\(^{-1}\). [3 marks]

A small sphere \(S\), of mass \(m\) kg is released from rest at the surface of a liquid in a right circular cylinder whose axis is vertical. When \(S\) is moving downwards with speed \(v\) ms\(^{-1}\), the viscous resistive force acting upwards on it has magnitude \(v^2\) N.

1. Write down a differential equation for the motion of \(S\), clearly defining any symbol(s) that you introduce. [4 marks]
2. Find, in terms of \(m\), the distance \(S\) has fallen when its speed is \(\sqrt{\frac{mg}{2}}\) ms\(^{-1}\). [9 marks]

A particle \(P\) of mass 3 kg moves in a straight line on a smooth horizontal plane. When the speed of \(P\) is \(v\) m s\(^{-1}\), the resultant force acting on \(P\) is a resistance to motion of magnitude \(2v\) N. Find the distance moved by \(P\) while slowing down from 5 m s\(^{-1}\) to 2 m s\(^{-1}\). [5]

A particle \(P\) of mass \(0.5\) kg is released from rest at time \(t = 0\) and falls vertically through a liquid. The motion of \(P\) is resisted by a force of magnitude \(2v\) N, where \(v\) m s\(^{-1}\) is the speed of \(v\) at time \(t\) seconds.

1. Show that \(5 \frac{\mathrm{d}v}{\mathrm{d}t} = 49 - 20v\). [2]
2. Find the speed of \(P\) when \(t = 1\). [5]

A lorry of mass \(M\) is moving along a straight horizontal road. The engine produces a constant driving force of magnitude \(F\). The total resistance to motion is modelled as having magnitude \(kv^2\), where \(k\) is a constant, and \(v\) is the speed of the lorry. Given the lorry moves with constant speed \(V\),

1. show that \(V = \sqrt{\frac{F}{k}}\). [2]

Given instead that the lorry starts from rest,

1. show that the distance travelled by the lorry in attaining a speed of \(\frac{1}{2}V\) is $$\frac{M}{2k}\ln\left(\frac{4}{3}\right).$$ [9]

The variables \(x\) and \(y\) are related by the differential equation $$x^3 \frac{dy}{dx} = xy + x + 1. \qquad (A)$$

1. Use the substitution \(y = u - \frac{1}{x}\), where \(u\) is a function of \(x\), to show that the differential equation may be written as $$x^2 \frac{du}{dx} = u.$$ [4]
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y = f(x)\). [5]