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

A particle \(P\) of mass \(m\) kg moves along a horizontal straight line with acceleration \(a\) ms\(^{-2}\) given by $$a = \frac{v(1-2t^2)}{t},$$ where \(v\) ms\(^{-1}\) is the velocity of \(P\) at time \(t\) s.

1. Find an expression for \(v\) in terms of \(t\) and an arbitrary constant. [3]
2. Given that \(a = 5\) when \(t = 1\), find an expression, in terms of \(m\) and \(t\), for the horizontal force acting on \(P\) at time \(t\). [3]

A particle \(P\) of mass \(2\) kg moves along a horizontal straight line. The point \(O\) is a fixed point on this line. At time \(t\) s the velocity of \(P\) is \(v\) m s\(^{-1}\) and the displacement of \(P\) from \(O\) is \(x\) m. A force of magnitude \(\left(8x - \frac{128}{x^3}\right)\) N acts on \(P\) in the direction \(OP\). When \(t = 0\), \(x = 8\) and \(v = -15\).

1. Show that \(v = -\frac{2}{3}(x^2 - 4)\). [5]
2. Find an expression for \(x\) in terms of \(t\). [4]

A particle \(P\) of mass \(5\) kg moves along a horizontal straight line. At time \(t\) s, the velocity of \(P\) is \(v\) m s\(^{-1}\) and its displacement from a fixed point \(O\) on the line is \(x\) m. The forces acting on \(P\) are a force of magnitude \(\frac{500}{v}\) N in the direction \(OP\) and a resistive force of magnitude \(\frac{1}{2}v^2\) N. When \(t = 0\), \(x = 0\) and \(v = 5\).

1. Find an expression for \(v\) in terms of \(x\). [6]
2. State the value that the speed approaches for large values of \(x\). [1]

A ball of mass \(2\) kg is projected vertically downwards with speed \(5\text{ ms}^{-1}\) through a liquid. At time \(t\) s after projection, the velocity of the ball is \(v\text{ ms}^{-1}\) and its displacement from its starting point is \(x\) m. The forces acting on the ball are its weight and a resistive force of magnitude \(0.2v^2\) N.

1. Find an expression for \(v\) in terms of \(t\). [6]
2. Deduce what happens to \(v\) for large values of \(t\). [1]

\includegraphics{figure_3} Figure 3 shows a large vertical cylindrical tank containing a liquid. The radius of the circular cross-section of the tank is 40 cm. At time \(t\) minutes, the depth of liquid in the tank is \(h\) centimetres. The liquid leaks from a hole \(P\) at the bottom of the tank. The liquid leaks from the tank at a rate of \(32\pi \sqrt{h}\) cm\(^3\) min\(^{-1}\).

1. Show that at time \(t\) minutes, the height \(h\) cm of liquid in the tank satisfies the differential equation $$\frac{dh}{dt} = -0.02\sqrt{h}$$ [4]
2. Find the time taken, to the nearest minute, for the depth of liquid in the tank to decrease from 100 cm to 50 cm. [5]

1. Express \(\frac{2}{P(P-2)}\) in partial fractions. [3]

A team of biologists is studying a population of a particular species of animal. The population is modelled by the differential equation $$\frac{dP}{dt} = \frac{1}{2}P(P-2)\cos 2t, \quad t \geqslant 0$$ where \(P\) is the population in thousands, and \(t\) is the time measured in years since the start of the study. Given that \(P = 3\) when \(t = 0\),

1. solve this differential equation to show that $$P = \frac{6}{3 - e^{\frac{1}{2}\sin 2t}}$$ [7]
2. find the time taken for the population to reach 4000 for the first time. Give your answer in years to 3 significant figures. [3]

In an experiment a scientist considered the loss of mass of a collection of picked leaves. The mass \(M\) grams of a single leaf was measured at times \(t\) days after the leaf was picked. The scientist attempted to find a relationship between \(M\) and \(t\). In a preliminary model she assumed that the rate of loss of mass was proportional to the mass \(M\) grams of the leaf.

1. Write down a differential equation for the rate of change of mass of the leaf, using this model. [2]
2. Show, by differentiation, that \(M = 10(0.98)^t\) satisfies this differential equation. [2]

Further studies implied that the mass \(M\) grams of a certain leaf satisfied a modified differential equation $$10 \frac{dM}{dt} = -k(10M - 1), \quad (1)$$ where \(k\) is a positive constant and \(t \geq 0\). Given that the mass of this leaf at time \(t = 0\) is 10 grams, and that its mass at time \(t = 10\) is 8.5 grams,

1. solve the modified differential equation (1) to find the mass of this leaf at time \(t = 15\). [9]

In a chemical reaction two substances combine to form a third substance. At time \(t\), \(t \geq 0\), the concentration of this third substance is \(x\) and the reaction is modelled by the differential equation $$\frac{dx}{dt} = k(1 - 2x)(1 - 4x), \text{ where } k \text{ is a positive constant.}$$

1. Solve this differential equation and hence show that $$\ln\left|\frac{1 - 2x}{1 - 4x}\right| = 2kt + c, \text{ where } c \text{ is an arbitrary constant.}$$ [7]
2. Given that \(x = 0\) when \(t = 0\), find an expression for \(x\) in terms of \(k\) and \(t\). [4]
3. Find the limiting value of the concentration \(x\) as \(t\) becomes very large. [2]

Liquid is poured into a container at a constant rate of 30 cm\(^3\) s\(^{-1}\). At time \(t\) seconds liquid is leaking from the container at a rate of \(\frac{1}{5}V\) cm\(^3\) s\(^{-1}\), where \(V\) cm\(^3\) is the volume of liquid in the container at that time.

1. Show that $$-15 \frac{dV}{dt} = 2V - 450.$$ [3]

Given that \(V = 1000\) when \(t = 0\),

1. find the solution of the differential equation, in the form \(V = f(t)\). [7]
2. Find the limiting value of \(V\) as \(t \to \infty\). [1]

It is given that \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\).

1. By writing \(\theta = \tan^{-1}\left(\frac{3x}{2}\right)\) as \(2\tan\theta = 3x\), use implicit differentiation to show that $$\frac{d\theta}{dx} = \frac{k}{4 + 9x^2}$$, where \(k\) is an integer. [3 marks]
2. Hence solve the differential equation $$9y(4 + 9x^2)\frac{dy}{dx} = \cosec 3y$$ given that \(x = 0\) when \(y = \frac{\pi}{3}\). Give your answer in the form \(\mathbf{g}(y) = \mathbf{h}(x)\). [7 marks]

Fluid flows out of a cylindrical tank with constant cross section. At time \(t\) minutes, \(t \geq 0\), the volume of fluid remaining in the tank is \(V\) m\(^3\). The rate at which the fluid flows, in m\(^3\) min\(^{-1}\), is proportional to the square root of \(V\).

1. Show that the depth \(h\) metres of fluid in the tank satisfies the differential equation $$\frac{dh}{dt} = -k\sqrt{h}, \quad \text{where } k \text{ is a positive constant.}$$ [3]
2. Show that the general solution of the differential equation may be written as $$h = (A - Bt)^2, \quad \text{where } A \text{ and } B \text{ are constants.}$$ [4] Given that at time \(t = 0\) the depth of fluid in the tank is 1 m, and that 5 minutes later the depth of fluid has reduced to 0.5 m,
3. find the time, \(T\) minutes, which it takes for the tank to empty. [3]
4. Find the depth of water in the tank at time \(0.5T\) minutes. [2]

In a chemical reaction two substances combine to form a third substance. At time \(t\), \(t \geq 0\), the concentration of this third substance is \(x\) and the reaction is modelled by the differential equation $$\frac{dx}{dt} = k(1 - 2x)(1 - 4x), \text{ where } k \text{ is a positive constant.}$$

1. Solve this differential equation and hence show that $$\ln \left| \frac{1 - 2x}{1 - 4x} \right| = 2kt + c, \text{ where } c \text{ is an arbitrary constant.}$$ [7]
2. Given that \(x = 0\) when \(t = 0\), find an expression for \(x\) in terms of \(k\) and \(t\). [4]
3. Find the limiting value of the concentration \(x\) as \(t\) becomes very large. [2]

Newton's law of cooling states that the rate at which the temperature of an object is falling at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of hot liquid is placed in a room which has a constant temperature of \(20°C\). At time \(t\) minutes later, the temperature of the liquid is \(\theta°C\).

1. Explain how the information above leads to the differential equation $$\frac{d\theta}{dt} = -k(\theta - 20),$$ where \(k\) is a positive constant. [2]
2. The liquid is initially at a temperature of \(100°C\). It takes 5 minutes for the liquid to cool from \(100°C\) to \(68°C\). Show that $$\theta = 20 + 80e^{-(\frac{k}{5} \ln \frac{5}{3})t}.$$ [8]
3. Calculate how much longer it takes for the liquid to cool by a further \(32°C\). [3]

A forest is burning so that, \(t\) hours after the start of the fire, the area burnt is \(A\) hectares. It is given that, at any instant, the rate at which this area is increasing is proportional to \(A^2\).

1. Write down a differential equation which models this situation. [2]
2. After 1 hour, 1000 hectares have been burnt; after 2 hours, 2000 hectares have been burnt. Find after how many hours 3000 hectares have been burnt. [6]

\includegraphics{figure_9} 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 \(k\) 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 \(3T\) seconds. [6]

A curve satisfies the differential equation \(\frac{dy}{dx} = 3x^2y\), and passes through the point \((1, 1)\). Find \(y\) in terms of \(x\). [4]

Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x\) cm, and the volume of water is \(V\) cm\(^3\), where \(V = \frac{1}{3}x^3\). The rate at which water is lost is proportional to \(x\), so that \(\frac{dV}{dt} = -kx\), where \(k\) is a constant.

1. Show that \(x \frac{dx}{dt} = -k\). [3]

Initially, the depth of water in the container is 10 cm.

1. Show by integration that \(x = \sqrt{100 - 2kt}\). [4]
2. Given that the container empties after 50 seconds, find \(k\). [2]

Once the container is empty, water is poured into it at a constant rate of 1 cm\(^3\) per second. The container continues to lose water as before.

1. Show that, \(t\) seconds after starting to pour the water in, \(\frac{dx}{dt} = \frac{1-x}{x^2}\). [2]
2. Show that \(\frac{1}{1-x} - x - 1 = \frac{x^2}{1-x}\). Hence solve the differential equation in part (iv) to show that $$t = \ln\left(\frac{1}{1-x}\right) - \frac{1}{2}x^2 - x.$$ [6]
3. Show that the depth cannot reach 1 cm. [1]

The total value of the sales made by a new company in the first \(t\) years of its existence is denoted by \(£V\). A model is proposed in which the rate of increase of \(V\) is proportional to the square root of \(V\). The constant of proportionality is \(k\).

1. Express the model as a differential equation. Verify by differentiation that \(V = (\frac{1}{2}kt + c)^2\), where \(c\) is an arbitrary constant, satisfies this differential equation. [4]
2. The value of the company's sales in its first year is £10000, and the total value of the sales in the first two years is £40000. Find \(V\) in terms of \(t\). [4]

Solve the differential equation \(\frac{dy}{dx} = \frac{y}{x(x+1)}\), given that when \(x = 1\), \(y = 1\). Your answer should express \(y\) explicitly in terms of \(x\). [8]

The motion of a particle is modelled by the differential equation $$v \frac{dv}{dt} + 4x = 0,$$ where \(x\) is its displacement from a fixed point, and \(v\) is its velocity. Initially \(x = 1\) and \(v = 4\).

1. Solve the differential equation to show that \(v^2 = 20 - 4x^2\). [4]

Now consider motion for which \(x = \cos 2t + 2 \sin 2t\), where \(x\) is the displacement from a fixed point at time \(t\).

1. Verify that, when \(t = 0\), \(x = 1\). Use the fact that \(v = \frac{dx}{dt}\) to verify that when \(t = 0\), \(v = 4\). [4]
2. Express \(x\) in the form \(R \cos(2t - \alpha)\), where \(R\) and \(\alpha\) are constants to be determined, and obtain the corresponding expression for \(v\). Hence or otherwise verify that, for this motion too, \(v^2 = 20 - 4x^2\). [7]
3. Use your answers to part (iii) to find the maximum value of \(x\), and the earliest time at which \(x\) reaches this maximum value. [3]

Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. \includegraphics{figure_8.1} The height of the water surface above the hole \(t\) seconds after opening the hole is \(h\) metres, where $$\frac{dh}{dt} = -A\sqrt{h}$$ and where \(A\) is a positive constant. Initially the water surface is 1 metre above the hole.

1. Verify that the solution to this differential equation is $$h = \left(1 - \frac{1}{2}At\right)^2.$$ [3]

The water stops leaking when \(h = 0\). This occurs after 20 seconds.

1. Find the value of \(A\), and the time when the height of the water surface above the hole is 0.5 m. [4]

Fig. 8.2 shows a similar situation with a different barrel; \(h\) is in metres. \includegraphics{figure_8.2} For this barrel, $$\frac{dh}{dt} = -B\frac{\sqrt{h}}{(1+h)^2},$$ where \(B\) is a positive constant. When \(t = 0\), \(h = 1\).

1. Solve this differential equation, and hence show that $$h^{\frac{1}{2}}(30 + 20h + 6h^2) = 56 - 15Bt.$$ [7]
2. Given that \(h = 0\) when \(t = 20\), find \(B\). Find also the time when the height of the water surface above the hole is 0.5 m. [4]

A mathematician is selling goods at a car boot sale. She believes that the rate at which she makes sales depends on the length of time since the start of the sale, \(t\) hours, and the total value of sales she has made up to that time, £\(x\). She uses the model $$\frac{dx}{dt} = \frac{k(5-t)}{x},$$ where \(k\) is a constant. Given that after two hours she has made sales of £96 in total,

1. solve the differential equation and show that she made £72 in the first hour of the sale. [8]

The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than £10 per hour.

1. Verify that at 3 hours and 5 minutes after the start of the sale, she should have already left. [4]

\includegraphics{figure_2} Figure 2 shows a hemispherical bowl of radius 5 cm. The bowl is filled with water but the water leaks from a hole at the base of the bowl. At time \(t\) minutes, the depth of water is \(h\) cm and the volume of water in the bowl is \(V\) cm³, where $$V = \frac{1}{3}\pi h^2(15 - h).$$ In a model it is assumed that the rate at which the volume of water in the bowl decreases is proportional to \(V\).

1. Show that $$\frac{dh}{dt} = -\frac{kh(15-h)}{3(10-h)},$$ where \(k\) is a positive constant. [5]
2. Express \(\frac{3(10-h)}{h(15-h)}\) in partial fractions. [3]

Given that when \(t = 0\), \(h = 5\),

1. show that $$h^2(15-h) = 250e^{-kt}.$$ [6]

Given also that when \(t = 2\), \(h = 4\),

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