4.10a General/particular solutions: of differential equations

The current, \(x\) amps, at time \(t\) seconds after a switch is closed in a particular electric circuit is modelled by the equation $$\frac{dx}{dt} = k - 3x$$ where \(k\) is a constant. Initially there is zero current in the circuit.

1. Solve the differential equation to find an equation, in terms of \(k\), for the current in the circuit at time \(t\) seconds. Give your answer in the form \(x = f(t)\). [6]

Given that in the long term the current in the circuit approaches \(7\) amps,

1. find the value of \(k\). [2]
2. Hence find the time in seconds it takes for the current to reach \(5\) amps, giving your answer to \(2\) significant figures. [3]

Liquid is pouring into a container at a constant rate of \(20\text{ cm}^3\text{s}^{-1}\) and is leaking out at a rate proportional to the volume of the liquid already in the container.

1. Explain why, at time \(t\) seconds, the volume, \(V\text{ cm}^3\), of liquid in the container satisfies the differential equation $$\frac{dV}{dt} = 20 - kV,$$ where \(k\) is a positive constant. [2]

The container is initially empty.

1. By solving the differential equation, show that $$V = A + Be^{-kt},$$ giving the values of \(A\) and \(B\) in terms of \(k\). [6]

Given also that \(\frac{dV}{dt} = 10\) when \(t = 5\),

1. find the volume of liquid in the container at 10 s after the start. [5]

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]

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]

A particle of mass \(m\) kg moves in a horizontal straight line. Its initial speed is \(u\) ms\(^{-1}\) and the only force acting on it is a variable resistance of magnitude \(mkv\) N, where \(v\) ms\(^{-1}\) is the speed of the particle after \(t\) seconds and \(k\) is a constant. Show that \(v = ue^{-kt}\). [7 marks]

A rocket, with initial mass 1500 kg, including 600 kg of fuel, is launched vertically upwards from rest. The rocket burns fuel at a rate of 15 kg s\(^{-1}\) and the burnt fuel is ejected vertically downwards with a speed of 1000 m s\(^{-1}\) relative to the rocket. At time \(t\) seconds after launch \((t \leqslant 40)\) the rocket has mass \(m\) kg and velocity \(v\) m s\(^{-1}\).

1. Show that $$\frac{dv}{dt} + \frac{1000}{m} \frac{dm}{dt} = -9.8$$ [5]

1. Find \(v\) at time \(t\), \(0 \leqslant t \leqslant 40\) [5]

A space-ship is moving in a straight line in deep space and needs to reduce its speed from \(U\) to \(V\). This is done by ejecting fuel from the front of the space-ship at a constant speed \(k\) relative to the space-ship. When the speed of the space-ship is \(v\), its mass is \(m\).

1. Show that, while the space-ship is ejecting fuel, \(\frac{\mathrm{d}m}{\mathrm{d}v} = -\frac{m}{k}\). [6]

The initial mass of the space-ship is \(M\).

1. Find, in terms of \(U\), \(V\), \(k\) and \(M\), the amount of fuel which needs to be used to reduce the speed of the space-ship from \(U\) to \(V\). [6]

A rocket, with initial mass 1500 kg, including 600 kg of fuel, is launched vertically upwards from rest. The rocket burns fuel at a rate of 15 kg s\(^{-1}\) and the burnt fuel is ejected vertically downwards with a speed of 1000 m s\(^{-1}\) relative to the rocket. At time \(t\) seconds after launch \((t \leqslant 40)\) the rocket has mass \(m\) kg and velocity \(v\) m s\(^{-1}\).

1. Show that $$\frac{dv}{dt} + \frac{1000}{m}\frac{dm}{dt} = -9.8$$ [5]
2. Find \(v\) at time \(t\), \(0 \leqslant t \leqslant 40\) [5]

A spacecraft is travelling in a straight line in deep space where all external forces can be assumed to be negligible. The spacecraft decelerates by ejecting fuel at a constant speed \(k\) relative to the spacecraft, in the direction of motion of the spacecraft. At time \(t\), the spacecraft has speed \(v\) and mass \(m\).

1. Show, from first principles, that while the spacecraft is ejecting fuel, $$\frac{dv}{dm} - \frac{k}{m} = 0$$ [5]

At time \(t = 0\), the spacecraft has speed \(U\) and mass \(M\).

1. Find the mass of the spacecraft when it comes to rest. [6]

Given that \(m = Me^{-\alpha t^2}\), where \(\alpha\) is a positive constant, and that the spacecraft comes to rest at time \(t = T\),

1. find, in terms of \(U\) and \(T\) only, the distance travelled by the spacecraft in decelerating from speed \(U\) to rest. [6]

As a hailstone falls under gravity in still air, its mass increases. At time \(t\) the mass of the hailstone is \(m\). The hailstone is modelled as a uniform sphere of radius \(r\) such that $$\frac{dr}{dt} = kr,$$ where \(k\) is a positive constant.

1. Show that \(\frac{dm}{dt} = 3km\). [2]

Assuming that there is no air resistance,

1. show that the speed \(v\) of the hailstone at time \(t\) satisfies $$\frac{dv}{dt} = g - 3kv.$$ [4]

Given that the speed of the hailstone at time \(t = 0\) is \(u\),

1. find an expression for \(v\) in terms of \(t\). [5]
2. Hence show that the speed of the hailstone approaches the limiting value \(\frac{g}{3k}\). [1]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{dy}{dx} + 4y = 5 \cos 3x.$$

1. Find the complementary function. [2]
2. Hence, or otherwise, find the general solution. [7]
3. Find the approximate range of values of \(y\) when \(x\) is large and positive. [2]

The substitution \(y = u^k\), where \(k\) is an integer, is to be used to solve the differential equation $$x \frac{dy}{dx} + 3y = x^2 y^2 \qquad (A)$$ by changing it into an equation (B) in the variables \(u\) and \(x\).

1. Show that equation (B) may be written in the form $$\frac{du}{dx} + \frac{3}{kx} u = \frac{1}{k} x u^{k+1}.$$ [4]
2. Write down the value of \(k\) for which the integrating factor method may be used to solve equation (B). [1]
3. Using this value of \(k\), solve equation (B) and hence find the general solution of equation (A), giving your answer in the form \(y = f(x)\). [4]

A curve passes through the point \((-2, 4.73)\) and satisfies the differential equation $$\frac{dy}{dx} = \frac{y^2 - x^2}{2x + 3y}$$ Use Euler's step by step method once, and then the midpoint formula $$y_{r+1} = y_{r-1} + 2hf(x_r, y_r), \quad x_{r+1} = x_r + h$$ once, each with a step length of \(0.02\), to estimate the value of \(y\) when \(x = -1.96\) Give your answer to five significant figures. [4 marks]

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]

A financial institution models the repayment of a loan to a client in the following way.

• An amount, \(£C\), is loaned to the client at the start of the repayment period.
• The amount owed \(n\) years after the start of the repayment period is \(£L_n\), so that \(L_0 = C\).
• At the end of each year, interest of \(\alpha\%\) (\(\alpha > 0\)) of the amount owed at the start of that year is added to the amount owed.
• Immediately after interest has been added to the amount owed a repayment of \(£R\) is made by the client.
• Once \(L_n\) becomes negative the repayment is finished and the overpayment is refunded to the client.

1. Show that during the repayment period, \(L_{n+1} = aL_n + b\), giving \(a\) and \(b\) in terms of \(\alpha\) and \(R\). [2]
2. Find the solution of the recurrence relation \(L_{n+1} = aL_n + b\) with \(L_0 = C\), giving your solution in terms of \(a\), \(b\), \(C\) and \(n\). [5]
3. Deduce from parts (a) and (b) that, for the repayment scheme to terminate, \(R > \frac{\alpha C}{100}\). [2]

A client takes out a £30000 loan at 8% interest and agrees to repay £3000 at the end of each year.

1. 1. Use an algebraic method to find the number of years it will take for the loan to be repaid. [3]
   2. Taking into account the refund of overpayment, find the total amount that the client repays over the lifetime of the loan. [3]

Solve the differential equation $$2\cot x \frac{dy}{dx} = (4 - y^2)$$ for which \(y = 0\) at \(x = \frac{\pi}{3}\), giving your answer in the form \(\sec^2 x = g(y)\). [8]

The function \(y\) satisfies \(\frac{d^2y}{dx^2} + x^2y = x\), and is such that \(y = 1\) and \(\frac{dy}{dx} = 1\) when \(x = 1\).

1. Using the given differential equation
   1. state the value of \(\frac{d^2y}{dx^2}\) when \(x = 1\), [1]
   2. find, by differentiation, the value of \(\frac{d^3y}{dx^3}\) when \(x = 1\). [2]
2. Hence determine the Taylor series for \(y\) about \(x = 1\) up to and including the term in \((x-1)^3\) and deduce, correct to 4 decimal places, an approximation for \(y\) when \(x = 1.1\). [3]

A stone of mass \(m\) is projected vertically upwards with initial velocity \(u\). At time \(t\), the height risen above the point of projection is \(x\) and the resistance to motion is \(kv\) when the velocity of the stone is \(v\).

1. Write down a first-order differential equation relating \(v\) and \(t\) and hence find \(t\) in terms of \(v\). [5]
2. Write down a first-order differential equation relating \(v\) and \(x\) and hence find \(x\) in terms of \(v\). [6]