4.10e Second order non-homogeneous: complementary + particular integral

Find the general solution of the differential equation $$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = 65 \sin 2x.$$ [9]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = e^{3x}.$$

1. Find the complementary function. [3]
2. Explain briefly why there is no particular integral of either of the forms \(y = ke^{3x}\) or \(y = kxe^{3x}\). [1]
3. Given that there is a particular integral of the form \(y = kx^2e^{3x}\), find the value of \(k\). [5]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{\text{d}^2y}{\text{d}x^2} + 16y = 8\cos 4x.$$

1. Find the complementary function of the differential equation. [2]
2. Given that there is a particular integral of the form \(y = px\sin 4x\), where \(p\) is a constant, find the general solution of the equation. [6]
3. Find the solution of the equation for which \(y = 2\) and \(\frac{\text{d}y}{\text{d}x} = 0\) when \(x = 0\). [4]

Find the general solution of the differential equation $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 3\frac{\mathrm{d}y}{\mathrm{d}x} - 4y = \cos 2x + 5x$$ [9 marks]

In this question use \(g = 9.8\) m s\(^{-2}\) Two light elastic strings each have one end attached to a small ball \(B\) of mass 0.5 kg The other ends of the strings are attached to the fixed points \(A\) and \(C\), which are 8 metres apart with \(A\) vertically above \(C\) The whole system is in a thin tube of oil, as shown in the diagram below. \includegraphics{figure_18} The string connecting \(A\) and \(B\) has natural length 2 metres, and the tension in this string is \(7e\) newtons when the extension is \(e\) metres. The string connecting \(B\) and \(C\) has natural length 3 metres, and the tension in this string is \(3e\) newtons when the extension is \(e\) metres.

1. Find the extension of each string when the system is in equilibrium. [3 marks]
2. It is known that in a large bath of oil, the oil causes a resistive force of magnitude \(4.5v\) newtons to act on the ball, where \(v\) m s\(^{-1}\) is the speed of the ball. Use this model to answer part (b)(i) and part (b)(ii).
   1. The ball is pulled a distance of 0.6 metres downwards from its equilibrium position towards \(C\), and released from rest. Show that during the subsequent motion the particle satisfies the differential equation $$\frac{d^2x}{dt^2} + 9\frac{dx}{dt} + 20x = 0$$ where \(x\) metres is the displacement of the particle below the equilibrium position at time \(t\) seconds after the particle is released. [3 marks]
   2. Find \(x\) in terms of \(t\) [5 marks]
3. State one limitation of the model used in part (b) [1 mark]

Charlotte is trying to solve this mathematical problem: Find the general solution of the differential equation $$\frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 10e^{-2x}$$ Charlotte's solution starts as follows: Particular integral: \(y = \lambda e^{-2x}\) so $$\frac{dy}{dx} = -2\lambda e^{-2x}$$ and $$\frac{d^2y}{dx^2} = 4\lambda e^{-2x}$$

1. Show that Charlotte's method will fail to find a particular integral for the differential equation. [2 marks]
2. Explain how Charlotte should have started her solution differently and find the general solution of the differential equation. [8 marks]

Solve the differential equation $$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} - 45y = 21e^{5x} - 0.3x + 27x^2$$ given that \(y = \frac{37}{225}\) and \(\frac{dy}{dx} = 0\) when \(x = 0\) [10 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]

The sequence \(u_0, u_1, u_2, \ldots\) satisfies the recurrence relation \(u_{n+2} - 3u_{n+1} - 10u_n = 24n - 10\).

1. Determine the general solution of the recurrence relation. [6]
2. Hence determine the particular solution of the recurrence relation for which \(u_0 = 6\) and \(u_1 = 10\). [3]
3. Show, by direct calculation, that your solution in part (b) gives the correct value for \(u_2\). [1]

The sequence \(v_0, v_1, v_2, \ldots\) is defined by \(v_n = \frac{u_n}{p^n}\) for some constant \(p\), where \(u_n\) denotes the particular solution found in part (b). You are given that \(v_n\) converges to a finite non-zero limit, \(q\), as \(n \to \infty\).

1. Determine \(p\) and \(q\). [4]

Find the solution of the differential equation $$3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 8 + 6x - 2x^2,$$ where \(y = 6\) and \(\frac{dy}{dx} = 5\) when \(x = 0\). [12]

Find the general solution of the differential equation \(\frac{d^2 y}{dx^2} + y = 8x^2\). [7]

The differential equation \((\star)\) is $$\frac{\text{d}^2 u}{\text{d}x^2} + 4u = 8x + 1.$$

1. Find the general solution of \((\star)\). [5]
2. The differential equation \((\star \star)\) is $$x \frac{\text{d}^2 v}{\text{d}x^2} + 2 \frac{\text{d}v}{\text{d}x} + 4xv = 8x + 1.$$ By using the substitution \(u = xv\), show that \((\star)\) becomes \((\star \star)\) and deduce the general solution of \((\star \star)\). [4]

Find the general solution of the differential equation $$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 6\frac{\mathrm{d}y}{\mathrm{d}x} + 9y = 72\mathrm{e}^{3x}.$$ [7]

11 Answer only one of the following two alternatives. EITHER State the fifth roots of unity in the form \(\cos \theta + \mathrm { i } \sin \theta\), where \(- \pi < \theta \leqslant \pi\). Simplify $$\left( x - \left[ \cos \frac { 2 } { 5 } \pi + i \sin \frac { 2 } { 5 } \pi \right] \right) \left( x - \left[ \cos \frac { 2 } { 5 } \pi - i \sin \frac { 2 } { 5 } \pi \right] \right) .$$ Hence find the real factors of $$x ^ { 5 } - 1$$ Express the six roots of the equation $$x ^ { 6 } - x ^ { 3 } + 1 = 0$$ as three conjugate pairs, in the form \(\cos \theta \pm \mathrm { i } \sin \theta\). Hence find the real factors of $$x ^ { 6 } - x ^ { 3 } + 1$$ OR Given that $$y ^ { 2 } \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } - 6 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) ^ { 2 } + 3 y ^ { 3 } = 25 \mathrm { e } ^ { - 2 x }$$ and that \(v = y ^ { 3 }\), show that $$\frac { \mathrm { d } ^ { 2 } v } { \mathrm {~d} x ^ { 2 } } - 6 \frac { \mathrm {~d} v } { \mathrm {~d} x } + 9 v = 75 \mathrm { e } ^ { - 2 x }$$ Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0 , y = 2\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\). \end{document}