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

$$\frac{d^2 y}{dt^2} - 6\frac{dy}{dt} + 9y = 4e^{3t}, \quad t \geq 0.$$

1. Show that \(Kte^{3t}\) is a particular integral of the differential equation, where \(K\) is a constant to be found. [4]
2. Find the general solution of the differential equation. [3] Given that a particular solution satisfies \(y = 3\) and \(\frac{dy}{dt} = 1\) when \(t = 0\),
3. find this solution. [4] Another particular solution which satisfies \(y = 1\) and \(\frac{dy}{dt} = 0\) when \(t = 0\), has equation $$y = (1 - 3t + 2t^2)e^{3t}.$$
4. For this particular solution draw a sketch graph of \(y\) against \(t\), showing where the graph crosses the \(t\)-axis. Determine also the coordinates of the minimum of the point on the sketch graph. [5]

$$\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + 5y = 65 \sin 2x, \quad x > 0.$$

1. Find the general solution of the differential equation. [9]
2. Show that for large values of \(x\) this general solution may be approximated by a sine function and find this sine function. [2]

\includegraphics{figure_2} In a simple model of a shock absorber, a particle \(P\) of mass \(m\) kg is attached to one end of a light elastic horizontal spring. The other end of the spring is fixed at \(A\) and the motion of \(P\) takes place along a fixed horizontal line through \(A\). The spring has natural length \(L\) metres and modulus of elasticity \(2mL\) newtons. The whole system is immersed in a fluid which exerts a resistance on \(P\) of magnitude \(3mv\) newtons, where \(v\) m s\(^{-1}\) is the speed of \(P\) at time \(t\) seconds. The compression of the spring at time \(t\) seconds is \(x\) metres, as shown in Fig. 2.

1. Show that $$\frac{\text{d}^2 x}{\text{d}t^2} + 3\frac{\text{d}x}{\text{d}t} + 2x = 0.$$ [4]

Given that when \(t = 0\), \(x = 2\) and \(\frac{\text{d}x}{\text{d}t} = -4\),

1. find \(x\) in terms of \(t\). [8]
2. Sketch the graph of \(x\) against \(t\). [2]
3. State, with a reason, whether the model is realistic. [1]

A particle \(P\) moves in a straight line. At time \(t\) seconds its displacement from a fixed point \(O\) on the line is \(x\) metres. The motion of \(P\) is modelled by the differential equation $$\frac{\text{d}^2 x}{\text{d}t^2} + 2\frac{\text{d}x}{\text{d}t} + 2x = 12\cos 2t - 6\sin 2t.$$ When \(t = 0\), \(P\) is at rest at \(O\).

1. Find, in terms of \(t\), the displacement of \(P\) from \(O\). [11]
2. Show that \(P\) comes to instantaneous rest when \(t = \frac{\pi}{4}\). [2]
3. Find, in metres to 3 significant figures, the displacement of \(P\) from \(O\) when \(t = \frac{\pi}{4}\). [2]
4. Find the approximate period of the motion for large values of \(t\). [2]

A particle \(P\) of mass \(m\) is attached to the mid-point of a light elastic string, of natural length \(2L\) and modulus of elasticity \(2mk^2L\), where \(k\) is a positive constant. The ends of the string are attached to points \(A\) and \(B\) on a smooth horizontal surface, where \(AB = 3L\). The particle is released from rest at the point \(C\), where \(AC = 2L\) and \(ACB\) is a straight line. During the subsequent motion \(P\) experiences air resistance of magnitude \(2mkv\), where \(v\) is the speed of \(P\). At time \(t\), \(AP = 1.5L + x\).

1. Show that \(\frac{d^2x}{dt^2} + 2k\frac{dx}{dt} + 4k^2x = 0\). [6]
2. Find an expression, in terms of \(t\), \(k\) and \(L\), for the distance \(AP\) at time \(t\). [8]

A particle of mass \(m\) is attached to one end \(P\) of a light elastic spring \(PQ\), of natural length \(a\) and modulus of elasticity \(man^2\). At time \(t = 0\), the particle and the spring are at rest on a smooth horizontal table, with the spring straight but unstretched and uncompressed. The end \(Q\) of the spring is then moved in a straight line, in the direction \(PQ\), with constant acceleration \(f\). At time \(t\), the displacement of the particle in the direction \(PQ\) from its initial position is \(x\) and the length of the spring is \((a + y)\).

1. Show that \(x + y = \frac{1}{2}ft^2\). [2]
2. Hence show that $$\frac{d^2x}{dt^2} + n^2x = \frac{1}{2}n^2ft^2.$$ [6]

You are given that the general solution of this differential equation is $$x = A\cos nt + B\sin nt + \frac{1}{2}ft^2 - \frac{f}{n^2},$$ where \(A\) and \(B\) are constants.

1. Find the values of \(A\) and \(B\). [6]
2. Find the maximum tension in the spring. [4]

A particle \(P\) of mass \(m\) is suspended from a fixed point by a light elastic spring. The spring has natural length \(a\) and modulus of elasticity \(2m\omega^2a\), where \(\omega\) is a positive constant. At time \(t = 0\) the particle is projected vertically downwards with speed \(U\) from its equilibrium position. The motion of the particle is resisted by a force of magnitude \(2m\omega v\), where \(v\) is the speed of the particle. At time \(t\), the displacement of \(P\) downwards from its equilibrium position is \(x\).

1. Show that \(\frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega \frac{\mathrm{d}x}{\mathrm{d}t} + 2\omega^2x = 0\). [5] Given that the solution of this differential equation is \(x = e^{-\omega t}(A \cos \omega t + B \sin \omega t)\), where \(A\) and \(B\) are constants,
2. find \(A\) and \(B\). [4]
3. Find an expression for the time at which \(P\) first comes to rest. [3]

A light elastic string, of natural length \(a\) and modulus of elasticity \(5ma\omega^2\), lies unstretched along a straight line on a smooth horizontal plane. A particle of mass \(m\) is attached to one end of the spring. At time \(t = 0\), the other end of the spring starts to move with constant speed \(U\) along the line of the spring and away from the particle. As the particle moves along the plane it is subject to a resistance of magnitude \(2m\omega v\), where \(v\) is its speed. At time \(t\), the extension of the spring is \(x\) and the displacement of the particle from its initial position is \(y\). Show that

1. \(x + y = Ut\), [2]
2. \(\frac{d^2x}{dt^2} + 2\omega \frac{dx}{dt} + 5\omega^2 x = 2\omega U\). [7]

1. Find \(x\) in terms of \(\omega\), \(U\) and \(t\). [8]

\includegraphics{figure_4} A light elastic spring has natural length \(l\) and modulus of elasticity \(4mg\). One end of the spring is attached to a point \(A\) on a plane that is inclined to the horizontal at an angle \(\alpha\), where \(\tan\alpha = \frac{3}{4}\). The other end of the spring is attached to a particle \(P\) of mass \(m\). The plane is rough and the coefficient of friction between \(P\) and the plane is \(\frac{1}{4}\). The particle \(P\) is held at a point \(B\) on the plane where \(B\) is below \(A\) and \(AB = l\), with the spring lying along a line of greatest slope of the plane, as shown in Figure 4. At time \(t = 0\), the particle is projected up the plane towards \(A\) with speed \(\frac{1}{2}\sqrt{gl}\). At time \(t\), the compression of the spring is \(x\).

1. Show that $$\frac{d^2x}{dt^2} + 4\omega^2x = -g, \text{ where } \omega = \sqrt{\frac{g}{l}}.$$ [6]

1. Find \(x\) in terms of \(l\), \(\omega\) and \(t\). [7]

1. Find the distance that \(P\) travels up the plane before first coming to rest. [4]

A small ball is attached to one end of a spring. The ball is modelled as a particle of mass 0.1 kg and the spring is modelled as a light elastic spring \(AB\), of natural length 0.5 m and modulus of elasticity 2.45 N. The particle is attached to the end \(B\) of the spring. Initially, at time \(t = 0\), \(A\) is held at rest and the particle hangs at rest in equilibrium below \(A\) at the point \(E\). The end \(A\) then begins to move along the line of the spring in such a way that, at time \(t\) seconds, \(t \leq 1\), the downward displacement of \(A\) from its initial position is \(2 \sin 2t\) metres. At time \(t\) seconds, the extension of the spring is \(x\) metres and the displacement of the particle below \(E\) is \(y\) metres.

1. Show, by referring to a simple diagram, that \(y + 0.2 = x + 2 \sin 2t\). [3]
2. Hence show that \(\frac{d^2y}{dt^2} + 49y = 98 \sin 2t\). [5]

Given that \(y = \frac{98}{45} \sin 2t\) is a particular integral of this differential equation,

1. find \(y\) in terms of \(t\). [5]
2. Find the time at which the particle first comes to instantaneous rest. [4]

A particle \(P\) of mass \(m\) kg is attached to the end \(A\) of a light elastic string \(AB\), of natural length \(a\) metres and modulus of elasticity \(9ma\) newtons. Initially the particle and the string lie at rest on a smooth horizontal plane with \(AB = a\) metres. At time \(t = 0\) the end \(B\) of the string is set in motion and moves at a constant speed \(U\) m s\(^{-1}\) in the direction \(AB\). The air resistance acting on \(P\) has magnitude \(6mv\) newtons, where \(v\) m s\(^{-1}\) is the speed of \(P\). At time \(t\) seconds, the extension of the string is \(x\) metres and the displacement of \(P\) from its initial position is \(y\) metres. Show that, while the string is taut,

1. \(x + y = Ut\) [2]
2. \(\frac{d^2x}{dt^2} + 6\frac{dx}{dt} + 9x = 6U\) [5]

You are given that the general solution of the differential equation in (b) is $$x = (A + Bt)e^{-3t} + \frac{2U}{3}$$ where \(A\) and \(B\) are arbitrary constants.

1. Find the value of \(A\) and the value of \(B\). [5]
2. Find the speed of \(P\) at time \(t\) seconds. [2]

\includegraphics{figure_2} A railway truck of mass \(M\) approaches the end of a straight horizontal track and strikes a buffer. The buffer is parallel to the track, as shown in Figure 2. The buffer is modelled as a light horizontal spring \(PQ\), which is fixed at the end \(P\). The spring has a natural length \(a\) and modulus of elasticity \(Mn^2a\), where \(n\) is a positive constant. At time \(t = 0\), the spring has length \(a\) and the truck strikes the end \(Q\) with speed \(U\). A resistive force whose magnitude is \(Mkv\), where \(v\) is the speed of the truck at time \(t\), and \(k\) is a positive constant, also opposes the motion of the truck. At time \(t\), the truck is in contact with the buffer and the compression of the buffer is \(x\).

1. Show that, while the truck is compressing the buffer $$\frac{\text{d}^2x}{\text{d}t^2} + k\frac{\text{d}x}{\text{d}t} + n^2x = 0$$ (4)

It is given that \(k = \frac{5n}{2}\)

1. Find \(x\) in terms of \(U\), \(n\) and \(t\). (7)

1. Find, in terms of \(U\) and \(n\), the greatest value of \(x\). (5)

A particle of mass \(m\) kg is attached to one end of a light elastic string of natural length \(a\) metres and modulus of elasticity \(5ma\) newtons. The other end of the string is attached to a fixed point \(O\) on a smooth horizontal plane. The particle is held at rest on the plane with the string stretched to a length \(2a\) metres and then released at time \(t = 0\). During the subsequent motion, when the particle is moving with speed \(v\) m s\(^{-1}\), the particle experiences a resistance of magnitude \(4mv\) newtons. At time \(t\) seconds after the particle is released, the length of the string is \((a + x)\) metres, where \(0 \leqslant x \leqslant a\).

1. Show that, from \(t = 0\) until the string becomes slack, $$\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 4\frac{\mathrm{d}x}{\mathrm{d}t} + 5x = 0$$ [3]
2. Hence express \(x\) in terms of \(a\) and \(t\). [6]
3. Find the speed of the particle at the instant when the string first becomes slack, giving your answer in the form \(ka\), where \(k\) is a constant to be found correct to 2 significant figures. [4]

At time \(t\) seconds, the position vector of a particle \(P\) is \(\mathbf{r}\) metres, where \(\mathbf{r}\) satisfies the vector differential equation $$\frac{d^2\mathbf{r}}{dt^2} + 4\mathbf{r} = e^{2t} \mathbf{j}.$$ When \(t = 0\), \(P\) has position vector \((i + j)\) m and velocity \(2i\) m s\(^{-1}\). Find an expression for \(\mathbf{r}\) in terms of \(t\). [11]

A particle \(P\) moves in the \(x\)-\(y\) plane so that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation $$\frac{d^2\mathbf{r}}{dt^2} - 4\mathbf{r} = -3e^t\mathbf{j}$$ When \(t = 0\), the particle is at the origin and is moving with velocity \((2i + j)\) ms\(^{-1}\). Find \(\mathbf{r}\) in terms of \(t\). [10]

A particle \(P\) moves in the \(x\)-\(y\) plane so that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation $$\frac{d^2\mathbf{r}}{dt^2} - 4\mathbf{r} = -3e^t\mathbf{j}$$ When \(t = 0\), the particle is at the origin and is moving with velocity \((2\mathbf{i} + \mathbf{j})\) ms\(^{-1}\). Find \(\mathbf{r}\) in terms of \(t\). [10]

A particle \(P\) moves in the \(x\)-\(y\) plane and has position vector \(\mathbf{r}\) metres relative to a fixed origin \(O\) at time \(t\) s. Given that \(\mathbf{r}\) satisfies the vector differential equation $$\frac{d^2\mathbf{r}}{dt^2} + 9\mathbf{r} = 8\sin t \mathbf{i}$$ and that when \(t = 0\) s, \(P\) is at \(O\) and moving with velocity \((\mathbf{i} + 3\mathbf{j})\) m s\(^{-1}\),

1. find \(\mathbf{r}\) at time \(t\). [11]
2. Hence find when \(P\) next returns to \(O\). [2]

Find the general solution of the differential equation $$\frac{d^2y}{dx^2} - c\frac{dy}{dx} + 8y = e^{3x}.$$ [6]

Find the general solution of the differential equation $$\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 16y = 4x.$$ [7]