Particular solution with initial conditions

1

1. Find the roots of the equation \(m ^ { 2 } + 2 m + 2 = 0\) in the form \(a + i b\).
   (2 marks)
2. 1. Find the general solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 2 \frac { \mathrm {~d} y } { \mathrm {~d} x } + 2 y = 4 x$$
   2. Hence express \(y\) in terms of \(x\), given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2\) when \(x = 0\).

3 The equation of a plane is \(4 x + 2 y + z = 7\).
The point \(A\) has coordinates \(( 9,6,1 )\) and the point \(B\) is the reflection of \(A\) in the plane.
Find the coordinates of the point \(B\). You are given the matrix \(\mathbf { A }\) where \(\mathbf { A } = \left( \begin{array} { l l l } a & 2 & 0 \\ 0 & a & 2 \\ 4 & 5 & 1 \end{array} \right)\).

1. Find, in terms of \(a\), the determinant of \(\mathbf { A }\), simplifying your answer.
2. Hence find the values of \(a\) for which \(\mathbf { A }\) is singular. You are given the following equations which are to be solved simultaneously. $$\begin{aligned} a x + 2 y & = 6 \\ a y + 2 z & = 8 \\ 4 x + 5 y + z & = 16 \end{aligned}$$
3. For each of the values of \(a\) found in part (b) determine whether the equations have
   A particle is suspended in a resistive medium from one end of a light spring. The other end of the spring is attached to a point which is made to oscillate in a vertical line. The displacement of the particle may be modelled by the differential equation \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 5 x = 10 \sin t\) where \(x\) is the displacement of the particle below the equilibrium position at time \(t\).
   When \(t = 0\) the particle is stationary and its displacement is 2 .
   1. Find the particular solution of the differential equation.
   2. Write down an approximate equation for the displacement when \(t\) is large.

10

1. Find the particular solution of the differential equation $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 10 x = 37 \sin 3 t$$ given that \(x = 3\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 0\) when \(t = 0\).
2. Show that, for large positive values of \(t\) and for any initial conditions, $$x \approx \sqrt { } ( 37 ) \sin ( 3 t - \phi ) ,$$ where the constant \(\phi\) is such that \(\tan \phi = 6\).

7

1. Find the value of the constant \(k\) for which \(y = k x \sin 2 x\) is a particular integral of the differential equation \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 8 \cos 2 x\).
2. Solve \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + 4 y = 8 \cos 2 x\), given that \(y = 1\) and \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 1\) when \(x = 0\).

Solve the differential equation $$\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 24e^{2x},$$ given that \(y = 1\) and \(\frac{dy}{dx} = 9\) when \(x = 0\). [7]

Answer only one of the following two alternatives. **EITHER** It is given that \(w = \cos y\) and $$\tan y \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 + 2 \tan y \frac{dy}{dx} = 1 + e^{-2x} \sec y.$$

1. Show that $$\frac{d^2 w}{dx^2} + 2 \frac{dw}{dx} + w = -e^{-2x}.$$ [4]
2. Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0\), \(y = \frac{1}{4}\pi\) and \(\frac{dy}{dx} = \frac{1}{\sqrt{3}}\). [10]

**OR** The curves \(C_1\) and \(C_2\) have polar equations, for \(0 \leqslant \theta \leqslant \frac{1}{2}\pi\), as follows: \begin{align} C_1 : r &= 2(e^\theta + e^{-\theta}),
C_2 : r &= e^{2\theta} - e^{-2\theta}. \end{align} The curves intersect at the point \(P\) where \(\theta = \alpha\).

1. Show that \(e^{2\alpha} - 2e^\alpha - 1 = 0\). Hence find the exact value of \(\alpha\) and show that the value of \(r\) at \(P\) is \(4\sqrt{2}\). [6]
2. Sketch \(C_1\) and \(C_2\) on the same diagram. [3]
3. Find the area of the region enclosed by \(C_1\), \(C_2\) and the initial line, giving your answer correct to 3 significant figures. [5]

Answer only one of the following two alternatives. **EITHER** It is given that \(w = \cos y\) and $$\tan y \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 2\tan y \frac{dy}{dx} = 1 + e^{-2x} \sec y.$$

1. Show that $$\frac{d^2 w}{dx^2} + 2\frac{dw}{dx} + w = -e^{-2x}.$$ [4]
2. Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0\), \(y = \frac{1}{4}\pi\) and \(\frac{dy}{dx} = \frac{1}{\sqrt{3}}\). [10]

**OR** The curves \(C_1\) and \(C_2\) have polar equations, for \(0 \leq \theta \leq \frac{1}{2}\pi\), as follows: \begin{align} C_1: r &= 2(e^\theta + e^{-\theta}),
C_2: r &= e^{2\theta} - e^{-2\theta}. \end{align} The curves intersect at the point \(P\) where \(\theta = \alpha\).

1. Show that \(e^{2\alpha} - 2e^\alpha - 1 = 0\). Hence find the exact value of \(\alpha\) and show that the value of \(r\) at \(P\) is \(4\sqrt{2}\). [6]
2. Sketch \(C_1\) and \(C_2\) on the same diagram. [3]
3. Find the area of the region enclosed by \(C_1\), \(C_2\) and the initial line, giving your answer correct to 3 significant figures. [5]

Find the particular solution of the differential equation $$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 3y = 27x^2,$$ given that, when \(x = 0\), \(y = 2\) and \(\frac{dy}{dx} = -8\). [10]

Find the particular solution of the differential equation $$3\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = x^2,$$ given that, when \(x = 0\), \(y = \frac{dy}{dx} = 0\). [10]

$$\frac{d^2 x}{dt^2} + 6 \frac{dx}{dt} + 6x = 2e^{-t}.$$ Given that \(x = 0\) and \(\frac{dx}{dt} = 2\) at \(t = 0\),

1. find \(x\) in terms of \(t\). [8]

The solution to part (a) is used to represent the motion of a particle \(P\) on the \(x\)-axis. At time \(t\) seconds, where \(t > 0\), \(P\) is \(x\) metres from the origin \(O\).

1. Show that the maximum distance between \(O\) and \(P\) is \(\frac{2\sqrt{3}}{9}\) m and justify that this distance is a maximum. [7]

\includegraphics{figure_1} The differential equation $$\frac{d^2 x}{dt^2} + 6 \frac{dx}{dt} + 9x = \cos 3t, \quad t \geq 0,$$ describes the motion of a particle along the \(x\)-axis.

1. Find the general solution of this differential equation. [8]
2. Find the particular solution of this differential equation for which, at \(t = 0\), \(x = \frac{1}{2}\) and \(\frac{dx}{dt} = 0\). [5]

On the graph of the particular solution defined in part (b), the first turning point for \(T > 30\) is the point \(A\).

1. Find approximate values for the coordinates of \(A\). [2]

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]

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 moves in a plane in such a way that its position vector \(\mathbf{r}\) metres at time \(t\) seconds satisfies the differential equation $$\frac{d^2\mathbf{r}}{dt^2} - 2\frac{d\mathbf{r}}{dt} = 0$$ When \(t = 0\), the particle is at the origin and is moving with velocity \((4i + 2j)\) m s\(^{-1}\). Find \(\mathbf{r}\) in terms of \(t\). [7]

A particle \(P\) moves in the \(x\)-\(y\) plane and has position vector \(\mathbf{r}\) metres at time \(t\) seconds. It is given that \(\mathbf{r}\) satisfies the differential equation $$\frac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2} = 2\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}.$$ When \(t = 0\), \(P\) is at the point with position vector \(3\mathbf{i}\) metres and is moving with velocity \(\mathbf{j}\) m s\(^{-1}\).

1. Find \(\mathbf{r}\) in terms of \(t\). [8]
2. Describe the path of \(P\), giving its cartesian equation. [2]

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]

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]