| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Challenging +1.3 This is a standard M4 second-order differential equation with forcing, requiring auxiliary equation solution, particular integral by undetermined coefficients, and applying initial conditions. While it involves multiple steps (11 marks for part a), the techniques are routine for M4 students with no novel insight required. The complementary function and particular integral follow textbook methods, making it moderately above average difficulty but well within expected M4 scope. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| Auxiliary Equation: \(m^2 + 2m + 2 = 0, \Rightarrow m = -1 \pm i\) | M1, A1 | |
| \(\therefore\) Complementary Function is: \(x = e^{-t}(A\cos t + B\sin t)\) | M1 ft | |
| Let \(x = p\cos 2t + q\sin 2t\), \(\dot{x} = -2p\sin 2t + 2q\cos 2t\), \(\ddot{x} = -4x\) | M1 | |
| Sub. in D.E. | ||
| \(-2p\cos 2t - 2q\sin 2t - 4p\sin 2t + 4q\cos 2t = 12\cos 2t - 6\sin 2t\) | M1 | |
| \(-2p + 4q = 12, \quad -4p - 2q = -6\) | A1 | |
| \(-10p = 0 \Rightarrow p = 0, q = 3\) | M1 | |
| \(\therefore \quad x = 3\sin 2t + e^{-t}(A\cos t + B\sin t)\) | A1 | |
| \(t = 0, x = 0 \Rightarrow 0 = A\) | B1 | |
| \(\dot{x} = 6\cos 2t - e^{-t}B\sin t + e^{-t}B\cos t\) | M1 | |
| \(t = 0, \dot{x} = 0 \Rightarrow 0 = 6 + B \quad \therefore B = -6\) | M1 | |
| \(\therefore \quad x = 3\sin 2t - 6e^{-t}\sin t\) | A1 | (11) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(x = 6[\cos 2t + e^{-t}\sin t - e^{-t}\cos t]\) | ||
| Sub \(t = \frac{\pi}{4}\): \(\dot{x} = 6[\cos 2t + e^{-t} - 6e^{-t}\cos t]\) | ||
| \(\dot{x} = 6[0 + e^{-t} \times \frac{1}{\sqrt{2}} - e^{-t} \times \frac{1}{\sqrt{2}}] = 0\) | M1 | |
| \(\therefore P\) comes to instantaneous rest when \(t = \frac{\pi}{4}\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| sub \(t = \frac{\pi}{4}\) in \(x\): \(= 3\sin\frac{\pi}{2} - 6e^{-\pi/4} \times \frac{1}{\sqrt{2}} = 1.07\) | M1, A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Content | Marks | Notes |
| \(t \to \infty\): \(x \approx 3\sin 2t\), approximate period is \(\pi\) | M1, A1 | (2) |
| (17 marks) |
## Part (a)
| Content | Marks | Notes |
|---------|-------|-------|
| Auxiliary Equation: $m^2 + 2m + 2 = 0, \Rightarrow m = -1 \pm i$ | M1, A1 | |
| $\therefore$ Complementary Function is: $x = e^{-t}(A\cos t + B\sin t)$ | M1 ft | |
| Let $x = p\cos 2t + q\sin 2t$, $\dot{x} = -2p\sin 2t + 2q\cos 2t$, $\ddot{x} = -4x$ | M1 | |
| Sub. in D.E. | | |
| $-2p\cos 2t - 2q\sin 2t - 4p\sin 2t + 4q\cos 2t = 12\cos 2t - 6\sin 2t$ | M1 | |
| $-2p + 4q = 12, \quad -4p - 2q = -6$ | A1 | |
| $-10p = 0 \Rightarrow p = 0, q = 3$ | M1 | |
| $\therefore \quad x = 3\sin 2t + e^{-t}(A\cos t + B\sin t)$ | A1 | |
| $t = 0, x = 0 \Rightarrow 0 = A$ | B1 | |
| $\dot{x} = 6\cos 2t - e^{-t}B\sin t + e^{-t}B\cos t$ | M1 | |
| $t = 0, \dot{x} = 0 \Rightarrow 0 = 6 + B \quad \therefore B = -6$ | M1 | |
| $\therefore \quad x = 3\sin 2t - 6e^{-t}\sin t$ | A1 | (11) |
## Part (b)
| Content | Marks | Notes |
|---------|-------|-------|
| $x = 6[\cos 2t + e^{-t}\sin t - e^{-t}\cos t]$ | | |
| Sub $t = \frac{\pi}{4}$: $\dot{x} = 6[\cos 2t + e^{-t} - 6e^{-t}\cos t]$ | | |
| $\dot{x} = 6[0 + e^{-t} \times \frac{1}{\sqrt{2}} - e^{-t} \times \frac{1}{\sqrt{2}}] = 0$ | M1 | |
| $\therefore P$ comes to instantaneous rest when $t = \frac{\pi}{4}$ | A1 | (2) |
## Part (c)
| Content | Marks | Notes |
|---------|-------|-------|
| sub $t = \frac{\pi}{4}$ in $x$: $= 3\sin\frac{\pi}{2} - 6e^{-\pi/4} \times \frac{1}{\sqrt{2}} = 1.07$ | M1, A1 | (2) |
## Part (d)
| Content | Marks | Notes |
|---------|-------|-------|
| $t \to \infty$: $x \approx 3\sin 2t$, approximate period is $\pi$ | M1, A1 | (2) |
| | **(17 marks)** | |
---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 $O$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $t$, the displacement of $P$ from $O$.
[11]
\item Show that $P$ comes to instantaneous rest when $t = \frac{\pi}{4}$.
[2]
\item Find, in metres to 3 significant figures, the displacement of $P$ from $O$ when $t = \frac{\pi}{4}$.
[2]
\item Find the approximate period of the motion for large values of $t$.
[2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2003 Q5 [17]}}