| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| 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 second-order vector differential equation requiring complementary function (solving auxiliary equation for SHM), particular integral (using trial solution with sine/cosine), and applying two initial conditions to find constants. While it involves multiple steps and vector notation, it follows a completely standard M5 procedure with no novel insight required—harder than routine C3 calculus but typical for Further Maths mechanics. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks |
|---|---|
| i-component ODE: \(\ddot{x} + 4x = 3\sin t\), CF: \(A\cos 2t + B\sin 2t\), PI: try \(x = k\sin t\): \(-k\sin t + 4k\sin t = 3\sin t \Rightarrow k=1\) | M1 A1 |
| General solution i: \(x = A\cos 2t + B\sin 2t + \sin t\) | A1 |
| j-component ODE: \(\ddot{y} + 4y = 0\), solution: \(y = C\cos 2t + D\sin 2t\) | M1 A1 |
| Velocity: \(\dot{x} = -2A\sin 2t + 2B\cos 2t + \cos t\); \(\dot{y} = -2C\sin 2t + 2D\cos 2t\) | M1 |
| Conditions at \(t=\frac{\pi}{2}\): position \(= \mathbf{i}-\mathbf{j}\), velocity \(= \mathbf{0}\) | M1 |
| \(x(\frac{\pi}{2}): -A + \sin\frac{\pi}{2} = 1 \Rightarrow A = 0\); \(\dot{x}(\frac{\pi}{2}): -2B\sin\pi + \cos\frac{\pi}{2}=0 \Rightarrow\) check: \(-2B(0)+0=0\) ✓, need more: \(\dot{x}=2B\cos\pi + 0 = -2B = 0 \Rightarrow B=0\) | A1 |
| \(y(\frac{\pi}{2}): -C = -1 \Rightarrow C=1\); \(\dot{y}(\frac{\pi}{2}): -2C\sin\pi + 2D\cos\pi = -2D = 0 \Rightarrow D=0\) | A1 |
| Answer: \(\mathbf{v} = \cos t\,\mathbf{i} + (-2\sin 2t)\,\mathbf{j}\) | A1 A1 |
## Question 2:
**i-component ODE:** $\ddot{x} + 4x = 3\sin t$, CF: $A\cos 2t + B\sin 2t$, PI: try $x = k\sin t$: $-k\sin t + 4k\sin t = 3\sin t \Rightarrow k=1$ | M1 A1 |
**General solution i:** $x = A\cos 2t + B\sin 2t + \sin t$ | A1 |
**j-component ODE:** $\ddot{y} + 4y = 0$, solution: $y = C\cos 2t + D\sin 2t$ | M1 A1 |
**Velocity:** $\dot{x} = -2A\sin 2t + 2B\cos 2t + \cos t$; $\dot{y} = -2C\sin 2t + 2D\cos 2t$ | M1 |
**Conditions at $t=\frac{\pi}{2}$:** position $= \mathbf{i}-\mathbf{j}$, velocity $= \mathbf{0}$ | M1 |
$x(\frac{\pi}{2}): -A + \sin\frac{\pi}{2} = 1 \Rightarrow A = 0$; $\dot{x}(\frac{\pi}{2}): -2B\sin\pi + \cos\frac{\pi}{2}=0 \Rightarrow$ check: $-2B(0)+0=0$ ✓, need more: $\dot{x}=2B\cos\pi + 0 = -2B = 0 \Rightarrow B=0$ | A1 |
$y(\frac{\pi}{2}): -C = -1 \Rightarrow C=1$; $\dot{y}(\frac{\pi}{2}): -2C\sin\pi + 2D\cos\pi = -2D = 0 \Rightarrow D=0$ | A1 |
**Answer:** $\mathbf{v} = \cos t\,\mathbf{i} + (-2\sin 2t)\,\mathbf{j}$ | A1 A1 |
---\begin{enumerate}
\item A particle $P$ is moving in a plane. At time $t$ seconds the position vector of $P$ is $\mathbf { r }$ metres and the velocity of $P$ is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = \frac { \pi } { 2 } , P$ is instantaneously at rest at the point with position vector $( \mathbf { i } - \mathbf { j } ) \mathrm { m }$.
\end{enumerate}
Given that $\mathbf { r }$ satisfies the differential equation
$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = ( 3 \sin t ) \mathbf { i }$$
find $\mathbf { v }$ in terms of $t$.\\
(13)\\
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel M5 2016 Q2 [13]}}