| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2003 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Standard linear first order - vector form |
| Difficulty | Standard +0.8 This is a vector differential equation requiring integrating factor method (IF = e^t), component-wise integration with exponentials, and finding a cartesian path equation by eliminating the parameter. While the technique is standard for Further Maths, the vector context, exponential manipulation, and multi-step nature make it moderately challenging—above average but not exceptionally difficult. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks |
|---|---|
| Multiplying through: \(\frac{d}{dt}(re^t) = (i - j)e^{-t}\) | M1A1 |
| Integrating: \(re^t = -(i - j)e^{-t} + c\) | M1 A1 ft |
| Using \(\mathbf{r} = 0, t = 0\) to find \(c\): \(c = i - j\) | M1 |
| \(\mathbf{r} = -(i - j)e^{-2t} + (i - j)e^{-t}\) | A1 (7) |
| (b) Writing \(\mathbf{r} = f(t)i + g(t)j\) or \(x = f(t), y = g(t)\) and attempt to eliminate \(t\) | M1 |
| \(y = -x\) | A1 (2) |
| Answer | Marks |
|---|---|
| (a) AE: \(m + 1 = 0 \Rightarrow \mathbf{r} = Ae^{-t}\) [Form of PI: \(\mathbf{r} = Be^{-2t}\)] | B1 |
| Equation for PI: \(-2e^{-2t}B + Be^{-2t} = (i - j)e^{-2}\) | M1 A1 |
| \(\mathbf{B} = -(i - j)\) | A1 ft |
| General Solution: \(\mathbf{r} = Ae^{-t} + (-i + j)e^{-2t}\) | M1 |
| Using \(\mathbf{r} = 0, t = 0\) to find \(A\) | M1 |
| \(\mathbf{r} = (i - j)e^{-t} + (-i + j)e^{-2t}\) | A1 (7) |
**(a)** Integrating factor approach: $IF = e^{\int i \, dt} = e^t$
Multiplying through: $\frac{d}{dt}(re^t) = (i - j)e^{-t}$ | M1A1 |
Integrating: $re^t = -(i - j)e^{-t} + c$ | M1 A1 ft |
Using $\mathbf{r} = 0, t = 0$ to find $c$: $c = i - j$ | M1 |
$\mathbf{r} = -(i - j)e^{-2t} + (i - j)e^{-t}$ | A1 (7) |
**(b)** Writing $\mathbf{r} = f(t)i + g(t)j$ or $x = f(t), y = g(t)$ and attempt to eliminate $t$ | M1 |
$y = -x$ | A1 (2) |
**Total: (9 marks)**
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# Question 2 (Alt.)
**(a)** AE: $m + 1 = 0 \Rightarrow \mathbf{r} = Ae^{-t}$ [Form of PI: $\mathbf{r} = Be^{-2t}$] | B1 |
Equation for PI: $-2e^{-2t}B + Be^{-2t} = (i - j)e^{-2}$ | M1 A1 |
$\mathbf{B} = -(i - j)$ | A1 ft |
General Solution: $\mathbf{r} = Ae^{-t} + (-i + j)e^{-2t}$ | M1 |
Using $\mathbf{r} = 0, t = 0$ to find $A$ | M1 |
$\mathbf{r} = (i - j)e^{-t} + (-i + j)e^{-2t}$ | A1 (7) |
**Total: (9 marks)**
---2. With respect to a fixed origin $O$, the position vector, $\mathbf { r }$ metres, of a particle $P$ at time $t$ seconds satisfies
$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + \mathbf { r } = ( \mathbf { i } - \mathbf { j } ) \mathrm { e } ^ { - 2 t } .$$
Given that $P$ is at $O$ when $t = 0$, find
\begin{enumerate}[label=(\alph*)]
\item $\mathbf { r }$ in terms of $t$,
\item a cartesian equation of the path of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2003 Q2 [9]}}