Edexcel M5 2005 June — Question 2 7 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2005
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeStandard linear first order - vector form
DifficultyStandard +0.3 This is a straightforward application of the integrating factor method to a vector differential equation. While it involves vectors, the technique is standard: find integrating factor e^(2t), multiply through, integrate, and apply initial conditions. The exponential integration is routine and the vector nature doesn't add significant complexity since components can be handled separately. Slightly above average difficulty due to the vector context and being from M5, but still a textbook exercise requiring no novel insight.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.10c Integrating factor: first order equations

2. At time \(t\) seconds the position vector of a particle \(P\), relative to a fixed origin \(O\), is \(\mathbf { r }\) metres, where \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 3 \mathrm { e } ^ { - t } \mathbf { j }$$ Given that \(\mathbf { r } = 2 \mathbf { i } - \mathbf { j }\) when \(t = 0\), find \(\mathbf { r }\) in terms of \(t\).
(Total 7 marks)

Question 2:
AnswerMarks Guidance
Working/AnswerMark Guidance
Integrating factor: \(e^{\int 2\,dt} = e^{2t}\)M1
\(\frac{d}{dt}(e^{2t}\mathbf{r}) = 3e^{-t} \cdot e^{2t}\,\mathbf{j} = 3e^{t}\mathbf{j}\)A1 Correct equation after multiplying through
\(e^{2t}\mathbf{r} = 3e^{t}\mathbf{j} + \mathbf{c}\)M1 Integrating both sides
\(\mathbf{r} = 3e^{-t}\mathbf{j} + e^{-2t}\mathbf{c}\)A1
At \(t=0\): \(\mathbf{r} = 2\mathbf{i} - \mathbf{j}\), so \(2\mathbf{i} - \mathbf{j} = 3\mathbf{j} + \mathbf{c}\)M1 Applying initial condition
\(\mathbf{c} = 2\mathbf{i} - 4\mathbf{j}\)A1
\(\mathbf{r} = 2e^{-2t}\mathbf{i} + (3e^{-t} - 4e^{-2t})\mathbf{j}\)A1 Fully correct answer
# Question 2:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Integrating factor: $e^{\int 2\,dt} = e^{2t}$ | M1 | |
| $\frac{d}{dt}(e^{2t}\mathbf{r}) = 3e^{-t} \cdot e^{2t}\,\mathbf{j} = 3e^{t}\mathbf{j}$ | A1 | Correct equation after multiplying through |
| $e^{2t}\mathbf{r} = 3e^{t}\mathbf{j} + \mathbf{c}$ | M1 | Integrating both sides |
| $\mathbf{r} = 3e^{-t}\mathbf{j} + e^{-2t}\mathbf{c}$ | A1 | |
| At $t=0$: $\mathbf{r} = 2\mathbf{i} - \mathbf{j}$, so $2\mathbf{i} - \mathbf{j} = 3\mathbf{j} + \mathbf{c}$ | M1 | Applying initial condition |
| $\mathbf{c} = 2\mathbf{i} - 4\mathbf{j}$ | A1 | |
| $\mathbf{r} = 2e^{-2t}\mathbf{i} + (3e^{-t} - 4e^{-2t})\mathbf{j}$ | A1 | Fully correct answer |

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2. At time $t$ seconds the position vector of a particle $P$, relative to a fixed origin $O$, is $\mathbf { r }$ metres, where $\mathbf { r }$ satisfies the differential equation

$$\frac { \mathrm { d } \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = 3 \mathrm { e } ^ { - t } \mathbf { j }$$

Given that $\mathbf { r } = 2 \mathbf { i } - \mathbf { j }$ when $t = 0$, find $\mathbf { r }$ in terms of $t$.\\
(Total 7 marks)\\

\hfill \mbox{\textit{Edexcel M5 2005 Q2 [7]}}