Edexcel M5 2011 June — Question 2 10 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2011
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyChallenging +1.8 This is a second-order vector differential equation requiring complementary function (solving auxiliary equation with roots ±2), particular integral (trying e^t form), and applying two initial conditions to find four constants. While methodical, it demands facility with vector DEs, exponential solutions, and careful algebraic manipulation—significantly above average A-level difficulty but standard for M5.
Spec1.10a Vectors in 2D: i,j notation and column vectors4.10e Second order non-homogeneous: complementary + particular integral

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]

AnswerMarks
\(m^2 - 4 = 0 \Rightarrow m = 2\text{or } -2\)M1
CF is \(r = \mathbf{Ae}^{2t} + \mathbf{B e}^{-2t}\)A1
PI try \(\mathbf{r} = \mathbf{Ce}^t\)
\(\mathbf{r} = \mathbf{Ce}^t\)B1
\(\mathbf{\dot{r}} = \mathbf{Ce}^t\)
\(\mathbf{\ddot{r}} = \mathbf{Ce}^t\)
\(\mathbf{Ce}^t - 4\mathbf{Ce}^t = -3\mathbf{e}^t\mathbf{j}\)M1
\(\mathbf{C} = \mathbf{j}\)A1
GS is \(\mathbf{r} = \mathbf{A e}^{2t} + \mathbf{B e}^{-2t} + \mathbf{je}^t\)A1
\(\mathbf{v} = 2\mathbf{A e}^{2t} - 2\mathbf{B e}^{-2t} + \mathbf{je}^t\)M1
\(t = 0, \mathbf{r} = \mathbf{0}, \mathbf{v} = 2\mathbf{i} + \mathbf{j}\)M1
\(\mathbf{0} = \mathbf{A} + \mathbf{B} + \mathbf{j}\)
\(2\mathbf{i} + \mathbf{j} = 2\mathbf{A} - 2\mathbf{B} + \mathbf{j}\)A1
\(\mathbf{i} = \mathbf{A} - \mathbf{B}\)
\(\mathbf{A} = \frac{1}{2}(\mathbf{i} - \mathbf{j}); \mathbf{B} = -\frac{1}{2}(\mathbf{i} + \mathbf{j})\)
\(\mathbf{r} = \frac{1}{2}(\mathbf{i} - \mathbf{j}\mathbf{e}^{2t} - \frac{1}{2}(\mathbf{i} + \mathbf{j}\mathbf{e}^{-2t} + \mathbf{je}^t\)A1

Total: 10 marks

$m^2 - 4 = 0 \Rightarrow m = 2\text{or } -2$ | M1 |
CF is $r = \mathbf{Ae}^{2t} + \mathbf{B e}^{-2t}$ | A1 |
PI try $\mathbf{r} = \mathbf{Ce}^t$ | |
$\mathbf{r} = \mathbf{Ce}^t$ | B1 |
$\mathbf{\dot{r}} = \mathbf{Ce}^t$ | |
$\mathbf{\ddot{r}} = \mathbf{Ce}^t$ | |
$\mathbf{Ce}^t - 4\mathbf{Ce}^t = -3\mathbf{e}^t\mathbf{j}$ | M1 |
$\mathbf{C} = \mathbf{j}$ | A1 |
GS is $\mathbf{r} = \mathbf{A e}^{2t} + \mathbf{B e}^{-2t} + \mathbf{je}^t$ | A1 |
$\mathbf{v} = 2\mathbf{A e}^{2t} - 2\mathbf{B e}^{-2t} + \mathbf{je}^t$ | M1 |
$t = 0, \mathbf{r} = \mathbf{0}, \mathbf{v} = 2\mathbf{i} + \mathbf{j}$ | M1 |
$\mathbf{0} = \mathbf{A} + \mathbf{B} + \mathbf{j}$ | |
$2\mathbf{i} + \mathbf{j} = 2\mathbf{A} - 2\mathbf{B} + \mathbf{j}$ | A1 |
$\mathbf{i} = \mathbf{A} - \mathbf{B}$ | |
$\mathbf{A} = \frac{1}{2}(\mathbf{i} - \mathbf{j}); \mathbf{B} = -\frac{1}{2}(\mathbf{i} + \mathbf{j})$ | |
$\mathbf{r} = \frac{1}{2}(\mathbf{i} - \mathbf{j}\mathbf{e}^{2t} - \frac{1}{2}(\mathbf{i} + \mathbf{j}\mathbf{e}^{-2t} + \mathbf{je}^t$ | A1 |

Total: 10 marks

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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]

\hfill \mbox{\textit{Edexcel M5 2011 Q2 [10]}}