Question 2 - A-Level Maths

Edexcel M5 2009 June — Question 2 11 marks

Exam Board	Edexcel
Module	M5 (Mechanics 5)
Year	2009
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Particular solution with initial conditions
Difficulty	Challenging +1.2 This is a second-order vector differential equation requiring complementary function (CF) and particular integral (PI) with initial conditions. While it involves multiple steps (finding CF with auxiliary equation, finding PI for exponential forcing, applying two initial conditions), the techniques are standard for Further Maths mechanics. The vector form adds mild complexity but components separate independently. More challenging than routine integration but less demanding than proof-based or multi-concept synthesis questions.
Spec	1.10b Vectors in 3D: i,j,k notation 4.10e Second order non-homogeneous: complementary + particular integral

2. At time $t$ seconds, the position vector of a particle $P$ is $\mathbf { r }$ metres, where $\mathbf { r }$ satisfies the vector differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = \mathrm { e } ^ { 2 t } \mathbf { j }$$ When $t = 0 , P$ has position vector $( \mathbf { i } + \mathbf { j } ) \mathrm { m }$ and velocity $2 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find an expression for $\mathbf { r }$ in terms of $t$.

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Question 2:

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Consider j component: $\ddot{y} + 4y = e^{2t}$	M1	Separating into components
Auxiliary equation: $\lambda^2 + 4 = 0 \Rightarrow \lambda = \pm 2i$	M1
Complementary function: $y = A\cos 2t + B\sin 2t$	A1
Particular integral: try $y = ke^{2t}$, then $4ke^{2t} + 4ke^{2t} = e^{2t}$ so $k = \frac{1}{8}$	M1 A1
General solution (j): $y = A\cos 2t + B\sin 2t + \frac{1}{8}e^{2t}$	A1
Consider i component: $\ddot{x} + 4x = 0$	M1
$x = C\cos 2t + D\sin 2t$	A1
Apply initial conditions: $t=0$, $\mathbf{r} = \mathbf{i} + \mathbf{j}$, $\dot{\mathbf{r}} = 2\mathbf{i}$	M1
For x: $x(0)=1 \Rightarrow C=1$; $\dot{x}(0)=2 \Rightarrow 2D=2 \Rightarrow D=1$	A1
For y: $y(0)=1 \Rightarrow A + \frac{1}{8}=1 \Rightarrow A=\frac{7}{8}$; $\dot{y}(0)=0 \Rightarrow 2B + \frac{1}{4}=0 \Rightarrow B=-\frac{1}{8}$	A1
$\mathbf{r} = (\cos 2t + \sin 2t)\mathbf{i} + \left(\frac{7}{8}\cos 2t - \frac{1}{8}\sin 2t + \frac{1}{8}e^{2t}\right)\mathbf{j}$	A1	cao

# Question 2:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Consider **j** component: $\ddot{y} + 4y = e^{2t}$ | M1 | Separating into components |
| Auxiliary equation: $\lambda^2 + 4 = 0 \Rightarrow \lambda = \pm 2i$ | M1 | |
| Complementary function: $y = A\cos 2t + B\sin 2t$ | A1 | |
| Particular integral: try $y = ke^{2t}$, then $4ke^{2t} + 4ke^{2t} = e^{2t}$ so $k = \frac{1}{8}$ | M1 A1 | |
| General solution (j): $y = A\cos 2t + B\sin 2t + \frac{1}{8}e^{2t}$ | A1 | |
| Consider **i** component: $\ddot{x} + 4x = 0$ | M1 | |
| $x = C\cos 2t + D\sin 2t$ | A1 | |
| Apply initial conditions: $t=0$, $\mathbf{r} = \mathbf{i} + \mathbf{j}$, $\dot{\mathbf{r}} = 2\mathbf{i}$ | M1 | |
| For x: $x(0)=1 \Rightarrow C=1$; $\dot{x}(0)=2 \Rightarrow 2D=2 \Rightarrow D=1$ | A1 | |
| For y: $y(0)=1 \Rightarrow A + \frac{1}{8}=1 \Rightarrow A=\frac{7}{8}$; $\dot{y}(0)=0 \Rightarrow 2B + \frac{1}{4}=0 \Rightarrow B=-\frac{1}{8}$ | A1 | |
| $\mathbf{r} = (\cos 2t + \sin 2t)\mathbf{i} + \left(\frac{7}{8}\cos 2t - \frac{1}{8}\sin 2t + \frac{1}{8}e^{2t}\right)\mathbf{j}$ | A1 | cao |

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

$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 4 \mathbf { r } = \mathrm { e } ^ { 2 t } \mathbf { j }$$

When $t = 0 , P$ has position vector $( \mathbf { i } + \mathbf { j } ) \mathrm { m }$ and velocity $2 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.

Find an expression for $\mathbf { r }$ in terms of $t$.\\

\hfill \mbox{\textit{Edexcel M5 2009 Q2 [11]}}

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Q1 7 Q2 11 Q3 9 Q4 13 Q5 16 Q6 19