Question 7 - A-Level Maths

Edexcel FP2 2005 June — Question 7 14 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2005
Session	June
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Modeling context with interpretation
Difficulty	Standard +0.8 This is a standard FP2 second-order differential equation question requiring auxiliary equation solution, particular integral, and applying initial conditions. Part (c) adds mild complexity by requiring optimization (finding minimum via calculus) and justification, but the techniques are all routine for Further Maths students. The modeling context is superficial and doesn't add conceptual difficulty.
Spec	4.10a General/particular solutions: of differential equations 4.10e Second order non-homogeneous: complementary + particular integral

7. (a) Find the general solution of the differential equation $$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 2 t + 9$$ (b) Find the particular solution of this differential equation for which $x = 3$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = - 1$ when $t = 0$. The particular solution in part (b) is used to model the motion of a particle $P$ on the $x$-axis. At time $t$ seconds $( t \geq 0 ) , P$ is $x$ metres from the origin $O$.
(c) Show that the minimum distance between $O$ and $P$ is $\frac { 1 } { 2 } ( 5 + \ln 2 ) \mathrm { m }$ and justify that the distance is a minimum.
(4)(Total 14 marks)

Show mark scheme Show mark scheme source

Question 7:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$2m^2 + 5m + 2 = 0$	M1	Attempt auxiliary equation
$m = -\frac{1}{2}, -2$
$x_{CF} = Ae^{-2t} + Be^{-\frac{1}{2}t}$	A1	C.F.
Particular Integral: $x = pt + q$	B1	P.I.
$\dot{x} = p,\ \ddot{x} = 0$ and substitute	M1
$5p + 2q + 2pt = 2t + q \Rightarrow p = 1, q = 2$	A1
General solution $x = Ae^{-2t} + Be^{-\frac{1}{2}t} + t + 2$	A1ft	ft on $m$, $p$, $q$

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x = 3, t = 0 \Rightarrow 3 = A + B + 2$ (or $A + B = 1$)	M1
$\dot{x} = -2Ae^{-2t} - \frac{1}{2}Be^{-\frac{1}{2}t} + 1$	M1	Attempt $\dot{x}$
$\dot{x} = -1, t = 0 \Rightarrow -1 = -2A - \frac{1}{2}B + 1$ (or $4A + B = 4$)	A1	2 correct equations
Solving $\Rightarrow A = 1, B = 0$ and $x = e^{-2t} + t + 2$	A1	4 marks total

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\dot{x} = -2e^{-2t} + 1 = 0$	M1	Set $\dot{x} = 0$
$t = \frac{1}{2}\ln 2$	A1
$\ddot{x} = 4e^{-2t} > 0\ (\forall t)\ \therefore$ min	M1
$\min x = e^{-\ln 2} + \frac{1}{2}\ln 2 + 2 = \frac{1}{2} + \frac{1}{2}\ln 2 + 2$
$= \frac{1}{2}(5 + \ln 2)$	A1	c.s.o.; 4 marks total

# Question 7:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2m^2 + 5m + 2 = 0$ | M1 | Attempt auxiliary equation |
| $m = -\frac{1}{2}, -2$ | |  |
| $x_{CF} = Ae^{-2t} + Be^{-\frac{1}{2}t}$ | A1 | C.F. |
| Particular Integral: $x = pt + q$ | B1 | P.I. |
| $\dot{x} = p,\ \ddot{x} = 0$ and substitute | M1 | |
| $5p + 2q + 2pt = 2t + q \Rightarrow p = 1, q = 2$ | A1 | |
| General solution $x = Ae^{-2t} + Be^{-\frac{1}{2}t} + t + 2$ | A1ft | ft on $m$, $p$, $q$ |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 3, t = 0 \Rightarrow 3 = A + B + 2$ (or $A + B = 1$) | M1 | |
| $\dot{x} = -2Ae^{-2t} - \frac{1}{2}Be^{-\frac{1}{2}t} + 1$ | M1 | Attempt $\dot{x}$ |
| $\dot{x} = -1, t = 0 \Rightarrow -1 = -2A - \frac{1}{2}B + 1$ (or $4A + B = 4$) | A1 | 2 correct equations |
| Solving $\Rightarrow A = 1, B = 0$ and $x = e^{-2t} + t + 2$ | A1 | 4 marks total |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dot{x} = -2e^{-2t} + 1 = 0$ | M1 | Set $\dot{x} = 0$ |
| $t = \frac{1}{2}\ln 2$ | A1 | |
| $\ddot{x} = 4e^{-2t} > 0\ (\forall t)\ \therefore$ min | M1 | |
| $\min x = e^{-\ln 2} + \frac{1}{2}\ln 2 + 2 = \frac{1}{2} + \frac{1}{2}\ln 2 + 2$ | | |
| $= \frac{1}{2}(5 + \ln 2)$ | A1 | c.s.o.; 4 marks total |

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7. (a) Find the general solution of the differential equation

$$2 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 x = 2 t + 9$$

(b) Find the particular solution of this differential equation for which $x = 3$ and $\frac { \mathrm { d } x } { \mathrm {~d} t } = - 1$ when $t = 0$.

The particular solution in part (b) is used to model the motion of a particle $P$ on the $x$-axis. At time $t$ seconds $( t \geq 0 ) , P$ is $x$ metres from the origin $O$.\\
(c) Show that the minimum distance between $O$ and $P$ is $\frac { 1 } { 2 } ( 5 + \ln 2 ) \mathrm { m }$ and justify that the distance is a minimum.\\
(4)(Total 14 marks)\\

\hfill \mbox{\textit{Edexcel FP2 2005 Q7 [14]}}

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