Question 17 - A-Level Maths

OCR MEI Further Pure Core 2023 June — Question 17 24 marks

Exam Board	OCR MEI
Module	Further Pure Core (Further Pure Core)
Year	2023
Session	June
Marks	24
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	First order differential equations (integrating factor)
Type	Deriving the differential equation
Difficulty	Challenging +1.2 This is a structured coupled differential equations problem with clear guidance through each step. While it involves Further Maths content (systems of DEs), the question provides the form of the solution to show, making it primarily verification rather than discovery. The algebraic manipulation required is moderate but systematic, and part (c) is straightforward substitution. Harder than average A-level due to Further Maths content, but the scaffolding keeps it accessible.
Spec	4.10h Coupled systems: simultaneous first order DEs

17 Two similar species, X and Y , of a small mammal compete for food and habitat. A model of this competition assumes, in a particular area, the following.

In the absence of the other species, each species would increase at a rate proportional to the number present with the same constant of proportionality in each case.
The competition reduces the rate of increase of each species by an amount proportional to the number of the other species present.

So if the numbers of species X and Y present at time \(t\) years are \(x\) and \(y\) respectively, the model gives the differential equations \(\frac { d x } { d t } = k x - a y\) and \(\frac { d y } { d t } = k y - b x\),
where \(k , a\) and \(b\) are positive constants.

1. Show that the general solution for \(x\) is \(x = A e ^ { ( k + n ) t } + B e ^ { ( k - n ) t }\), where \(n = \sqrt { a b }\) and \(A\) and \(B\) are arbitrary constants.
2. Hence find the general solution for \(y\) in terms of \(A , B , k , n , a\) and \(t\). Observations suggest that suitable values for the model are \(k = 0.015 , a = 0.04\) and \(b = 0.01\). You should use these values in the rest of this question.
When \(t = 0\), the numbers present of species X and Y in this area are \(x _ { 0 }\) and \(y _ { 0 }\) respectively.
1. Show that \(\mathrm { x } = \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } - 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { 0.035 \mathrm { t } } + \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } + 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { - 0.005 \mathrm { t } }\).
2. Hence show that \(y = \frac { 1 } { 4 } \left( x _ { 0 } + 2 y _ { 0 } \right) e ^ { - 0.005 t } - \frac { 1 } { 4 } \left( x _ { 0 } - 2 y _ { 0 } \right) e ^ { 0.035 t }\).
Use initial values \(x _ { 0 } = 500\) and \(y _ { 0 } = 300\) with the results in part (b) to determine what the model predicts for each of the following questions.
1. What numbers of each species will be present after 25 years?
2. In this question you must show detailed reasoning. When will the numbers of the two species be equal?
3. Does either species ever disappear from the area? Justify your answer.
Different initial values will apply in other areas where the two species compete, but previous studies indicate that one species or the other will eventually dominate in any given area.
1. Identify a relationship between \(x _ { 0 }\) and \(y _ { 0 }\) where the model does not predict this outcome.
2. Explain what the model predicts in the long term for this exceptional case.

Show mark scheme Show mark scheme source

Question 17:

Answer	Marks	Guidance
17	(a)	(i)

= 𝑘 −𝑎

d𝑡(cid:2870) d𝑡 d𝑡

(cid:2914)(cid:3051)

= 𝑘 −𝑎(𝑘𝑦−𝑏𝑥)

(cid:2914)(cid:3047)

(cid:2914)(cid:3051) (cid:2914)(cid:3051)

= 𝑘 +𝑎𝑏𝑥−𝑘(cid:4672)𝑘𝑥− (cid:4673)

(cid:2914)(cid:3047) (cid:2914)(cid:3047)

d(cid:2870)𝑥 d𝑥

−2𝑘 +(𝑘(cid:2870)−𝑛(cid:2870))𝑥 = 0

d𝑡(cid:2870) d𝑡

AE 𝑚(cid:2870)−2𝑘𝑚+(𝑘(cid:2870)−𝑛(cid:2870))= 0 ⇒ 𝑚 = 𝑘±𝑛

Answer	Marks
Hence GS is 𝑥 = 𝐴𝑒((cid:3038)(cid:2878)(cid:3041))(cid:3047) +𝐵𝑒((cid:3038)(cid:2879)(cid:3041))(cid:3047)	M1*

M1dep

A1

M1

Answer	Marks
A1	3.1a

1.1 3.1a

2.1

Answer	Marks
2.3	(cid:2914)(cid:3051)

differentiate wrt t

(cid:2914)(cid:3047)

(cid:2914)(cid:3052)

substitution for

(cid:2914)(cid:3047)

substitution for 𝑦

correct differential equation = 0. Could see 𝑎𝑏 instead of 𝑛(cid:2870).

giving and solving their AE

AG www

Alternative method

𝑘 1d𝑥 d𝑦 𝑘d𝑥 1 d(cid:2870)𝑥

𝑦 = 𝑥− ⇒ = −

Answer	Marks	Guidance
𝑎 𝑎d𝑡 d𝑡 𝑎d𝑡 𝑎 d𝑡(cid:2870)	M1*	differentiate 𝑦 wrt t

𝑘d𝑥 1d(cid:2870)𝑥

− = 𝑘𝑦−𝑏𝑥

Answer	Marks	Guidance
𝑎d𝑡 𝑎d𝑡(cid:2870)	M1dep	(cid:2914)(cid:3052)

substitution for

(cid:2914)(cid:3047)

𝑘d𝑥 1d(cid:2870)𝑥 𝑘 1d𝑥

− = 𝑘(cid:3436) 𝑥− (cid:3440)−𝑏𝑥

Answer	Marks	Guidance
𝑎d𝑡 𝑎d𝑡(cid:2870) 𝑎 𝑎d𝑡	M1dep	substitution for 𝑦

d(cid:2870)𝑥 d𝑥

−2𝑘 +(𝑘(cid:2870)−𝑛(cid:2870))𝑥 = 0

Answer	Marks	Guidance
d𝑡(cid:2870) d𝑡	A1	correct differential equation = 0. Could see 𝑎𝑏 instead of 𝑛(cid:2870).
AE 𝑚(cid:2870)−2𝑘𝑚+(𝑘(cid:2870)−𝑛(cid:2870))= 0 ⇒ 𝑚 = 𝑘±𝑛	M1	giving and solving their AE
Hence GS is 𝑥 = 𝐴e((cid:3038)(cid:2878)(cid:3041))(cid:3047) +𝐵e((cid:3038)(cid:2879)(cid:3041))(cid:3047)	A1	AG www

[6]

Answer	Marks	Guidance
17	(a)	(ii)

(cid:2914)(cid:3047)

𝑦 = (cid:3041) (cid:3435)−𝐴e((cid:3038)(cid:2878)(cid:3041))(cid:3047) +𝐵e((cid:3038)(cid:2879)(cid:3041))(cid:3047)(cid:3439)

Answer	Marks
(cid:3028)	M1

A1

Answer	Marks
[2]	2.1
2.2a	differentiation of 𝑥 must be seen

may not be factorised but must be simplified

Answer	Marks	Guidance
17	(b)	(i)

= −0.005

1 𝑥 = 𝐴+𝐵, 𝑦 = − (𝐴−𝐵)

(cid:2868) (cid:2868) 2

1 1

⇒ 𝐴 = (𝑥 −2𝑦 ), 𝐵 = (𝑥 +2𝑦 )

2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)

1 1

𝑥 = (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047) + (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)

Answer	Marks
2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)	M1

M1

Answer	Marks
A1	1.1

3.3

Answer	Marks
2.2a	all three values established

a method to find both A and B (ft their y) from expressions

for 𝑥 and 𝑦

(cid:2868) (cid:2868)

AG dependent upon both M1s

Alternative method

Answer	Marks	Guidance
𝑛 = √0.004= 0.02 ⇒𝑘+𝑛 = 0.035, 𝑘−𝑛	M1	all three values established

= −0.005

(cid:2914)(cid:3051) (cid:2914)(cid:3051)

= 0.015𝑥 −0.04𝑦 , = 0.035𝐴−0.005𝐵

(cid:2868) (cid:2868)

(cid:2914)(cid:3047) (cid:2914)(cid:3047)

1 1

⇒ 𝐴 = (𝑥 −2𝑦 ), 𝐵 = (𝑥 +2𝑦 )

2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)

(cid:2914)(cid:3051)

a method to find both A and B from two expressions for

(cid:2914)(cid:3047)

M1

1 1

𝑥 = (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047) + (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)

Answer	Marks	Guidance
2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)	A1	AG dependent upon both M1s

[3]

Answer	Marks	Guidance
17	(b)	(ii)

𝑦 = (cid:3436)− (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)+ (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)(cid:3440)

2 2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)

1 1

= (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)− (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)

Answer	Marks	Guidance
4 (cid:2868) (cid:2868) 4 (cid:2868) (cid:2868)	B1
[1]	2.2a	AG first step or equivalent substitution must be seen
17	(c)	(i)

𝑦 = 275e(cid:2879)(cid:2868).(cid:2869)(cid:2870)(cid:2873)+25e(cid:2868).(cid:2876)(cid:2875)(cid:2873)

Answer	Marks
𝑥 = 365,𝑦 = 302	M1

A1

Answer	Marks
[2]	1.1
2.2b	for either

may be implied by awrt 365 or 303

for both correct

allow 𝑦 = 303

Answer	Marks	Guidance
17	(c)	(ii)

−50e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)+550e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047) = 275e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)+25e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)

275e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047) = 75e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)

11 0.04𝑡 = ln(cid:3436) (cid:3440)

3

Answer	Marks
𝑡 = 32.5, so numbers equal after about 32 or 33 years	M1

M1

A1

Answer	Marks
[4]	3.1b

2.1 3.1a

Answer	Marks
2.2a	equating 𝒙 and 𝒚 with 𝒙 , 𝒚 substituted

𝟎 𝟎

collecting like terms

oe. Taking logs for a single term in 𝒕.

Answer	Marks	Guidance
17	(c)	(iii)

and −50e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:2930) is always negative

Answer	Marks
y is the sum of two positive terms, so is never zero	M1

A1

B1

Answer	Marks
[3]	3.4

3.4

Answer	Marks
3.5a	M1 for correct conclusion with some explanation

A1 for complete explanation (e.g. x = 0 at t = 59.94...)

reason must be given, e.g. 𝑦 = 0 when 𝑡 = 25ln(−11)

which is undefined or e(cid:2868).(cid:2868)(cid:2872)(cid:3047) = −11 has no solution. Implied

by M1A1 without further working seen unless incorrect

conclusion for 𝑦.

Answer	Marks	Guidance
17	(d)	(i)
(cid:2868) (cid:2868)	B1
[1]	3.5b	oe. Subscripts must be seen.
17	(d)	(ii)

(cid:2870)

Answer	Marks
so both population numbers tend to zero	M1

A1

Answer	Marks
[2]	3.3
2.4	𝟏

oe; e.g. C is (𝒙 +𝟐𝒚 ) or 𝟐𝒚 or 𝒙

𝟎 𝟎 𝟎 𝟎

𝟐 oe, e.g. both species disappear. Must be supported by

explanation.

PMT

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Question 17:
17 | (a) | (i) | d(cid:2870)𝑥 d𝑥 d𝑦
= 𝑘 −𝑎
d𝑡(cid:2870) d𝑡 d𝑡
(cid:2914)(cid:3051)
= 𝑘 −𝑎(𝑘𝑦−𝑏𝑥)
(cid:2914)(cid:3047)
(cid:2914)(cid:3051) (cid:2914)(cid:3051)
= 𝑘 +𝑎𝑏𝑥−𝑘(cid:4672)𝑘𝑥− (cid:4673)
(cid:2914)(cid:3047) (cid:2914)(cid:3047)
d(cid:2870)𝑥 d𝑥
−2𝑘 +(𝑘(cid:2870)−𝑛(cid:2870))𝑥 = 0
d𝑡(cid:2870) d𝑡
AE 𝑚(cid:2870)−2𝑘𝑚+(𝑘(cid:2870)−𝑛(cid:2870))= 0 ⇒ 𝑚 = 𝑘±𝑛
Hence GS is 𝑥 = 𝐴𝑒((cid:3038)(cid:2878)(cid:3041))(cid:3047) +𝐵𝑒((cid:3038)(cid:2879)(cid:3041))(cid:3047) | M1*
M1dep
M1dep
A1
M1
A1 | 3.1a
1.1
3.1a
2.1
2.1
2.3 | (cid:2914)(cid:3051)
differentiate wrt t
(cid:2914)(cid:3047)
(cid:2914)(cid:3052)
substitution for
(cid:2914)(cid:3047)
substitution for 𝑦
correct differential equation = 0. Could see 𝑎𝑏 instead of 𝑛(cid:2870).
giving and solving their AE
AG www
Alternative method
𝑘 1d𝑥 d𝑦 𝑘d𝑥 1 d(cid:2870)𝑥
𝑦 = 𝑥− ⇒ = −
𝑎 𝑎d𝑡 d𝑡 𝑎d𝑡 𝑎 d𝑡(cid:2870) | M1* | differentiate 𝑦 wrt t
𝑘d𝑥 1d(cid:2870)𝑥
− = 𝑘𝑦−𝑏𝑥
𝑎d𝑡 𝑎d𝑡(cid:2870) | M1dep | (cid:2914)(cid:3052)
substitution for
(cid:2914)(cid:3047)
𝑘d𝑥 1d(cid:2870)𝑥 𝑘 1d𝑥
− = 𝑘(cid:3436) 𝑥− (cid:3440)−𝑏𝑥
𝑎d𝑡 𝑎d𝑡(cid:2870) 𝑎 𝑎d𝑡 | M1dep | substitution for 𝑦
d(cid:2870)𝑥 d𝑥
−2𝑘 +(𝑘(cid:2870)−𝑛(cid:2870))𝑥 = 0
d𝑡(cid:2870) d𝑡 | A1 | correct differential equation = 0. Could see 𝑎𝑏 instead of 𝑛(cid:2870).
AE 𝑚(cid:2870)−2𝑘𝑚+(𝑘(cid:2870)−𝑛(cid:2870))= 0 ⇒ 𝑚 = 𝑘±𝑛 | M1 | giving and solving their AE
Hence GS is 𝑥 = 𝐴e((cid:3038)(cid:2878)(cid:3041))(cid:3047) +𝐵e((cid:3038)(cid:2879)(cid:3041))(cid:3047) | A1 | AG www
[6]
17 | (a) | (ii) | (cid:2914)(cid:3051) = 𝐴(𝑘+𝑛)e((cid:3038)(cid:2878)(cid:3041))(cid:3047) +𝐵(𝑘−𝑛)e((cid:3038)(cid:2879)(cid:3041))(cid:3047)
(cid:2914)(cid:3047)
𝑦 = (cid:3041) (cid:3435)−𝐴e((cid:3038)(cid:2878)(cid:3041))(cid:3047) +𝐵e((cid:3038)(cid:2879)(cid:3041))(cid:3047)(cid:3439)
(cid:3028) | M1
A1
[2] | 2.1
2.2a | differentiation of 𝑥 must be seen
may not be factorised but must be simplified
17 | (b) | (i) | 𝑛 = √0.004= 0.02⇒ 𝑘+𝑛 = 0.035 and 𝑘−𝑛
= −0.005
1
𝑥 = 𝐴+𝐵, 𝑦 = − (𝐴−𝐵)
(cid:2868) (cid:2868) 2
1 1
⇒ 𝐴 = (𝑥 −2𝑦 ), 𝐵 = (𝑥 +2𝑦 )
2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)
1 1
𝑥 = (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047) + (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)
2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868) | M1
M1
A1 | 1.1
3.3
2.2a | all three values established
a method to find both A and B (ft their y) from expressions
for 𝑥 and 𝑦
(cid:2868) (cid:2868)
AG dependent upon both M1s
Alternative method
𝑛 = √0.004= 0.02 ⇒𝑘+𝑛 = 0.035, 𝑘−𝑛 | M1 | all three values established
= −0.005
(cid:2914)(cid:3051) (cid:2914)(cid:3051)
= 0.015𝑥 −0.04𝑦 , = 0.035𝐴−0.005𝐵
(cid:2868) (cid:2868)
(cid:2914)(cid:3047) (cid:2914)(cid:3047)
1 1
⇒ 𝐴 = (𝑥 −2𝑦 ), 𝐵 = (𝑥 +2𝑦 )
2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)
(cid:2914)(cid:3051)
a method to find both A and B from two expressions for
(cid:2914)(cid:3047)
M1
1 1
𝑥 = (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047) + (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)
2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868) | A1 | AG dependent upon both M1s
[3]
17 | (b) | (ii) | 1 1 1
𝑦 = (cid:3436)− (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)+ (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)(cid:3440)
2 2 (cid:2868) (cid:2868) 2 (cid:2868) (cid:2868)
1 1
= (𝑥 +2𝑦 )e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)− (𝑥 −2𝑦 )e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)
4 (cid:2868) (cid:2868) 4 (cid:2868) (cid:2868) | B1
[1] | 2.2a | AG first step or equivalent substitution must be seen
17 | (c) | (i) | 𝑥 = −50e(cid:2868).(cid:2876)(cid:2875)(cid:2873)+550e(cid:2879)(cid:2868).(cid:2869)(cid:2870)(cid:2873),
𝑦 = 275e(cid:2879)(cid:2868).(cid:2869)(cid:2870)(cid:2873)+25e(cid:2868).(cid:2876)(cid:2875)(cid:2873)
𝑥 = 365,𝑦 = 302 | M1
A1
[2] | 1.1
2.2b | for either
may be implied by awrt 365 or 303
for both correct
allow 𝑦 = 303
17 | (c) | (ii) | DR
−50e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)+550e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047) = 275e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)+25e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)
275e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047) = 75e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:3047)
11
0.04𝑡 = ln(cid:3436) (cid:3440)
3
𝑡 = 32.5, so numbers equal after about 32 or 33 years | M1
M1
M1
A1
[4] | 3.1b
2.1
3.1a
2.2a | equating 𝒙 and 𝒚 with 𝒙 , 𝒚 substituted
𝟎 𝟎
collecting like terms
oe. Taking logs for a single term in 𝒕.
17 | (c) | (iii) | x does become zero, as 550e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047) → 0 as t increases,
and −50e(cid:2868).(cid:2868)(cid:2871)(cid:2873)(cid:2930) is always negative
y is the sum of two positive terms, so is never zero | M1
A1
B1
[3] | 3.4
3.4
3.5a | M1 for correct conclusion with some explanation
A1 for complete explanation (e.g. x = 0 at t = 59.94...)
reason must be given, e.g. 𝑦 = 0 when 𝑡 = 25ln(−11)
which is undefined or e(cid:2868).(cid:2868)(cid:2872)(cid:3047) = −11 has no solution. Implied
by M1A1 without further working seen unless incorrect
conclusion for 𝑦.
17 | (d) | (i) | 𝑥 = 2𝑦
(cid:2868) (cid:2868) | B1
[1] | 3.5b | oe. Subscripts must be seen.
17 | (d) | (ii) | 𝑥 = 𝐶e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047), 𝑦 = (cid:2869) 𝐶e(cid:2879)(cid:2868).(cid:2868)(cid:2868)(cid:2873)(cid:3047)
(cid:2870)
so both population numbers tend to zero | M1
A1
[2] | 3.3
2.4 | 𝟏
oe; e.g. C is (𝒙 +𝟐𝒚 ) or 𝟐𝒚 or 𝒙
𝟎 𝟎 𝟎 𝟎
𝟐
oe, e.g. both species disappear. Must be supported by
explanation.
PMT
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17 Two similar species, X and Y , of a small mammal compete for food and habitat. A model of this competition assumes, in a particular area, the following.

\begin{itemize}
  \item In the absence of the other species, each species would increase at a rate proportional to the number present with the same constant of proportionality in each case.
  \item The competition reduces the rate of increase of each species by an amount proportional to the number of the other species present.
\end{itemize}

So if the numbers of species X and Y present at time $t$ years are $x$ and $y$ respectively, the model gives the differential equations\\
$\frac { d x } { d t } = k x - a y$ and $\frac { d y } { d t } = k y - b x$,\\
where $k , a$ and $b$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the general solution for $x$ is $x = A e ^ { ( k + n ) t } + B e ^ { ( k - n ) t }$, where $n = \sqrt { a b }$ and $A$ and $B$ are arbitrary constants.
\item Hence find the general solution for $y$ in terms of $A , B , k , n , a$ and $t$.

Observations suggest that suitable values for the model are $k = 0.015 , a = 0.04$ and $b = 0.01$. You should use these values in the rest of this question.
\end{enumerate}\item When $t = 0$, the numbers present of species X and Y in this area are $x _ { 0 }$ and $y _ { 0 }$ respectively.
\begin{enumerate}[label=(\roman*)]
\item Show that $\mathrm { x } = \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } - 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { 0.035 \mathrm { t } } + \frac { 1 } { 2 } \left( \mathrm { x } _ { 0 } + 2 \mathrm { y } _ { 0 } \right) \mathrm { e } ^ { - 0.005 \mathrm { t } }$.
\item Hence show that $y = \frac { 1 } { 4 } \left( x _ { 0 } + 2 y _ { 0 } \right) e ^ { - 0.005 t } - \frac { 1 } { 4 } \left( x _ { 0 } - 2 y _ { 0 } \right) e ^ { 0.035 t }$.
\end{enumerate}\item Use initial values $x _ { 0 } = 500$ and $y _ { 0 } = 300$ with the results in part (b) to determine what the model predicts for each of the following questions.
\begin{enumerate}[label=(\roman*)]
\item What numbers of each species will be present after 25 years?
\item In this question you must show detailed reasoning.

When will the numbers of the two species be equal?
\item Does either species ever disappear from the area? Justify your answer.
\end{enumerate}\item Different initial values will apply in other areas where the two species compete, but previous studies indicate that one species or the other will eventually dominate in any given area.
\begin{enumerate}[label=(\roman*)]
\item Identify a relationship between $x _ { 0 }$ and $y _ { 0 }$ where the model does not predict this outcome.
\item Explain what the model predicts in the long term for this exceptional case.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q17 [24]}}

This paper (17 questions)

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Q1 7 Q2 5 Q3 6 Q4 6 Q5 7 Q6 9 Q7 6 Q8 5 Q9 6 Q10 7 Q11 7 Q12 7 Q13 14 Q14 13 Q15 5 Q16 10 Q17 24