A2 June 2023 Paper 1 Q17

OCR MEICurrent spec24 marksFirst Order Differentials Second Order Differentials

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

\[\dfrac{\mathrm{d}x}{\mathrm{d}t} = kx - ay \quad \text{and} \quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = ky - bx,\]

where \(k\), \(a\) and \(b\) are positive constants.

(a)

(i) Show that the general solution for \(x\) is \(x = A\mathrm{e}^{(k+n)t} + B\mathrm{e}^{(k-n)t}\), where \(n = \sqrt{ab}\) and \(A\) and \(B\) are arbitrary constants. [6]

(ii) Hence find the general solution for \(y\) in terms of \(A\), \(B\), \(k\), \(n\), \(a\) and \(t\). [2]

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.

(b) When \(t = 0\), the numbers present of species X and Y in this area are \(x_0\) and \(y_0\) respectively.

(i) Show that \(x = \tfrac{1}{2}(x_0 - 2y_0)\mathrm{e}^{0.035t} + \tfrac{1}{2}(x_0 + 2y_0)\mathrm{e}^{-0.005t}\). [3]

(ii) Hence show that \(y = \tfrac{1}{4}(x_0 + 2y_0)\mathrm{e}^{-0.005t} - \tfrac{1}{4}(x_0 - 2y_0)\mathrm{e}^{0.035t}\). [1]

(c) 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.

(i) What numbers of each species will be present after 25 years? [2]

(ii) In this question you must show detailed reasoning.
When will the numbers of the two species be equal? [4]

(iii) Does either species ever disappear from the area? Justify your answer. [3]

(d) 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.

(i) Identify a relationship between \(x_0\) and \(y_0\) where the model does not predict this outcome. [1]

(ii) Explain what the model predicts in the long term for this exceptional case. [2]

Mark scheme (a)

Scheme	Marks	AO
(i) \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = k\dfrac{\mathrm{d}x}{\mathrm{d}t} - a\dfrac{\mathrm{d}y}{\mathrm{d}t}\)	M1*	3.1a
\(= k\dfrac{\mathrm{d}x}{\mathrm{d}t} - a(ky - bx)\)	M1dep	1.1
\(= k\dfrac{\mathrm{d}x}{\mathrm{d}t} + abx - k\left(kx - \dfrac{\mathrm{d}x}{\mathrm{d}t}\right)\)	M1dep	3.1a
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} - 2k\dfrac{\mathrm{d}x}{\mathrm{d}t} + (k^2 - n^2)x = 0\)	A1	2.1
AE \(m^2 - 2km + (k^2 - n^2) = 0 \Rightarrow m = k \pm n\)	M1	2.1
Hence GS is \(x = A\mathrm{e}^{(k+n)t} + B\mathrm{e}^{(k-n)t}\)	A1	2.3
	[6]
(ii) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = A(k + n)\mathrm{e}^{(k+n)t} + B(k - n)\mathrm{e}^{(k-n)t}\)	M1	2.1
\(y = \frac{n}{a}\left(-A\mathrm{e}^{(k+n)t} + B\mathrm{e}^{(k-n)t}\right)\)	A1	2.2a
	[2]

Notes

(a)(i)

M1*: differentiate \(\frac{\mathrm{d}x}{\mathrm{d}t}\) wrt \(t\)

M1dep: (1st) substitution for \(\frac{\mathrm{d}y}{\mathrm{d}t}\)

M1dep: (2nd) substitution for \(y\)

A1: (1st) correct differential equation \(= 0\). Could see \(ab\) instead of \(n^2\).

M1: giving and solving their AE

A1: (2nd) AG www

Alternative method

Scheme	Marks
\(y = \dfrac{k}{a}x - \dfrac{1}{a}\dfrac{\mathrm{d}x}{\mathrm{d}t} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{k}{a}\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{1}{a}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}\)	M1*
\(\dfrac{k}{a}\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{1}{a}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = ky - bx\)	M1dep
\(\dfrac{k}{a}\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{1}{a}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = k\left(\dfrac{k}{a}x - \dfrac{1}{a}\dfrac{\mathrm{d}x}{\mathrm{d}t}\right) - bx\)	M1dep
\(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} - 2k\dfrac{\mathrm{d}x}{\mathrm{d}t} + (k^2 - n^2)x = 0\)	A1
AE \(m^2 - 2km + (k^2 - n^2) = 0 \Rightarrow m = k \pm n\)	M1
Hence GS is \(x = A\mathrm{e}^{(k+n)t} + B\mathrm{e}^{(k-n)t}\)	A1

M1*: differentiate \(y\) wrt \(t\)

M1dep: (1st) substitution for \(\frac{\mathrm{d}y}{\mathrm{d}t}\)

M1dep: (2nd) substitution for \(y\)

A1: (1st) correct differential equation \(= 0\). Could see \(ab\) instead of \(n^2\).

M1: giving and solving their AE

A1: (2nd) AG www

(a)(ii)

M1: differentiation of \(x\) must be seen

A1: may not be factorised but must be simplified

Mark scheme (b)

Scheme	Marks	AO
(i) \(n = \sqrt{0.0004} = 0.02 \Rightarrow k + n = 0.035\) and \(k - n = -0.005\)	M1	1.1
\(x_0 = A + B, \quad y_0 = -\dfrac{1}{2}(A - B)\) \(\Rightarrow A = \dfrac{1}{2}(x_0 - 2y_0), \quad B = \dfrac{1}{2}(x_0 + 2y_0)\)	M1	3.3
\(x = \dfrac{1}{2}(x_0 - 2y_0)\mathrm{e}^{0.035t} + \dfrac{1}{2}(x_0 + 2y_0)\mathrm{e}^{-0.005t}\)	A1	2.2a
	[3]
(ii) \(y = \dfrac{1}{2}\left(-\dfrac{1}{2}(x_0 - 2y_0)\mathrm{e}^{0.035t} + \dfrac{1}{2}(x_0 + 2y_0)\mathrm{e}^{-0.005t}\right)\) \(= \dfrac{1}{4}(x_0 + 2y_0)\mathrm{e}^{-0.005t} - \dfrac{1}{4}(x_0 - 2y_0)\mathrm{e}^{0.035t}\)	B1	2.2a
	[1]

Notes

(b)(i)

M1: (1st) all three values established

(corrected from the printed mark scheme, which has \(n = \sqrt{0.004}\); \(n = \sqrt{ab} = \sqrt{0.04 \times 0.01} = \sqrt{0.0004} = 0.02\). The same correction applies in the alternative method.)

M1: (2nd) a method to find both A and B (ft their \(y\)) from expressions for \(x_0\) and \(y_0\)

A1: AG dependent upon both M1s

Alternative method

Scheme	Marks
\(n = \sqrt{0.0004} = 0.02 \Rightarrow k + n = 0.035, k - n = -0.005\)	M1
\(\frac{\mathrm{d}x}{\mathrm{d}t} = 0.015x_0 - 0.04y_0, \frac{\mathrm{d}x}{\mathrm{d}t} = 0.035A - 0.005B\) \(\Rightarrow A = \dfrac{1}{2}(x_0 - 2y_0), \quad B = \dfrac{1}{2}(x_0 + 2y_0)\)	M1
\(x = \dfrac{1}{2}(x_0 - 2y_0)\mathrm{e}^{0.035t} + \dfrac{1}{2}(x_0 + 2y_0)\mathrm{e}^{-0.005t}\)	A1

M1: (1st) all three values established

M1: (2nd) a method to find both A and B from two expressions for \(\frac{\mathrm{d}x}{\mathrm{d}t}\)

A1: AG dependent upon both M1s

(b)(ii)

B1: AG first step or equivalent substitution must be seen

Mark scheme (c)

Scheme	Marks	AO
(i) \(x = -50\mathrm{e}^{0.875} + 550\mathrm{e}^{-0.125}\), \(y = 275\mathrm{e}^{-0.125} + 25\mathrm{e}^{0.875}\)	M1	1.1
\(x = 365, y = 302\)	A1	2.2b
	[2]
(ii) DR \(-50\mathrm{e}^{0.035t} + 550\mathrm{e}^{-0.005t} = 275\mathrm{e}^{-0.005t} + 25\mathrm{e}^{0.035t}\)	M1	3.1b
\(275\mathrm{e}^{-0.005t} = 75\mathrm{e}^{0.035t}\)	M1	2.1
\(0.04t = \ln\left(\dfrac{11}{3}\right)\)	M1	3.1a
\(t = 32.5\), so numbers equal after about 32 or 33 years	A1	2.2a
	[4]
(iii) \(x\) does become zero, as \(550\mathrm{e}^{-0.005t} \to 0\) as \(t\) increases, and \(-50\mathrm{e}^{0.035t}\) is always negative	M1 A1	3.4 3.4
\(y\) is the sum of two positive terms, so is never zero	B1	3.5a
	[3]

Notes

(c)(i)

M1: for either
may be implied by awrt 365 or 303

A1: for both correct
allow \(y = 303\)

(c)(ii)

M1: (1st) equating \(x\) and \(y\) with \(x_0\), \(y_0\) substituted

M1: (2nd) collecting like terms

M1: (3rd) oe. Taking logs for a single term in \(t\).

(c)(iii)

M1: for correct conclusion with some explanation

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

B1: reason must be given, e.g. \(y = 0\) when \(t = 25\ln(-11)\) which is undefined or \(\mathrm{e}^{0.04t} = -11\) has no solution. Implied by M1A1 without further working seen unless incorrect conclusion for \(y\).

Mark scheme (d)

Scheme	Marks	AO
(i) \(x_0 = 2y_0\)	B1	3.5b
	[1]
(ii) \(x = C\mathrm{e}^{-0.005t}, \quad y = \frac{1}{2}C\mathrm{e}^{-0.005t}\)	M1	3.3
so both population numbers tend to zero	A1	2.4
	[2]

Notes

(d)(i)

B1: oe. Subscripts must be seen.

(d)(ii)

M1: oe; e.g. \(C\) is \(\frac{1}{2}(x_0 + 2y_0)\) or \(2y_0\) or \(x_0\)

A1: oe, e.g. both species disappear. Must be supported by explanation.