IB Mathematics AHL 5.18 Solutions of second order differential equation AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question examines a method for constructing phase portraits for non-linear coupled systems, using a predator-prey model as an example.
Gander Green wildlife park is home to populations of Czech geese (\( x \), measured in hundreds) and gray foxes (\( y \), measured in hundreds).
Studies indicate that the population dynamics of geese and foxes can be modeled by the following differential equations, where \( t \) is measured in years:
\[ \begin{cases} \dfrac{dx}{dt} = 2x – \dfrac{xy}{2} \\ \dfrac{dy}{dt} = -3y + xy \end{cases} \quad \text{for } x, y \geq 0 \]
(a)
At a given time, there are 500 geese and 500 foxes, represented by the coordinate pair (5, 5). Determine the rate of change of
(i) geese. [2]
(ii) foxes. [1]
At a given time, there are 500 geese and 500 foxes, represented by the coordinate pair (5, 5). Determine the rate of change of
(i) geese. [2]
(ii) foxes. [1]
The system has two equilibrium points: \( A(0,0) \) and \( B(p,q) \).
(b)
(i) Explain why \( A \) is an equilibrium point. [1]
(ii) Find the values of \( p \) and \( q \). [3]
(i) Explain why \( A \) is an equilibrium point. [1]
(ii) Find the values of \( p \) and \( q \). [3]
Near point \( A(0,0) \), the \( xy \) terms can be ignored, approximating the system as:
\[ \begin{cases} \dfrac{dx}{dt} = 2x \\ \dfrac{dy}{dt} = -3y \end{cases} \quad \text{for } x, y \geq 0 \]
(c)
By solving these differential equations,
(i) find an expression for \( x \) in terms of \( t \). [4]
(ii) find an expression for \( y \) in terms of \( t \). [1]
By solving these differential equations,
(i) find an expression for \( x \) in terms of \( t \). [4]
(ii) find an expression for \( y \) in terms of \( t \). [1]
(d)
(i) Using your answers from part (c), show that phase portrait trajectories near \( A \) are given by \( x^3 y^2 = k \), where \( k \) is a positive constant. [3]
(ii) Sketch one possible trajectory for small values of \( x \) and \( y \) on a phase portrait. [3]
(i) Using your answers from part (c), show that phase portrait trajectories near \( A \) are given by \( x^3 y^2 = k \), where \( k \) is a positive constant. [3]
(ii) Sketch one possible trajectory for small values of \( x \) and \( y \) on a phase portrait. [3]
Consider points \( (x,y) \) near \( B \) on the phase plane, expressed as \( x = p + X \), \( y = q + Y \), where \( p \) and \( q \) are from part (b)(ii).
(e)
By substituting into the original model, show that, for small \( X \) and \( Y \):
\[ \dot{X} \approx -\dfrac{3Y}{2} \] [3]
By substituting into the original model, show that, for small \( X \) and \( Y \):
\[ \dot{X} \approx -\dfrac{3Y}{2} \] [3]
It can be shown that \( \dot{Y} \approx 4X \).
(f)
Given that \[ \begin{pmatrix} \dot{X} \\ \dot{Y} \end{pmatrix} = M \begin{pmatrix} X \\ Y \end{pmatrix}, \] where \( M \) is a square matrix, write down \( M \). [1]
Given that \[ \begin{pmatrix} \dot{X} \\ \dot{Y} \end{pmatrix} = M \begin{pmatrix} X \\ Y \end{pmatrix}, \] where \( M \) is a square matrix, write down \( M \). [1]
(g)
By finding the eigenvalues of \( M \), describe the path of the trajectories near point \( B \). [4]
By finding the eigenvalues of \( M \), describe the path of the trajectories near point \( B \). [4]
(h)
Sketch a complete set of trajectories in the phase plane for the original model, clearly indicating both equilibrium points. [3]
Sketch a complete set of trajectories in the phase plane for the original model, clearly indicating both equilibrium points. [3]
(i)
At a given time, there are 500 Czech geese and 500 gray foxes in the park. The park manager assumes the geese will quickly die out based on the values from part (a). Suggest whether this assumption is supported by the model, with justification. [2]
At a given time, there are 500 Czech geese and 500 gray foxes in the park. The park manager assumes the geese will quickly die out based on the values from part (a). Suggest whether this assumption is supported by the model, with justification. [2]
▶️ Answer/Explanation
Markscheme
(a)(i)
Emma substitutes \( x = 5 \), \( y = 5 \) into \( \dfrac{dx}{dt} = 2x – \dfrac{xy}{2} \): M1
\( \dfrac{dx}{dt} = 2 \times 5 – \dfrac{5 \times 5}{2} = 10 – 12.5 = -2.5 \) (hundred geese per year) or \(-250\) (geese per year). A1
[2 marks]
Emma substitutes \( x = 5 \), \( y = 5 \) into \( \dfrac{dx}{dt} = 2x – \dfrac{xy}{2} \): M1
\( \dfrac{dx}{dt} = 2 \times 5 – \dfrac{5 \times 5}{2} = 10 – 12.5 = -2.5 \) (hundred geese per year) or \(-250\) (geese per year). A1
[2 marks]
(a)(ii)
Emma substitutes \( x = 5 \), \( y = 5 \) into \( \dfrac{dy}{dt} = -3y + xy \):
\( \dfrac{dy}{dt} = -3 \times 5 + 5 \times 5 = -15 + 25 = 10 \) (hundred foxes per year) or \(1000\) (foxes per year). A1
[1 mark]
Emma substitutes \( x = 5 \), \( y = 5 \) into \( \dfrac{dy}{dt} = -3y + xy \):
\( \dfrac{dy}{dt} = -3 \times 5 + 5 \times 5 = -15 + 25 = 10 \) (hundred foxes per year) or \(1000\) (foxes per year). A1
[1 mark]
(b)(i)
At \( A(0,0) \), Emma evaluates: \( \dfrac{dx}{dt} = 2 \times 0 – \dfrac{0 \times 0}{2} = 0 \), \( \dfrac{dy}{dt} = -3 \times 0 + 0 \times 0 = 0 \). Both rates are zero, so populations do not change. R1
[1 mark]
At \( A(0,0) \), Emma evaluates: \( \dfrac{dx}{dt} = 2 \times 0 – \dfrac{0 \times 0}{2} = 0 \), \( \dfrac{dy}{dt} = -3 \times 0 + 0 \times 0 = 0 \). Both rates are zero, so populations do not change. R1
[1 mark]
(b)(ii)
Emma sets \( \dfrac{dx}{dt} = 0 \): \( 2x – \dfrac{xy}{2} = 0 \Rightarrow x \left(2 – \dfrac{y}{2}\right) = 0 \Rightarrow x = 0 \) or \( y = 4 \). M1
Sets \( \dfrac{dy}{dt} = 0 \): \( -3y + xy = 0 \Rightarrow y (x – 3) = 0 \Rightarrow y = 0 \) or \( x = 3 \).
For \( B(p,q) \): \( p = 3 \), \( q = 4 \). A1 A1
[3 marks]
Emma sets \( \dfrac{dx}{dt} = 0 \): \( 2x – \dfrac{xy}{2} = 0 \Rightarrow x \left(2 – \dfrac{y}{2}\right) = 0 \Rightarrow x = 0 \) or \( y = 4 \). M1
Sets \( \dfrac{dy}{dt} = 0 \): \( -3y + xy = 0 \Rightarrow y (x – 3) = 0 \Rightarrow y = 0 \) or \( x = 3 \).
For \( B(p,q) \): \( p = 3 \), \( q = 4 \). A1 A1
[3 marks]
(c)(i)
Emma solves \( \dfrac{dx}{dt} = 2x \): M1
Separate variables: \( \dfrac{dx}{x} = 2 dt \).
Integrate: \( \int \dfrac{dx}{x} = \int 2 dt \). A1
\( \ln x = 2t + c \). A1
\( x = A e^{2t} \). A1
[4 marks]
Emma solves \( \dfrac{dx}{dt} = 2x \): M1
Separate variables: \( \dfrac{dx}{x} = 2 dt \).
Integrate: \( \int \dfrac{dx}{x} = \int 2 dt \). A1
\( \ln x = 2t + c \). A1
\( x = A e^{2t} \). A1
[4 marks]
(c)(ii)
Emma solves \( \dfrac{dy}{dt} = -3y \):
Separate variables: \( \dfrac{dy}{y} = -3 dt \).
Integrate: \( \ln y = -3t + c \).
\( y = B e^{-3t} \). A1
[1 mark]
Emma solves \( \dfrac{dy}{dt} = -3y \):
Separate variables: \( \dfrac{dy}{y} = -3 dt \).
Integrate: \( \ln y = -3t + c \).
\( y = B e^{-3t} \). A1
[1 mark]
(d)(i)
From (c)(i), \( x = A e^{2t} \), so \( x^3 = A^3 e^{6t} \). A1
From (c)(ii), \( y = B e^{-3t} \), so \( y^2 = B^2 e^{-6t} \). A1
Multiply: \( x^3 y^2 = A^3 e^{6t} \times B^2 e^{-6t} = A^3 B^2 = k \), where \( k > 0 \). A1
[3 marks]
From (c)(i), \( x = A e^{2t} \), so \( x^3 = A^3 e^{6t} \). A1
From (c)(ii), \( y = B e^{-3t} \), so \( y^2 = B^2 e^{-6t} \). A1
Multiply: \( x^3 y^2 = A^3 e^{6t} \times B^2 e^{-6t} = A^3 B^2 = k \), where \( k > 0 \). A1
[3 marks]
(d)(ii)
From \( x^3 y^2 = k \), solve: \( y = \sqrt{\dfrac{k}{x^3}} \). A1
Trajectory is a decreasing convex curve in the first quadrant. A1
Include direction arrows toward increasing \( x \), decreasing \( y \). A1
[3 marks]
(e)
Emma substitutes \( x = 3 + X \), \( y = 4 + Y \) into \( \dfrac{dx}{dt} = 2x – \dfrac{xy}{2} \): M1
\( \dot{X} = 2(3 + X) – \dfrac{(3 + X)(4 + Y)}{2} \).
Expand: \( 6 + 2X – \dfrac{12 + 4X + 3Y + XY}{2} = 6 + 2X – 6 – 2X – \dfrac{3Y}{2} – \dfrac{XY}{2} \). A1
For small \( X \), \( Y \), ignore \( XY \): \( \dot{X} \approx -\dfrac{3Y}{2} \). A1
[3 marks]
Emma substitutes \( x = 3 + X \), \( y = 4 + Y \) into \( \dfrac{dx}{dt} = 2x – \dfrac{xy}{2} \): M1
\( \dot{X} = 2(3 + X) – \dfrac{(3 + X)(4 + Y)}{2} \).
Expand: \( 6 + 2X – \dfrac{12 + 4X + 3Y + XY}{2} = 6 + 2X – 6 – 2X – \dfrac{3Y}{2} – \dfrac{XY}{2} \). A1
For small \( X \), \( Y \), ignore \( XY \): \( \dot{X} \approx -\dfrac{3Y}{2} \). A1
[3 marks]
(f)
Given \( \dot{X} \approx -\dfrac{3Y}{2} \), \( \dot{Y} \approx 4X \), Emma writes:
\[ M = \begin{pmatrix} 0 & -\dfrac{3}{2} \\ 4 & 0 \end{pmatrix} \] A1
[1 mark]
Given \( \dot{X} \approx -\dfrac{3Y}{2} \), \( \dot{Y} \approx 4X \), Emma writes:
\[ M = \begin{pmatrix} 0 & -\dfrac{3}{2} \\ 4 & 0 \end{pmatrix} \] A1
[1 mark]
(g)
Emma finds eigenvalues of \( M = \begin{pmatrix} 0 & -\dfrac{3}{2} \\ 4 & 0 \end{pmatrix} \): M1
Solve \( \det(M – \lambda I) = \det \begin{pmatrix} -\lambda & -\dfrac{3}{2} \\ 4 & -\lambda \end{pmatrix} = \lambda^2 + \dfrac{3 \times 4}{2} = \lambda^2 + 6 = 0 \). A1
Eigenvalues: \( \lambda = \pm i \sqrt{6} \) (or \( \pm 2.45i \)). A1
Purely imaginary eigenvalues indicate elliptical trajectories. R1
[4 marks]
Emma finds eigenvalues of \( M = \begin{pmatrix} 0 & -\dfrac{3}{2} \\ 4 & 0 \end{pmatrix} \): M1
Solve \( \det(M – \lambda I) = \det \begin{pmatrix} -\lambda & -\dfrac{3}{2} \\ 4 & -\lambda \end{pmatrix} = \lambda^2 + \dfrac{3 \times 4}{2} = \lambda^2 + 6 = 0 \). A1
Eigenvalues: \( \lambda = \pm i \sqrt{6} \) (or \( \pm 2.45i \)). A1
Purely imaginary eigenvalues indicate elliptical trajectories. R1
[4 marks]
(h)
Emma sketches at least one anti-clockwise elliptical loop in the first quadrant, centered near \( B(3,4) \). A1
Labels equilibrium points \( A(0,0) \) and \( B(3,4) \). A1
Includes trajectories near \( A \): diverging along \( x \)-axis, converging along \( y \)-axis. A1
[3 marks]
(i)
The park manager’s assumption is incorrect. Although \( \dfrac{dx}{dt} = -2.5 \) at (5,5), Emma’s phase portrait shows closed elliptical trajectories, indicating the geese population oscillates, not dies out. A1 R1
[2 marks]
The park manager’s assumption is incorrect. Although \( \dfrac{dx}{dt} = -2.5 \) at (5,5), Emma’s phase portrait shows closed elliptical trajectories, indicating the geese population oscillates, not dies out. A1 R1
[2 marks]
Total Marks: 31