IBDP MAI : Topic 5 Calculus - AHL 5.18 Solutions of second order differential equation AI HL Paper 3
Question : Predator-Prey Model Phase Portraits
The following question explores a possible method of drawing phase portraits for non-linear coupled systems, taking a predator-prey model as a particular example.
Gander Green wildlife park contains a population of Czech geese (x, measured in hundreds), and a population of gray foxes (y, measured in hundreds).
Research indicates that the population growth of both geese and foxes can be modelled by the following differential equations, in which t is measured in years.
(i) geese. Topic: AHL 5.18
(ii) foxes. Topic: AHL 5.18
Working space:
▶️Answer/Explanation
Correct answers:
(i) -2.5 hundred geese per year (-250 geese per year)
(ii) 10 hundred foxes per year (1000 foxes per year)
Working:
(i) Substitute \( x = 5 \), \( y = 5 \) into \( \frac{dx}{dt} = 2x – \frac{xy}{2} \):
\[ \frac{dx}{dt} = 2(5) – \frac{(5)(5)}{2} = 10 – \frac{25}{2} = 10 – 12.5 = -2.5 \]
So, -2.5 hundred geese per year.
(ii) Substitute \( x = 5 \), \( y = 5 \) into \( \frac{dy}{dt} = -3y + xy \):
\[ \frac{dy}{dt} = -3(5) + (5)(5) = -15 + 25 = 10 \]
So, 10 hundred foxes per year.
Key Concept:
Evaluating differential equations at specific points gives the instantaneous rate of change of populations in a predator-prey model.
(i) Explain why A is an equilibrium point. Topic: AHL 5.18
(ii) Find the value of p and the value of q. Topic: AHL 5.18
Working space:
▶️Answer/Explanation
Correct answers:
(i) At \( A(0,0) \), both \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} = 0 \), so populations remain constant.
(ii) \( p = 3 \), \( q = 4 \)
Working:
(i) At \( A(0,0) \), substitute \( x = 0 \), \( y = 0 \):
\[ \frac{dx}{dt} = 2(0) – \frac{(0)(0)}{2} = 0 \]
\[ \frac{dy}{dt} = -3(0) + (0)(0) = 0 \]
Since both derivatives are zero, there is no change in population, making \( A \) an equilibrium point.
(ii) Set \( \frac{dx}{dt} = 0 \):
\[ x \left( 2 – \frac{y}{2} \right) = 0 \Rightarrow x = 0 \text{ or } 2 – \frac{y}{2} = 0 \Rightarrow y = 4 \]
Set \( \frac{dy}{dt} = 0 \):
\[ y (-3 + x) = 0 \Rightarrow y = 0 \text{ or } x = 3 \]
Excluding \( x = 0 \), \( y = 0 \) (point \( A \)), the solution is \( x = 3 \), \( y = 4 \). So, \( p = 3 \), \( q = 4 \).
Key Concept:
Equilibrium points occur where the rates of change are zero, found by solving the system of differential equations set to zero.
By solving these two differential equations, Topic: AHL 5.18
(i) find an expression for x in terms of t.
(ii) find an expression for y in terms of t.
Working space:
▶️Answer/Explanation
Correct answers:
(i) \( x = A e^{2t} \)
(ii) \( y = B e^{-3t} \)
Working:
(i) Solve \( \frac{dx}{dt} = 2x \):
\[ \int \frac{dx}{x} = \int 2 \, dt \]
\[ \ln x = 2t + c_1 \]
\[ x = e^{2t + c_1} = A e^{2t} \]
(ii) Solve \( \frac{dy}{dt} = -3y \):
\[ \int \frac{dy}{y} = \int (-3) \, dt \]
\[ \ln y = -3t + c_2 \]
\[ y = e^{-3t + c_2} = B e^{-3t} \]
Key Concept:
Linear differential equations near equilibrium points can be solved using separation of variables to find population dynamics.
(ii) Hence sketch, on a phase portrait, one possible trajectory for small values of x and y.
Working space:
▶️Answer/Explanation
Correct answers:
(i) \( xy^2 = k \)
(ii) 
Working:
(i) From part (c):
\[ x = A e^{2t}, \quad y = B e^{-3t} \]
Compute \( x^3 \):
\[ x^3 = A^3 e^{6t} \]
Compute \( y^2 \):
\[ y^2 = B^2 e^{-6t} \]
Multiply:
\[ x^3 y^2 = (A^3 e^{6t})(B^2 e^{-6t}) = A^3 B^2 = k \]
Since \( A, B > 0 \), \( k \) is a positive constant. Thus, \( x^3 y^2 = k \), or equivalently, \( xy^2 = k / x^2 = k’ \), but we use \( xy^2 = k \).
(ii) The equation \( xy^2 = k \) gives \( y = \sqrt{\frac{k}{x^3}} \). For small \( x, y \), sketch the trajectory:
Key Concept:
Trajectories near an equilibrium point are found by eliminating time from the solutions, yielding a relationship between variables.
By substituting into the original model, show that, for small values of X and Y:
\[ \dot{X} \approx -\frac{3Y}{2} \] Topic: AHL 5.18
Working space:
▶️Answer/Explanation
Correct answer: \( \dot{X} \approx -\frac{3Y}{2} \)
Working:
Substitute \( x = 3 + X \), \( y = 4 + Y \) into \( \frac{dx}{dt} = 2x – \frac{xy}{2} \):
\[ \frac{dX}{dt} = 2(3 + X) – \frac{(3 + X)(4 + Y)}{2} \]
Expand:
\[ = 6 + 2X – \frac{(12 + 4X + 3Y + XY)}{2} \]
\[ = 6 + 2X – 6 – 2X – \frac{3Y}{2} – \frac{XY}{2} \]
For small \( X, Y \), neglect \( XY \):
\[ \dot{X} \approx -\frac{3Y}{2} \]
Key Concept:
Linearization near an equilibrium point approximates the system by neglecting higher-order terms, simplifying analysis.
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. Topic: AHL 5.18
Working space:
▶️Answer/Explanation
Correct answer:
\[ M = \begin{pmatrix} 0 & -\frac{3}{2} \\ 4 & 0 \end{pmatrix} \]
Working:
From part (e), \( \dot{X} = -\frac{3}{2} Y \), and given \( \dot{Y} = 4X \). Write the system:
\[ \begin{pmatrix} \dot{X} \\ \dot{Y} \end{pmatrix} = \begin{pmatrix} 0 & -\frac{3}{2} \\ 4 & 0 \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} \]
Thus, \( M = \begin{pmatrix} 0 & -\frac{3}{2} \\ 4 & 0 \end{pmatrix} \).
Key Concept:
The Jacobian matrix represents the linearized system near an equilibrium point, used for stability analysis.
Working space:
▶️Answer/Explanation
Correct answer: Trajectories near \( B(3,4) \) are closed orbits (ellipses or circles).
Working:
Find eigenvalues of \( M = \begin{pmatrix} 0 & -\frac{3}{2} \\ 4 & 0 \end{pmatrix} \):
\[ \det(M – \lambda I) = \det \begin{pmatrix} -\lambda & -\frac{3}{2} \\ 4 & -\lambda \end{pmatrix} = \lambda^2 + \frac{3}{2} \cdot 4 = \lambda^2 + 6 = 0 \]
\[ \lambda = \pm i \sqrt{6} \]
Pure imaginary eigenvalues indicate a center, so trajectories near \( B \) are closed orbits, representing stable periodic oscillations.
Key Concept:
Imaginary eigenvalues suggest oscillatory behavior, leading to closed trajectories in the phase plane.
Working space:
▶️Answer/Explanation
Correct answer:
Working:
\( A(0,0) \): Unstable node. Trajectories move away along \( y^2 = k x^{-3} \).
\( B(3,4) \): Center. Trajectories are closed loops around \( B \), indicating stable oscillations.
Sketch shows trajectories diverging from \( A \) and orbiting \( B \).
Key Concept:
The phase portrait combines local behaviors (unstable node at \( A \), center at \( B \)) to depict global dynamics.
Based on the values found in part (a), the park’s wildlife keeper is worried and assumes that the geese will quickly die out. Suggest whether this assumption is supported by the model. Justify your answer. Topic: AHL 5.18
Working space:
▶️Answer/Explanation
Correct answer: The keeper’s assumption is not supported; the geese will not die out quickly.
Working:
At \( (5,5) \), from part (a): \( \frac{dx}{dt} = -2.5 \) (geese decreasing), \( \frac{dy}{dt} = 10 \) (foxes increasing). However, \( (5,5) \) is near \( B(3,4) \), a center with closed orbits. The negative \( \frac{dx}{dt} \) suggests initial decline, but the oscillatory behavior around \( B \) indicates the geese population will recover in the cycle, not approach zero.
Key Concept:
Local rates of change must be interpreted with global phase portrait behavior to predict long-term dynamics.
Syllabus Reference
Calculus
- (a) AHL 5.18 – Rates of change in differential equations
- (b) AHL 5.18 – Equilibrium points in systems
- (c) AHL 5.18 – Solving linear differential equations
- (d) AHL 5.18 – Phase portrait trajectories
- (e) AHL 5.18 – Linearization near equilibrium
- (f) AHL 5.18 – Jacobian matrix formulation
- (g) AHL 5.18 – Eigenvalues and trajectory behavior
- (h) AHL 5.18 – Complete phase portraits
- (i) AHL 5.18 – Interpreting model predictions
Assessment Criteria: A (Knowledge), C (Communication), D (Scientific Thinking)