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Topic 5.16 · HL only

Phase portrait (for solutions to coupled systems) & sketching trajectories: notes and practice questions

Summary
  • Coupled differential equations: System where rates of change of variables depend on each other, typically linear in matrix form x˙=Mx\mathbf{\dot{x}} = M\mathbf{x}.
  • Phase portrait: Diagram illustrating how xx and yy values change over time, displaying typical solution trajectories.
  • Solution trajectory: Specific path on a phase portrait based on initial conditions.
  • Equilibrium point: Coordinate (x,y)(x, y) where dxdt=0\frac{dx}{dt} = 0 and dydt=0\frac{dy}{dt} = 0.
  • Stability of the origin: Depends on the eigenvalues of the matrix MM.
  • Matrix form of coupled differential equations:

(x˙y˙)=(abcd)(xy) \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}

  • Exact solution formula (HL): For real, distinct eigenvalues λ1,λ2\lambda_1, \lambda_2 with eigenvectors p1,p2\mathbf{p}_1, \mathbf{p}_2:

x=Aeλ1tp1+Beλ2tp2 \mathbf{x} = A e^{\lambda_1 t}\mathbf{p}_1 + B e^{\lambda_2 t}\mathbf{p}_2
where A,BA, B are constants found using initial conditions.

  • Second order differential equation conversion (HL): For d2xdt2+adxdt+bx=0\frac{d^2x}{dt^2} + a\frac{dx}{dt} + bx = 0, substitute y=dxdty = \frac{dx}{dt}.
  • Resulting coupled system: dxdt=y\frac{dx}{dt} = y and dydt=−bx−ay\frac{dy}{dt} = -bx - ay.
  • Corresponding matrix: M=(01−b−a)M = \begin{pmatrix} 0 & 1 \\ -b & -a \end{pmatrix}.
  • Finding an equilibrium point: Set dxdt=0\frac{dx}{dt} = 0 and dydt=0\frac{dy}{dt} = 0, then solve the resulting equations simultaneously for (x,y)(x, y).
  • Determining phase portrait shape (HL, based on eigenvalues of MM):
  • Both real, one positive, one negative: Saddle point; trajectories approach along negative eigenvector, curve away along positive.
  • Both real and negative: All trajectories converge to origin, curving towards the eigenvector of the least negative eigenvalue.
  • Purely imaginary: Circular or elliptical orbits around the origin (unstable).
  • Complex with a positive real part: Trajectories spiral away from the origin (unstable).
  • Complex with a negative real part: Trajectories spiral in towards the origin (stable).
  • Determining direction of spirals/orbits (HL):
  • Choose a simple point (e.g., (1,0)(1, 0) or (0,1)(0, 1)).
  • Substitute its coordinates into dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} to find the initial velocity vector.
  • Interpret the vector's direction to determine clockwise or anticlockwise motion.
  • Sketching a specific solution trajectory (HL):
  • Mark the initial point.
  • Determine the initial direction by substituting the initial (x,y)(x, y) into dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}.
  • Draw a continuous curve from the initial point, following the initial direction and conforming to the overall phase portrait shape.
  • GDC use: Use simultaneous equation solver to find constants AA and BB from initial conditions.
  • HL distinction: Coupled differential equations and phase portraits are exclusively HL topics (Topic 5.7).
  • Contextual interpretation: In a converted second-order system where y=dxdty = \frac{dx}{dt}, the yy-axis represents velocity.
  • Saddle point: Term for the origin when eigenvalues are real with opposite signs.
  • Exact solution formula limits: Applicable only when eigenvalues of MM are real and distinct.

How it is examined

A question gives the coefficient matrix, asks for the eigenvalues (AHL 1.15 machinery again), and then asks what kind of equilibrium the origin is and how trajectories behave near it, sketch included. The restriction to distinct, non-zero eigenvalues means a student never has to handle the repeated or zero-eigenvalue edge cases, and the qualitative classification table above is worth memorising rather than re-deriving under time pressure.

Given in the booklet

The eigenvalue classification above.

Key ideas
  • Sketch the phase portrait for solutions of coupled differential equations of the form dxdt=ax+by\dfrac{dx}{dt} = ax + by, dydt=cx+dy\dfrac{dy}{dt} = cx + dy.
  • Carry out a qualitative analysis of future paths for distinct real, complex and imaginary eigenvalues.
  • Sketch trajectories and use phase portraits to identify equilibrium points, stable populations and saddle points.

Linking questions

  • Other contexts: the Jacobian matrix, used to investigate the stability of equilibrium states for non-linear differential equations.

Practice questions

6 questions · 4 medium · 2 hard
Showing 6 of 6

Question 1

MediumPaper 1 · calculator14 marks
(a)

The populations of two interacting species, A and B, in a controlled environment are modelled by the following system of linear differential equations:

x˙=x+2y\dot{x} = x + 2y

y˙=3x+2y\dot{y} = 3x + 2y

where x(t)x(t) represents the population of species A and y(t)y(t) represents the population of species B at time tt. Find the general solution for this system.

[7]
(b)

Sketch the phase portrait for the system, clearly indicating the critical point, the equations of any asymptotes, and the general direction of trajectories. Briefly describe the long-term behaviour of the populations.

[7]

Question 2

HardPaper 2 · calculator22 marks
(a)(i)

The concentration of a chemical, CC (in mol dm−3^{-3}), in a reaction vessel at time tt seconds is modelled by the differential equation

d2Cdt2+7dCdt+10C=0\frac{\mathrm{d}^2C}{\mathrm{d}t^2} + 7\frac{\mathrm{d}C}{\mathrm{d}t} + 10C = 0

(a) (i) Use the substitution V=dCdtV = \frac{\mathrm{d}C}{\mathrm{d}t} to show that this equation can be written as

(dCdtdVdt)=(01−10−7)(CV)\begin{pmatrix} \frac{\mathrm{d}C}{\mathrm{d}t} \\ \frac{\mathrm{d}V}{\mathrm{d}t} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -10 & -7 \end{pmatrix} \begin{pmatrix} C \\ V \end{pmatrix}.

[5]
(a)(ii)

(ii) Find the eigenvalues for the matrix (01−10−7)\begin{pmatrix} 0 & 1 \\ -10 & -7 \end{pmatrix}.

[3]
(a)(iii)

(iii) Hence state the long-term rate of change of the chemical concentration.

[1]
(b)(i)

The chemical reaction is now subjected to an external input, and the equation for the concentration is amended to

d2Cdt2+7dCdt+10C=2t+5\frac{\mathrm{d}^2C}{\mathrm{d}t^2} + 7\frac{\mathrm{d}C}{\mathrm{d}t} + 10C = 2t + 5.

(b) (i) Use the substitution V=dCdtV = \frac{\mathrm{d}C}{\mathrm{d}t} to write the differential equation as a system of coupled, first order differential equations. State the initial conditions given that, at t=0t = 0, the concentration of chemical C is 11 mol dm−3^{-3} and its rate of change is 00 mol dm−3^{-3} s−1^{-1}.

[3]
(b)(ii)

(ii) Use Euler's method with a step length of 0.10.1 to find the concentration of the chemical when t=1t = 1 s. Give your answer to three significant figures.

[7]
(b)(iii)

(iii) Find the long-term rate of change of the chemical concentration.

[3]

Question 3

MediumPaper 2 · calculator20 marks
(a)

A population model for two interacting species, XX and YY, is described by the following system of coupled differential equations, where x(t)x(t) and y(t)y(t) represent the deviations from their equilibrium populations. (a) Determine (with a worked reason) whether the populations of both species tend towards their equilibrium, move away from their equilibrium, or exhibit another behaviour.

dxdt=−3x+y\frac{dx}{dt} = -3x + y

dydt=2x−4y\frac{dy}{dt} = 2x - 4y

[5]
(b)

(b) Determine (with a worked reason) whether the populations of both species tend towards their equilibrium, move away from their equilibrium, or exhibit another behaviour.

dxdt=x+2y\frac{dx}{dt} = x + 2y

dydt=3x\frac{dy}{dt} = 3x

[5]
(c)

(c) Determine (with a worked reason) whether the populations of both species tend towards their equilibrium, move away from their equilibrium, or exhibit another behaviour.

dxdt=−x−2y\frac{dx}{dt} = -x - 2y

dydt=2x−y\frac{dy}{dt} = 2x - y

[5]
(d)

(d) Determine (with a worked reason) whether the populations of both species tend towards their equilibrium, move away from their equilibrium, or exhibit another behaviour.

dxdt=2y\frac{dx}{dt} = 2y

dydt=−2x\frac{dy}{dt} = -2x

[5]

Question 4

HardPaper 3 · calculator31 marks
(a)(i)

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.

A forest ecosystem contains a population of rabbits (xx, measured in hundreds), and a population of wolves (yy, measured in hundreds).

Research indicates that the population dynamics of both rabbits and wolves can be modelled by the following differential equations, in which tt is measured in years.

dxdt=3x−xy2\frac{dx}{dt} = 3x - \frac{xy}{2}

dydt=−2y+2xy5\frac{dy}{dt} = -2y + \frac{2xy}{5}

for x,y≥0x, y \ge 0

At a specific time, there are 400 rabbits and 400 wolves, represented here by the coordinate pair (4, 4). At this time, determine the rate of change of

rabbits.

[2]
(a)(ii)

wolves.

[1]
(b)(i)

There are two equilibrium points for the populations: A(0, 0) and B (p,qp, q).

Explain why A is an equilibrium point.

[1]
(b)(ii)

Find the value of pp and the value of qq.

[3]
(c)(i)

At points close to A(0, 0), we can ignore the xyxy terms, so that the system can be approximated by:

dxdt=3x\frac{dx}{dt} = 3x

dydt=−2y\frac{dy}{dt} = -2y

for x,y≥0x, y \ge 0.

By solving these two differential equations,

find an expression for xx in terms of tt.

[4]
(c)(ii)

find an expression for yy in terms of tt.

[1]
(d)(i)

Using your answers from part (c), show that phase portrait trajectories close to A may be given by the equation x2y3=kx^2y^3 = k, where kk is a positive constant.

[3]
(d)(ii)

Hence sketch, on a phase portrait, one possible trajectory for small values of xx and yy.

[3]
(e)

Now consider points (x,y)(x, y) close to B on the phase plane. These coordinates can be rewritten as x=p+Xx = p + X and y=q+Yy = q + Y, where pp and qq are the values from part (b)(ii).

By substituting into the original model, show that, for small values of XX and YY:

X˙≈−5Y2\dot{X} \approx -\frac{5Y}{2}

Similarly, it can be shown that Y˙≈12X5\dot{Y} \approx \frac{12X}{5}.

[3]
(f)

Given that (X˙Y˙)=M(XY)\begin{pmatrix} \dot{X} \\ \dot{Y} \end{pmatrix} = M \begin{pmatrix} X \\ Y \end{pmatrix}, where MM is a square matrix, write down MM.

[1]
(g)

By finding the eigenvalues of MM, describe the path of the trajectories close to point B.

[4]
(h)

Hence sketch a complete set of trajectories in the phase plane for the original model, clearly indicating both equilibrium points.

[3]
(i)

In this forest ecosystem, at a specific time, there are 400 rabbits and 400 wolves.

Based on the values found in part (a), the wildlife keeper is worried and assumes that the wolves will quickly die out. Suggest whether this assumption is supported by the model. Justify your answer.

[2]

Question 5

MediumPaper 1 · calculator6 marks
(a)

The populations of two interacting bacterial strains, Alpha and Beta, in a controlled environment are modelled by a system of coupled differential equations. Let AA be the population of strain Alpha (in thousands) and BB be the population of strain Beta (in thousands).

The system is given by:

dAdt=1.5A−2.0B\frac{dA}{dt} = 1.5A - 2.0B

dBdt=2.5A−1.5B\frac{dB}{dt} = 2.5A - 1.5B

At time t=0t = 0 hours, the population of strain Alpha is A=6A = 6 thousand and the population of strain Beta is B=3B = 3 thousand.

Find the value of dBdA\frac{dB}{dA} at t=0t = 0.

[3]
(b)

The eigenvalues for this system are ±4i\pm 4i.

On the following phase portrait, sketch the trajectory that passes through the point (6,3)(6, 3). Clearly indicate the direction of this trajectory.

Phase portrait with point (6,3)
[3]

Question 6

MediumPaper 2 · calculator15 marks
(a)

A microscopic particle is observed moving in a fluid. Its position (x,y)(x, y) metres, relative to a fixed origin O, at time tt seconds (t≥0t \ge 0), is modelled by the system of differential equations:

dxdt=−0.1x+3y\frac{dx}{dt} = -0.1x + 3y

dydt=−3x−0.1y\frac{dy}{dt} = -3x - 0.1y

Find the eigenvalues of the matrix A=(−0.13−3−0.1)A = \begin{pmatrix} -0.1 & 3 \\ -3 & -0.1 \end{pmatrix}, giving your answers in the form a+bia + bi, where a≠0,b≠0a \ne 0, b \ne 0.

[4]
(b)(i)

State what a≠0a \ne 0 indicates about the path of the particle.

[1]
(b)(ii)

State what the sign of aa indicates about the path of the particle.

[1]
(c)(i)

At time t=0t = 0, the particle is at position (10,0)(10, 0).

At time t=0t = 0, find the value of dydt\frac{dy}{dt}.

[2]
(c)(ii)

Find the value of dydx\frac{dy}{dx} at time t=0t = 0.

[3]
(d)

Use your answers to parts (b) and (c) to sketch the path of the particle.

[4]

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What does Phase portrait (for solutions to coupled systems) & sketching trajectories cover in IB Maths AI?

Coupled differential equations: System where rates of change of variables depend on each other, typically linear in matrix form dotx = Mx. Phase portrait: Diagram illustrating how x and y values change over time, displaying typical solution trajectories. Solution trajectory: Specific path on a phase portrait based on initial conditions.

Is Phase portrait (for solutions to coupled systems) & sketching trajectories SL or HL?

Phase portrait (for solutions to coupled systems) & sketching trajectories is HL only. SL students are not examined on it.

How do I revise Phase portrait (for solutions to coupled systems) & sketching trajectories for IB Maths AI?

Start from the core idea: coupled differential equations: System where rates of change of variables depend on each other, typically linear in matrix form dotx = Mx. In the exam: a question gives the coefficient matrix, asks for the eigenvalues (AHL 1.15 machinery again), and then asks what kind of equilibrium the origin is and how trajectories behave near it, sketch included. The restriction to distinct, non-zero eigenvalues means a student never has to handle the repeated or zero-eigenvalue edge cases, and the qualitative classification table above is worth memorising rather than re-deriving under time pressure. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

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