IB DP Mathematics: Applications and Interpretation – Differential equations, slope fields, Euler's method and phase portraits (HL) Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the HL differential equations unit of IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 5.14–5.18, and every question is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

All three HL papers are “technology required” in the guide, so every question here is labelled “(calculator allowed)”. There is no calculator-free paper in this course. Show your method anyway: a GDC value on its own can lose the method marks. Give answers exactly or to 3 significant figures unless the question says otherwise.

Learn the methods in the differential equations study guide and the revision notes. The IB DP Maths AI course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (HL, calculator allowed) The value V of a machine decreases at a rate proportional to V.

(a) Write down a differential equation for V in terms of t. [1] (b) Find the general solution of your equation. [2]

2. (HL, calculator allowed) Solve dy/dx = 3x²/y, given that y = 2 when x = 1 and y > 0. Hence find y when x = 2. [5]

3. (HL, calculator allowed) A slope field is drawn for dy/dx = y − x².

(a) Find the gradient of the segments at (0, 1), (1, 3) and (2, 1). [2] (b) Write down the equation of the curve on which solution curves have horizontal tangents, and state the region in which solution curves are increasing. [2]

4. (HL, calculator allowed) Use Euler’s method with step length 0.1 to estimate y(1.3) for dy/dx = x − y², y(1) = 0.5. [4]

5. (HL, calculator allowed) Consider dx/dt = 2x + y, dy/dt = −5x − 2y.

(a) Find the eigenvalues of the matrix of this system. [2] (b) Describe the trajectories in the phase portrait, including the direction of motion. [2]

6. (HL, calculator allowed) Consider dx/dt = 0.2x − y, dy/dt = x + 0.2y.

(a) Find the eigenvalues of the matrix of this system. [2] (b) Describe the phase portrait, including the direction of motion. [2]

7. (HL, calculator allowed) Rabbits R and foxes F (in hundreds) satisfy dR/dt = 0.8R − 0.02RF and dF/dt = −0.6F + 0.01RF, with R = 50 and F = 30 at t = 0.

(a) Use Euler’s method with h = 0.05 to estimate R and F at t = 0.1. [3] (b) Show that R = 60, F = 40 is an equilibrium point. [2]

8. (HL, calculator allowed) A driven spring satisfies d²x/dt² = cos t − 0.5 dx/dt − 4x, with x = 0 and dx/dt = 1 when t = 0.

(a) Write this as a pair of coupled first order equations. [2] (b) Use Euler’s method with h = 0.1 to estimate x when t = 0.3. [4]

9. (HL, calculator allowed) Consider dy/dx = x²/(1 + y), y(0) = 1.

(a) Use Euler’s method with h = 0.2 to estimate y(1). [3] (b) Show that y + y²/2 = x³/3 + 3/2. [3] (c) Hence find the exact value of y(1), giving it to 3 s.f. [2] (d) Find the percentage error in your answer to (a). [1]

10. (HL, calculator allowed) Two quantities satisfy dx/dt = x + 4y, dy/dt = x + y, with x = 4 and y = 0 when t = 0.

(a) Find the eigenvalues and corresponding eigenvectors of the matrix of this system. [3] (b) Find x and y in terms of t. [3] (c) Find x and y when t = 0.5. [1] (d) Classify the origin and describe the long-term behaviour of this solution. [2] (e) State the set of starting points, other than the origin, whose trajectories approach the origin. [1]

11. (HL, calculator allowed) A damped oscillator satisfies d²x/dt² + 5 dx/dt + 6x = 0, with x = 1 and dx/dt = 0 when t = 0.

(a) Write the equation as a coupled system with matrix M, and find the eigenvalues of M. [2] (b) Find x in terms of t. [3] (c) Find the time at which x = 0.5. [1] (d) Describe the phase portrait and what it means for the motion. [1] (e) The coefficient 5 is changed to 2. State the new eigenvalues and how the motion changes. [1]

Answers

1. (a) dV/dt = −kV with k > 0 [1] (b) ∫ (1/V) dV = −∫ k dt [1], ln V = −kt + c, so V = Ae^(−kt) [1] Examiner insight: The general solution must keep its arbitrary constant; writing V = V₀e^(−kt) is fine only if V₀ is defined.

2. y dy = 3x² dx [1]. y²/2 = x³ + c [1]. x = 1, y = 2: 2 = 1 + c, so c = 1 [1]. y = √(2x³ + 2) [1]. y(2) = √18 = 3√2 ≈ 4.24 [1] Examiner insight: A constant placed on both sides and never combined usually costs the accuracy mark for the particular solution.

3. (a) Gradient = y − x²: 1 − 0 = 1, 3 − 1 = 2 [1], 1 − 4 = −3; 1, 2, −3 [1] (b) y = x² [1]; increasing where y − x² > 0, that is above the parabola y = x² [1] Examiner insight: “State the region” needs an inequality or a clear description; “positive gradient” alone does not identify where.

4. y_(n+1) = y_n + 0.1(x_n − y_n²) [1]. y₁ = 0.5 + 0.1(0.75) = 0.575 [1]. y₂ = 0.575 + 0.1(0.769375) = 0.651938 [1]. y₃ = 0.651938 + 0.1(0.774977) = 0.729 [1] Examiner insight: The formula line earns the method mark; a GDC table with no formula shown risks losing it.

5. (a) M = [2 1; −5 −2], λ² − 0λ + (−4 + 5) = λ² + 1 = 0 [1], λ = ±i [1] (b) Imaginary eigenvalues: closed ellipses around the origin [1]. At (1, 0), dx/dt = 2 and dy/dt = −5, so the motion is clockwise [1] Examiner insight: The direction mark needs evidence, such as the velocity at a named test point, not just the word “clockwise”.

6. (a) λ² − 0.4λ + 1.04 = 0 [1], λ = 0.2 ± i [1] (b) Complex with positive real part: trajectories spiral away from the origin [1]. At (1, 0), dy/dt = 1 > 0, so anticlockwise [1] Examiner insight: “Spiral” without “outwards” or “away from the origin” does not earn the mark, because the sign of the real part is the point.

7. (a) At t = 0: dR/dt = 40 − 30 = 10, dF/dt = −18 + 15 = −3, so R₁ = 50.5, F₁ = 29.85 [1]. At t = 0.05: dR/dt = 10.2515, dF/dt = −2.83575 [1]. R ≈ 51.0, F ≈ 29.7 [1] (b) dR/dt = 0.8(60) − 0.02(60)(40) = 48 − 48 = 0 [1]; dF/dt = −0.6(40) + 0.01(60)(40) = −24 + 24 = 0, so neither population changes [1] Examiner insight: In a “show that”, both rates must be shown equal to zero with the substitution visible; one rate is not enough.

8. (a) dx/dt = y [1], dy/dt = cos t − 0.5y − 4x [1] (b) x_(n+1) = x_n + 0.1y_n, y_(n+1) = y_n + 0.1(cos t_n − 0.5y_n − 4x_n), t_(n+1) = t_n + 0.1 [1]. x₁ = 0.1, y₁ = 1.05 [1]. x₂ = 0.205, y₂ = 1.057 [1]. x₃ = 0.311 [1] Examiner insight: Forgetting to increase t inside cos t is a method error, so follow-through marks are lost from that step.

9. (a) y_(n+1) = y_n + 0.2 x_n²/(1 + y_n) [1]. y = 1, 1, 1.004, 1.01997, 1.05561 at x = 0 to 0.8 [1]. y(1) ≈ 1.12 [1] (b) (1 + y) dy = x² dx [1]. y + y²/2 = x³/3 + c [1]. x = 0, y = 1: c = 1 + 1/2 = 3/2 [1] (c) y²/2 + y − 11/6 = 0, so y² + 2y − 11/3 = 0 [1]. Positive root: y = −1 + √(14/3) = 1.16 [1] (d) |1.11788 − 1.16025| / 1.16025 × 100 = 3.65% [1] Examiner insight: In (c), choose the root that fits y(0) = 1 and say why; the negative root is not the solution through (0, 1).

10. (a) λ² − 2λ − 3 = 0 [1], λ = 3, −1 [1]. Eigenvectors (2, 1) for λ = 3, (2, −1) for λ = −1 [1] (b) (x, y) = A e^(3t)(2, 1) + B e^(−t)(2, −1) [1]. t = 0: 2A + 2B = 4, A − B = 0, so A = B = 1 [1]. x = 2e^(3t) + 2e^(−t), y = e^(3t) − e^(−t) [1] (c) x ≈ 10.2, y ≈ 3.88 [1] (d) Eigenvalues of opposite sign: the origin is a saddle point [1]. As t → ∞ the e^(3t) terms dominate, so x, y → ∞ and y/x → 1/2: the trajectory approaches the direction of y = x/2 [1] (e) Points on the line y = −x/2 [1] Examiner insight: Accuracy marks in (b) depend on pairing each eigenvector with its own eigenvalue; a swap gives a solution that fails the initial check.

11. (a) dx/dt = y, dy/dt = −6x − 5y, M = [0 1; −6 −5], λ² + 5λ + 6 = 0 [1], λ = −2, −3 [1] (b) Eigenvectors (1, −2) and (1, −3) [1]. x = A e^(−2t) + B e^(−3t) with A + B = 1 and −2A − 3B = 0 [1]. A = 3, B = −2: x = 3e^(−2t) − 2e^(−3t) [1] (c) t = ln 2 ≈ 0.693 [1] (d) Both eigenvalues negative: all trajectories go to the origin, so x returns to 0 without oscillating [1] (e) λ² + 2λ + 6 = 0 gives λ = −1 ± i√5: trajectories spiral into the origin, so x oscillates with decreasing amplitude [1] Examiner insight: In (b) the initial condition dx/dt = 0 applies to the y-component, −2A − 3B = 0; using it on x loses the method mark.

Where marks are usually lost

• No arbitrary constant in a general solution (question 1).
• Constants of integration on both sides, never combined (question 2).
• Describing slope field regions without an inequality (question 3).
• GDC-only Euler answers with no update formula shown (questions 4, 7, 8, 9).
• Direction of rotation stated with no test point (questions 5 and 6).
• “Spiral” with no in or out (question 6).
• Coupled Euler updates that use the new value of one variable (questions 7 and 8).
• Taking the wrong root of a quadratic in an exact solution (question 9).
• Eigenvectors matched to the wrong eigenvalues (questions 10 and 11).
• Calling a saddle point “stable” (question 10).

Next steps

• Recap the key facts in the differential equations revision notes.
• Re-learn any weak topic in the differential equations study guide.
• See the rest of the course on the IB DP Maths AI course hub and tick off the printable syllabus checklist.
• Plan your final weeks with the AI exam preparation guide.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021.