Free Differential equations & Euler’s method questions for IB Mathematics: Applications & Interpretation (SL & HL), HL only, part of Calculus. Each question has the answer and the marking points an examiner would look for.
Generate a full Differential Equations & Euler’s Method worksheet →All Calculus questionsConsider the differential equation dydx = x − 0.5y, with y = 2 when x = 0.
Use Euler’s method with step length 0.2 to estimate y when x = 0.8.
| Part | Answer and marking points | Marks |
|---|---|---|
| 1.54 use of Euler’s formula (M1) y1 = 1.80 (A1) y2 = 1.66 (A1) y ≈ 1.54 A1 | 4 |
The rate of growth of a population P is proportional to P, so that dPdt = kP. Initially P = 500, and after 20 years P = 2477.
By solving the differential equation, show that P = 500ekt.
Find the value of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | shown separating variables: ∫1PdP = ∫k dt M1 ln P = kt + c A1 using P(0) = 500 AG A1 | 3 |
| (b) | 0.0800 2477 = 500e20k (M1) k = 0.0800 A1 | 2 |
Consider the differential equation dydx = e−x + y, with y = 2 when x = 0.
Use Euler’s method with step length 0.25 to estimate y when x = 1.
| Part | Answer and marking points | Marks |
|---|---|---|
| 5.98 use of Euler’s formula (M1) y1 = 2.75 (A1) y2 = 3.63 (A1) y ≈ 5.98 A1 | 4 |
The rate of growth of a population P is proportional to P, so that dPdt = kP. Initially P = 1200, and after 20 years P = 3262.
By solving the differential equation, show that P = 1200ekt.
Find the value of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | shown separating variables: ∫1PdP = ∫k dt M1 ln P = kt + c A1 using P(0) = 1200 AG A1 | 3 |
| (b) | 0.0500 3262 = 1200e20k (M1) k = 0.0500 A1 | 2 |
Consider the differential equation dydx = xy + 1, with y = 2 when x = 0.
Use Euler’s method with step length 0.1 to estimate y when x = 0.4.
| Part | Answer and marking points | Marks |
|---|---|---|
| 2.43 use of Euler’s formula (M1) y1 = 2.10 (A1) y2 = 2.20 (A1) y ≈ 2.43 A1 | 4 |
The rate of growth of a population P is proportional to P, so that dPdt = kP. Initially P = 3000, and after 20 years P = 8155.
By solving the differential equation, show that P = 3000ekt.
Find the value of k.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | shown separating variables: ∫1PdP = ∫k dt M1 ln P = kt + c A1 using P(0) = 3000 AG A1 | 3 |
| (b) | 0.0500 8155 = 3000e20k (M1) k = 0.0500 A1 | 2 |
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Meet Riyaz →Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in