OCR A Level Mathematics A: Differential equations: separable equations and their interpretation (H240) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OCR A Level Mathematics A specification (Version 3.1, August 2026, for first assessment in 2018), sections 1.08k and 1.08l. Both are Stage 2, so they are A Level only. Learn the methods from the study guide or the revision notes first. The course page is OCR A Level Mathematics and the printable list is the checklist. The free diagnostics can show where to start.

Questions

1. Find the general solution of dy/dx = 6√x − 2/x². [3]

2. Given that x > 0 and y > 0, find the general solution of dy/dx = y/x². [4]

3. Solve dy/dx = 3eˣ/y², given that y = 3 when x = 0. Hence find y when x = ln 2, giving your answer to 3 significant figures. [5]

4. Solve dy/dx = 4x − 2xy, given that y = 5 when x = 0. Give y in terms of x. [5]

5. Solve dy/dx = e⁻ʸ sin x, given that y = 0 when x = π/2. State the values of x for which your solution is valid on the interval 0 ≤ x ≤ 2π. [5]

6. A radioactive sample has mass m grams after t days. The rate of decrease of m is proportional to m, so dm/dt = −km with k > 0. Initially m = 80.

(a) Solve the equation to show that m = 80e^(−kt). [2] (b) After 5 days the mass is 62 g. Find k to 3 significant figures. [2] (c) Find the half-life of the sample. [2]

7. The population P thousand of a town t years after 2020 is modelled by dP/dt = 0.03P, with P = 120 when t = 0.

(a) Find P in terms of t. [2] (b) Find the time for the population to double. [2] (c) Use the model to predict the population in 2070 and give one reason the model may be unreliable for such a long period. [2]

8. A thermometer at 5 °C is placed in a room kept at 21 °C. Its temperature T °C after t minutes satisfies dT/dt = k(21 − T), where k is a positive constant.

(a) Solve the equation to show that T = 21 − 16e^(−kt). [3] (b) The reading is 12 °C when t = 4. Find k, then the time for the reading to reach 20 °C. [3]

9. A trolley moves in a straight line. Its velocity v m s⁻¹ at time t seconds satisfies dv/dt = −kv², where k > 0. When t = 0, v = 12, and when t = 5, v = 8.

(a) Show that 1/v = kt + 1/12. [3] (b) Find k and hence show that v = 120/(t + 10). [2] (c) Find the velocity when t = 15 and the distance travelled in the first 15 seconds. [3]

10. A small sphere falls through a liquid. Its speed v m s⁻¹ after t seconds satisfies dv/dt = 9.8 − 0.35v, and v = 0 when t = 0.

(a) Show that v = 28(1 − e^(−0.35t)). [6] (b) State the terminal velocity. [1] (c) Find the time for the sphere to reach 20 m s⁻¹. [2] (d) Does the sphere ever reach its terminal velocity according to this model? [1]

11. A tank empties through a hole in its base. Its depth h metres at time t seconds satisfies dh/dt = −0.25√h, with h = 25 when t = 0.

(a) Solve the equation to find h in terms of t. [4] (b) Find the time at which the tank is empty. [1] (c) Find the depth when t = 16. [1] (d) Explain why the solution cannot describe the tank for t greater than your answer to part (b). [2]

Answers

1. 6x^(1/2) integrates to 4x^(3/2) [1]; −2x⁻² integrates to 2x⁻¹ [1]; add the constant: y = 4x^(3/2) + 2/x + c [1] [3] Examiner insight: the constant is easy to forget when there is no initial condition; a general solution without it is incomplete.

2. Separate: (1/y) dy = x⁻² dx [1]. Integrate: ln y = −1/x + c [1]. Exponentiate: y = eᶜe^(−1/x) [1]. So y = Ae^(−1/x), where A is a positive constant [1] [4] Examiner insight: writing y = e^(−1/x) + c is a common slip; the constant must multiply, so keep it as A.

3. y² dy = 3eˣ dx [1]. y³/3 = 3eˣ + c [1]. At (0, 3): 9 = 3 + c, so c = 6 [1]. y³ = 9eˣ + 18, so y = ∛(9eˣ + 18) [1]. At x = ln 2: y = ∛36 = 3.30 [1] [5] Examiner insight: substituting the condition before integrating, or forgetting the factor 3 when multiplying through by 3, are the two likeliest places to go wrong.

4. Factorise: dy/dx = 2x(2 − y) [1]. ∫ 1/(2 − y) dy = ∫ 2x dx [1]. −ln|2 − y| = x² + c [1]. So y − 2 = Ae^(−x²), and at (0, 5), A = 3 [1]. y = 2 + 3e^(−x²) [1] [5] Examiner insight: if you do not take out the common factor you cannot separate, so the whole method is lost, not just one step.

5. e^y dy = sin x dx [1]. e^y = −cos x + c [1]. At (π/2, 0): 1 = 0 + c, so c = 1 [1]. y = ln(1 − cos x) [1]. Since 1 − cos x = 0 at x = 0 and 2π, the solution is valid for 0 < x < 2π [1] [5] Examiner insight: a request to state where the solution holds is looking for a reason tied to the formula, such as the logarithm needing a positive argument.

6. (a) ∫ 1/m dm = ∫ −k dt gives ln m = −kt + c [1]. At t = 0, m = 80, so m = 80e^(−kt) [1] (b) 62 = 80e^(−5k), so k = ln(80/62)/5 = 0.0510 [2] (c) 40 = 80e^(−kt), so t = ln 2/k = 13.6 days [2] [6] Examiner insight: carry k unrounded into part (c); using a rounded k can change the third significant figure.

7. (a) ln P = 0.03t + c [1], so P = 120e^(0.03t) [1] (b) 240 = 120e^(0.03t), so t = ln 2/0.03 = 23.1 years [2] (c) In 2070, t = 50 and P = 120e^(1.5) = 538 thousand (about 538 000) [1]. The model allows unlimited growth, but resources and space limit a real town [1] [6] Examiner insight: the limitation must refer to the situation (resources, space, migration), not just say the model is unrealistic.

8. (a) ∫ 1/(21 − T) dT = ∫ k dt [1]. −ln|21 − T| = kt + c, so 21 − T = Ae^(−kt) [1]. At t = 0, T = 5, so A = 16 and T = 21 − 16e^(−kt) [1] (b) 9 = 16e^(−4k), so k = ln(16/9)/4 = 0.144 [1]. For T = 20: 1 = 16e^(−kt) [1], so t = ln 16/k = 19.3 minutes [1] [6] Examiner insight: in a “show that” question, every line from the separation to the final form should be on the page.

9. (a) ∫ v⁻² dv = ∫ −k dt [1]. −1/v = −kt + c [1]. At t = 0, v = 12: c = −1/12, so 1/v = kt + 1/12 [1] (b) At t = 5: 1/8 = 5k + 1/12, so 5k = 1/24 and k = 1/120 [1]. Then 1/v = (t + 10)/120, so v = 120/(t + 10) [1] (c) v = 120/25 = 4.8 m s⁻¹ [1]. Distance = ∫₀¹⁵ 120/(t + 10) dt = [120 ln(t + 10)] from 0 to 15 [1] = 120 ln 2.5 = 110 m (3 s.f.) [1] [8] Examiner insight: distance needs integration of v with respect to t; substituting t = 15 into v gives a velocity, not a distance.

10. (a) ∫ 1/(9.8 − 0.35v) dv = ∫ dt [1]. −(1/0.35) ln|9.8 − 0.35v| = t + c [2]. So 9.8 − 0.35v = Ae^(−0.35t) [1]. At t = 0, v = 0: A = 9.8 [1]. Then 0.35v = 9.8(1 − e^(−0.35t)), so v = 28(1 − e^(−0.35t)) [1] (b) As t → ∞, e^(−0.35t) → 0, so the terminal velocity is 28 m s⁻¹ [1] (c) 20 = 28(1 − e^(−0.35t)), so e^(−0.35t) = 2/7 [1]. t = ln 3.5/0.35 = 3.58 s [1] (d) No: e^(−0.35t) is always positive, so v is always less than 28 and only approaches it [1] [10] Examiner insight: the coefficient −1/0.35 from integrating 1/(a − bv) is where answers most often go wrong; check it by differentiating your answer.

11. (a) h^(−1/2) dh = −0.25 dt [1]. 2√h = −0.25t + c [1]. At t = 0, h = 25: c = 10 [1]. √h = 5 − t/8, so h = (5 − t/8)² [1] (b) h = 0 when t = 40 s [1] (c) h = (5 − 2)² = 9 m [1] (d) After t = 40 the formula gives h > 0 and increasing again [1], but a draining tank cannot refill, and the original equation has no meaning once the tank is empty. So the model holds for 0 ≤ t ≤ 40 only [1] [8] Examiner insight: squaring √h = 5 − t/8 loses the condition 5 − t/8 ≥ 0, which is the root of the limitation in part (d).

Where marks are usually lost

• Putting a constant on both sides of the integrated equation.
• Failing to factorise first, as in question 4, so the variables cannot be separated.
• Writing y − a = e^(g(x)) + c rather than y − a = Ae^(g(x)).
• Substituting the initial condition into the differential equation instead of the integrated result.
• Using a rounded value of k in a later calculation.
• A sign error for decay: “rate of decrease” needs the minus sign in the equation.
• Vague limitations with no link to the context.
• Stopping at the solution when the question asks for a time, a limit or a distance.

Next steps

• Revise from the revision notes.
• Re-read the study guide.
• Course page: OCR A Level Mathematics and the checklist.
• Try the free diagnostics to find your weakest topics.
• Book a free trial class.

Official syllabus

OCR, A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. Sections 1.08k and 1.08l.