OxfordAQA International AS and A-level Mathematics: Forming and solving first-order differential equations (9660) – 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 cover section P2.8 (Differential equations) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. It belongs to International A-level (A2) Unit P2 and is not examined at International AS. Read the study guide or the revision notes first. See also the course hub, the printable checklist and the free 10-minute diagnostics. Give decimal answers to 3 significant figures unless told otherwise.

Questions

1. Write a differential equation for each statement. Define any constant you introduce.

(a) The number N of bacteria in a culture grows at a rate proportional to N. [1] (b) The mass m of a chemical decreases at a rate proportional to m². [1] (c) The depth h cm of water in a pond falls at a rate proportional to h, and rain adds 3 cm per day. [2]

2. Show that y = 2 + Ce^(−3x), where C is a constant, is a solution of dy/dx = 6 − 3y. [3]

3. Solve dy/dx = (2x + 1)y, given that y = 3 when x = 0. Give y in terms of x. [4]

4. For x > 0 and y > −2, solve x dy/dx = 3(y + 2), given y = 6 when x = 2. [5]

5. Given that y > 0, solve y dy/dx = x e^(−x²), where y = 1 when x = 0. Give y in terms of x. [5]

6. Solve dy/dx = y² sin x, given that y = 1/2 when x = 0. Give y in terms of x. [5]

7. The number P of bacteria in a colony grows at a rate proportional to P. At time t = 0 hours there are 2000 bacteria. At t = 5 there are 3500. Calculate how long it takes for the colony to grow to 10 000 bacteria, to 3 significant figures. [6]

8. A metal block at 150 °C is placed in a room at 25 °C. Its temperature T °C decreases at a rate proportional to (T − 25). After 8 minutes the block is at 90 °C. Find its temperature after a total of 20 minutes. [6]

9. Water drains from a tank. The depth h metres satisfies dh/dt = −k√h, where t is in minutes and k is a positive constant. Initially h = 16. After 5 minutes h = 9.

(a) Find k. [4] (b) Find the time taken for the tank to empty. [1] (c) Find the depth after 12 minutes. [1]

10. The volume of a spherical balloon increases at a constant 40 cm³ per second. The radius is r cm after t seconds. The volume of a sphere is V = (4/3)πr³.

(a) Show that dr/dt = 10/(πr²). [3] (b) When t = 0, r = 2. Find the time taken for the radius to reach 5 cm. [4]

11. A tank holds 100 litres of pure water. A salt solution of 0.5 kg per litre flows in at 4 litres per minute. The mixture is stirred well and leaves at 4 litres per minute. Let x kg be the salt in the tank after t minutes.

(a) Show that dx/dt = 2 − 0.04x. [3] (b) Solve the equation to find x in terms of t. [4] (c) Find the time for the tank to hold 25 kg of salt. [2]

12. A population P is modelled by dP/dt = P(400 − P)/2000, with P = 50 when t = 0.

(a) Express 1/(P(400 − P)) in partial fractions. [2] (b) By integrating, show that P/(400 − P) = Ae^(t/5) and find the value of A. [5] (c) Find the time at which P = 200. [2]

Answers

1. (a) dN/dt = kN, k > 0 [1] [1] (b) dm/dt = −km², k > 0 [1] [1] (c) Rate of change is a rise of 3 from the rain [1] minus a fall proportional to h: dh/dt = 3 − kh, k > 0 [1] [2] Examiner insight: A constant must be defined and its sign stated. In (c) two separate contributions need to appear in the equation.

2. dy/dx = −3Ce^(−3x) [1]. 6 − 3y = 6 − 6 − 3Ce^(−3x) [1] = −3Ce^(−3x), which equals dy/dx [1]. [3] Examiner insight: A “show that” question needs both sides worked out and compared. Substitute y into the right side and state that the two sides match.

3. ∫(1/y) dy = ∫(2x + 1) dx [1]. ln y = x² + x + c [1]. y = 3, x = 0: c = ln 3 [1]. ln y = x² + x + ln 3, so y = 3e^(x² + x) [1] [4] Examiner insight: Find the constant straight after integrating, then exponentiate the whole of the right side. Writing y = x² + x + ln 3 is a common slip.

4. ∫1/(y + 2) dy = ∫3/x dx [1]. ln(y + 2) = 3 ln x + c [1]. y + 2 = Ax³ [1]. x = 2, y = 6: 8 = 8A, A = 1 [1]. y = x³ − 2 [1] [5] Examiner insight: Combining logs via c = ln A keeps the working short. Check the answer: at x = 2, y = 8 − 2 = 6.

5. y dy = x e^(−x²) dx. ∫y dy = y²/2 [1]. ∫x e^(−x²) dx = −½e^(−x²) [1]. y²/2 = −½e^(−x²) + c. x = 0, y = 1: ½ = −½ + c, c = 1 [1]. y² = 2 − e^(−x²) [1]. y = √(2 − e^(−x²)), positive root as y > 0 [1] [5] Examiner insight: A reverse-chain-rule integral hides in the x side. Differentiate your answer to check it. State the root chosen when the question gives y > 0.

6. ∫y⁻² dy = ∫sin x dx [1]. −1/y = −cos x + c [2]. x = 0, y = ½: −2 = −1 + c, c = −1 [1]. 1/y = cos x + 1, so y = 1/(1 + cos x) [1] [5] Examiner insight: Take care with the sign when integrating y⁻². Each side of the integration is a separate piece of working that can be credited.

7. dP/dt = kP, so ln P = kt + c [1]. t = 0, P = 2000: P = 2000e^(kt) [1]. 3500 = 2000e^(5k), k = (1/5) ln 1.75 [1] = 0.112 [1]. 10 000 = 2000e^(kt): e^(kt) = 5 [1]. t = ln 5/k = 14.4 hours [1] [6] Examiner insight: Using the exact k (or ln 1.75 over 5) gives a more reliable final answer than using a rounded 0.112. Say what t means in the context.

8. dT/dt = −k(T − 25). ln(T − 25) = −kt + c [1]. T − 25 = Be^(−kt) [1]. t = 0: B = 125 [1]. t = 8: 65 = 125e^(−8k), k = (1/8) ln(125/65) = 0.0817 [1]. t = 20: T = 25 + 125e^(−20k) [1] = 49.4 °C [1] [6] Examiner insight: Apply the starting condition to T − 25, not to T. A temperature below the room temperature signals an error in the setup.

9. (a) h^(−1/2) dh = −k dt [1]. 2√h = −kt + c [1]. t = 0, h = 16: c = 8 [1]. t = 5, h = 9: 6 = 8 − 5k, k = 0.4 [1] [4] (b) Empty at √h = 0: 8 − 0.4t = 0, t = 20 minutes [1] [1] (c) √h = 8/2 − 0.2 × 12 = 1.6, so h = 2.56 m [1] [1] Examiner insight: Write the solution as √h in terms of t. Both later parts follow quickly from it, and a clear formula makes them easy to check.

10. (a) dV/dr = 4πr² [1]. dV/dt = (dV/dr)(dr/dt), so 40 = 4πr² dr/dt [1]. dr/dt = 10/(πr²) [1] [3] (b) πr² dr = 10 dt [1]. πr³/3 = 10t + c [1]. t = 0, r = 2: c = 8π/3 [1]. r = 5: 125π/3 = 10t + 8π/3, so t = 39π/10 = 12.3 seconds [1] [4] Examiner insight: Keep π as a symbol until the end. Show the chain rule line in (a), since the result is given.

11. (a) Salt in: 0.5 × 4 = 2 kg per minute [1]. Salt out: 4 litres per minute carrying x/100 kg per litre = 0.04x [1]. dx/dt = 2 − 0.04x [1] [3] (b) dx/(2 − 0.04x) = dt [1]. −25 ln(2 − 0.04x) = t + c [1]. t = 0, x = 0: c = −25 ln 2. So 2 − 0.04x = 2e^(−t/25) [1]. x = 50(1 − e^(−t/25)) [1] [4] (c) 25 = 50(1 − e^(−t/25)), e^(−t/25) = ½ [1]. t = 25 ln 2 = 17.3 minutes [1] [2] Examiner insight: Explain both the inflow and outflow terms in words as you build the equation. The tank never exceeds 50 kg, which is a useful check on part (b).

12. (a) 1/(P(400 − P)) = (1/400)(1/P + 1/(400 − P)) [2] [2] (b) ∫2000/(P(400 − P)) dP = ∫dt, so 5∫(1/P + 1/(400 − P)) dP = ∫dt [1]. 5[ln P − ln(400 − P)] = t + c [2]. P/(400 − P) = Ae^(t/5) [1]. P = 50: A = 50/350 = 1/7 [1] [5] (c) P = 200 gives 1 = (1/7)e^(t/5) [1]. t = 5 ln 7 = 9.73 [1] [2] Examiner insight: The integral of 1/(400 − P) has a minus sign from the chain rule. Check by differentiating, or by testing P = 50 at the end.

Where marks are usually lost

• Not separating: integrating the whole equation with respect to x only.
• Putting a constant on both sides, or leaving it out entirely.
• Using the initial condition on the wrong quantity (T instead of T − 25).
• Turning ln y = kt + c into y = kt + c.
• Rounding k early, then giving answers that drift from 3 significant figures.
• Wrong sign for k in decay, cooling or draining models.
• Missing the minus sign when integrating terms such as 1/(400 − P) or y⁻².
• Forgetting the chain rule when the rate of change is of V but the variable is r.
• Not saying what t, x or P stands for when giving a final answer in context.

Next steps

• Condense the topic with the revision notes.
• Go back to the study guide for method and worked examples.
• See the course hub and the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA. These questions cover section P2.8 of Unit P2.