Free Differential equations questions for A Level Mathematics (9709), part of Pure Mathematics 3. Each question has the answer and the marking points an examiner would look for.
Generate a full Differential Equations worksheet →All Pure Mathematics 3 questionsThe variables x and y satisfy the differential equation dydx = x e−y, and y = 0 when x = 0.
Solve the differential equation, obtaining y in terms of x, and find the exact value of y when x = 4.
| Part | Answer and marking points | Marks |
|---|---|---|
| y = ln(½x2 + 1); ln 9 separate variables M1 ey = ½x2 + C A1 use y = 0 when x = 0 M1 C = 1 A1 y = ln(½x2 + 1) A1 ln 9 A1 | 6 |
A fish population, P thousand at time t years, is harvested at a constant rate and is modelled by dPdt = (P − 1)(5 − P)8. When t = 0, P = 2.
Solve the differential equation, obtaining an expression for P in terms of t, and state what happens to the population in the long term.
Find the time at which P = 4.5.
In a different lake the same model applies, but P = 0.5 when t = 0. Find the exact time at which the population dies out.
In the first lake the harvest is increased, so that dPdt = −(P − 3)28, with P = 2 when t = 0. Find P in terms of t and the time at which the population dies out.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | P = 5et/2 + 3et/2 + 3; P → 5 (5000 fish) separate variables and partial fractions 14(1P − 1 + 15 − P) M1 14 lnP − 15 − P = t8 + c A1 use t = 0, P = 2 M1 P − 15 − P = 13et/2 A1 P = 5et/2 + 3et/2 + 3 oe A1 P → 5 B1 | 6 |
| (b) | t = 2 ln 21 = 6.09 years t = 2 ln 21 = 6.09 B1 | 1 |
| (c) | t = 2 ln 95 (1.18 years) P − 15 − P = −19et/2, set P = 0 M1 t = 2 ln 95 ≈ 1.18 A1 | 2 |
| (d) | P = 3 + 8t − 8; t = 163 ∫ (P − 3)−2 dP = −∫ 18 dt M1 1P − 3 = t8 − 1 A1 P = 3 + 8t − 8 A1 P = 0 at t = 163 A1 | 4 |
Solve the differential equation dydx = 2xy2, given that y = 1 when x = 0. Express y in terms of x.
| Part | Answer and marking points | Marks |
|---|---|---|
| y = 22 − 2x2 separate variables M1 −1y = 22x2 + c A1 use initial condition M1 c = −1 A1 y = 22 − 2x2 A1 | 5 |
The number of fish, N, in a lake satisfies dNdt = 0.02N(1 − N1000), where t is in weeks. Initially N = 100.
Solve the differential equation to find N in terms of t, and find the time when N = 500, giving your answer correct to 3 significant figures.
| Part | Answer and marking points | Marks |
|---|---|---|
| N = 10001 + 9e−0.02t; t = 110 weeks separate variables M1 partial fractions M1 integrate A1 use initial condition M1 N = 10001 + 9e−0.02t oe A1 substitute N = 500 M1 110 (allow exact 50 ln 9) A1 | 7 |
A liquid is cooling. Its temperature above the room temperature is θ °C at time t minutes. The rate of decrease of θ is proportional to θ. When t = 0, θ = 60; when t = 5, θ = 36.4.
Write down a differential equation and solve it to find θ in terms of t.
Find the time at which θ = 10.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | θ = 60e−0.1t dθdt = −kθ B1 separate and integrate to ln θ = −kt + c M1 use t = 0 to find c M1 k = 0.1, θ = 60e−0.1t A1 | 4 |
| (b) | 17.9 minutes ln 6 ÷ k M1 17.9 A1 | 2 |
The variables x and y satisfy dydx = 0.17y, and y = 13 when x = 0.
(a) Solve the differential equation, obtaining y in terms of x.
(b) Find the value of x when y = 52.
| Part | Answer and marking points | Marks |
|---|---|---|
| y = 13e0.17x; x = 8.15 Separate variables M1 ln y = 0.17x + c A1 Use condition M1 y = 13e0.17x A1 x = 8.15 A1 | 5 |
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