Edexcel International A Level Maths · P4 · Pure Mathematics 4
Differential equations
28 free, high-quality past-paper-style questions on Differential equations for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 253 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P4-7-B-01P4 · A2Easy6 marks
The variables x and y, where y > 0, satisfy the differential equation dy/dx = 4x3 ÷ y.
(a)
Find the general solution, giving your answer in the form y2 = f(x).
(3)
(b)
Find the particular solution for which y = 3 when x = 1.
(2)
(c)
Hence find the value of y when x = 2, giving your answer to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
∫ y dy = ∫ 4x3 dx
M1
½y2 = x4 + c
A1
y2 = 2x4 + C
A1
Part (b)
9 = 2 + C
M1
y2 = 2x4 + 7
A1
Part (c)
y = √39 = 6.24 (6.2449…)
B1
Common error: Forgetting to double the constant: from ½y2 = x4 + c, multiplying by 2 gives 2x4 + 2c. Renaming 2c as C is fine; writing 2x4 + c and then using the old value of c is not.
A curve satisfies the differential equation dy/dx = 2xy.
(a)
Show that y = Aex2, where A is a constant, satisfies the differential equation.
(2)
(b)
Given that the curve passes through (0, 5), find the value of A.
(1)
(c)
Find the positive value of x for which y = 20, giving your answer to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
dy/dx = A × 2xex2 (chain rule)
M1
= 2x(Aex2) = 2xy as required
A1
Part (b)
5 = Ae0, so A = 5
B1
Part (c)
ex2 = 4 ⇒ x2 = ln 4
M1
x = √(ln 4) = 1.18 (1.1774…)
A1
Exam tip: To show that a given function is a solution, differentiate it and rewrite the result in terms of y. There is no need to solve the differential equation from scratch.
In each part of this question k is a positive constant and t is the time in minutes.
(a)
The number of cells, N, in a culture increases at a rate proportional to N. Write down a differential equation connecting N and t.
(1)
(b)
A steel rod is placed in a furnace kept at 900 °C. The temperature of the rod, θ °C, increases at a rate proportional to the difference between the furnace temperature and θ. Write down a differential equation connecting θ and t.
(2)
(c)
The radius, r cm, of a circular patch of oil on water increases at a rate inversely proportional to r. Write down a differential equation connecting r and t.
(1)
(d)
Given that r = 2 when t = 0, solve the differential equation in part (c), giving r2 in terms of k and t.
(2)
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Answer
Mark
Notes
Part (a)
dN/dt = kN
B1
Part (b)
rate proportional to the difference, i.e. k(900 − θ) or k(θ − 900)
M1
sign correct so that θ increases while θ < 900: dθ/dt = k(900 − θ)
A1
Part (c)
dr/dt = k ÷ r
B1
Part (d)
∫ r dr = ∫ k dt ⇒ ½r2 = kt + c
M1
c = 2, so r2 = 2kt + 4
A1
Exam tip: After writing an equation for an increasing or decreasing quantity, check the sign: with k > 0, does your right-hand side have the sign the description demands?
Figure 1 shows part of the curve C. At every point on C, dy/dx = x√y, where y > 0. The lowest point of C is the point A on the y-axis. The point B on C, with x > 0, has y-coordinate 16.
(a)
Find the general solution of the differential equation, giving your answer in the form √y = f(x).
(3)
(b)
Use the coordinates of A, read from Figure 1, to show that y = (x2 + 8)2 ÷ 16.
Common error: Squaring term by term. From √y = ¼x2 + 2 the curve is y = (¼x2 + 2)2, not y = 1⁄16x4 + 4: square the whole right-hand side as one bracket.
Figure 1 shows the direction field of the differential equation dy/dx = y(1 − x), y > 0, together with the solution curve that passes through the point P. Each short segment is centred on a grid point and has the gradient given by the differential equation at that point.
(a)
Find the gradient of the segment drawn at the point (2, 1), and explain why every segment drawn on the line x = 1 is horizontal.
(2)
(b)
Find the general solution, giving y in terms of x and a constant A.
(3)
(c)
Using the coordinates of P from Figure 1, show that the solution curve shown has equation y = 2e−½(x − 1)2.
(2)
(d)
Find the value of y on this curve when x = 3, giving your answer to 3 significant figures.
(1)
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Answer
Mark
Notes
Part (a)
gradient = 1 × (1 − 2) = −1
B1
on x = 1 the factor 1 − x is zero, so dy/dx = 0 for every y
B1
Part (b)
∫ 1 ÷ y dy = ∫ (1 − x) dx
M1
ln y = x − ½x2 + c
A1
y = Aex − ½x2
A1
Part (c)
P is (1, 2): 2 = Ae½, so A = 2e−½
M1
y = 2ex − ½x2 − ½ and x − ½x2 − ½ = −½(x − 1)2
A1
Part (d)
2e−2 = 0.271 (0.2706…)
B1
Exam tip: Where the segments of a direction field are horizontal, every solution curve crossing there has a stationary point. Here that is the line x = 1, so your solution should have its maximum at x = 1: a quick check on the algebra.
The curves C1, C2 and C3 in Figure 1 are three members of the family of solution curves of the differential equation dy/dx = −4x ÷ y, y ≠ 0. Each curve crosses the positive y-axis at a point with integer coordinates.
(a)
Show that the general solution can be written as 4x2 + y2 = k, where k is a constant.
(3)
(b)
Use Figure 1 to find the equation of C2, and hence find the coordinates of the points where C2 crosses the x-axis.
(2)
(c)
Find the exact coordinates of the point on C3 in the first quadrant at which the gradient is −1.
(3)
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Answer
Mark
Notes
Part (a)
∫ y dy = ∫ −4x dx
M1
½y2 = −2x2 + c
A1
4x2 + y2 = 2c = k
A1
Part (b)
C2 passes through (0, 6), so 4x2 + y2 = 36
B1
y = 0: x2 = 9, giving (3, 0) and (−3, 0)
B1
Part (c)
−4x ÷ y = −1 ⇒ y = 4x
M1
C3 passes through (0, 8): 4x2 + 16x2 = 64
M1
x = 4 ÷ √5, giving (4√5 ÷ 5, 16√5 ÷ 5)
A1
Common error: Using the constant before rearranging. ½y2 = −2x2 + c through (0, 6) gives c = 18, so the curve is 4x2 + y2 = 36, not 18. Rearrange first, then substitute the point.
The variables x and y satisfy the differential equation dy/dx = sin x cos2y, where −π⁄2 < y < π⁄2. Figure 1 shows the solution curve for 0 ≤ x ≤ 2π that passes through the point P(0, π⁄4). The point Q is the highest point on the curve.
(a)
Find the general solution, giving tan y in terms of x.
(3)
(b)
Show that the curve in Figure 1 has equation tan y = 2 − cos x.
(2)
(c)
Find the exact coordinates of Q.
(2)
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Answer
Mark
Notes
Part (a)
∫ sec2y dy = ∫ sin x dx
M1
tan y on the left
A1
tan y = −cos x + c
A1
Part (b)
x = 0, y = π⁄4: 1 = −1 + c
M1
c = 2, so tan y = 2 − cos x
A1
Part (c)
tan y increases with y, so y is greatest when cos x = −1, i.e. x = π
M1
Q(π, arctan 3)
A1
Common error: Integrating sin x as +cos x. Since d/dx(cos x) = −sin x, ∫ sin x dx = −cos x + c; with the wrong sign the curve would have its highest point at x = 0 instead of x = π.
Mould is growing on a slice of bread. The area covered by mould, A cm2, t days after it is first measured, increases at a rate proportional to A. When t = 0, A = 3, and when t = 2, A = 4.5.
(a)
Write down a differential equation connecting A and t.
(1)
(b)
Show that A = 3ekt.
(3)
(c)
Find the exact value of k.
(2)
(d)
Find the time taken for the area covered to reach 30 cm2, giving your answer in days to 3 significant figures.
(2)
(e)
Give one reason why this model will not be suitable for large values of t.
(1)
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Answer
Mark
Notes
Part (a)
dA/dt = kA, k > 0
B1
Part (b)
∫ 1 ÷ A dA = ∫ k dt
M1
ln A = kt + c ⇒ A = Bekt
M1
t = 0, A = 3 ⇒ B = 3, giving A = 3ekt
A1
Part (c)
3e2k = 4.5 ⇒ e2k = 1.5
M1
k = ½ ln 1.5
A1
Part (d)
ekt = 10 ⇒ t = ln 10 ÷ k = 2 ln 10 ÷ ln 1.5
M1
= 11.4 days (11.357…)
A1
Part (e)
the model predicts the area increases without limit, but the slice of bread has a finite area
B1
Exam tip: Keep k exact (½ ln 1.5) when you use it in the next part. Rounding k to 0.20 would give 11.5 days instead of 11.4.
A can of drink is taken from a fridge at 4 °C and left in a room kept at 25 °C. The temperature of the drink, θ °C, t minutes later is modelled by dθ/dt = k(25 − θ), where k is a positive constant. After 15 minutes the temperature of the drink is 10 °C.
(a)
Show that θ = 25 − 21e−kt.
(4)
(b)
Find the exact value of k.
(2)
(c)
Find the time at which the temperature of the drink reaches 18 °C, giving your answer to 3 significant figures.
(2)
(d)
Explain why, according to the model, the temperature of the drink never exceeds 25 °C.
(1)
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Answer
Mark
Notes
Part (a)
∫ 1 ÷ (25 − θ) dθ = ∫ k dt
M1
−ln(25 − θ) = kt + c
A1
25 − θ = Be−kt
M1
t = 0, θ = 4 ⇒ B = 21, giving θ = 25 − 21e−kt
A1
Part (b)
21e−15k = 15 ⇒ e15k = 7⁄5
M1
k = 1⁄15 ln 7⁄5
A1
Part (c)
21e−kt = 7 ⇒ ekt = 3 ⇒ t = 15 ln 3 ÷ ln 1.4
M1
= 49.0 minutes (48.98…)
A1
Part (d)
21e−kt > 0 for all t, so θ < 25; it approaches 25 °C as t → ∞
B1
Model sketch
Figure: the temperature approaches 25 °C but never reaches it
Common error: Integrating 1 ÷ (25 − θ) as +ln(25 − θ). The derivative of 25 − θ is −1, so the integral is −ln(25 − θ); with the wrong sign the drink would cool rather than warm.
Hence find y in terms of x, giving your answer as a single fraction.
(5)
(c)
Write down the limit of y as x → ∞.
(1)
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Answer
Mark
Notes
Part (a)
4 ≡ P(x + 3) + Q(x − 1); x = 1 and x = −3
M1
1 ÷ (x − 1) − 1 ÷ (x + 3)
A1
Part (b)
∫ 1 ÷ y dy = ∫ [1 ÷ (x − 1) − 1 ÷ (x + 3)] dx
M1
ln y = ln(x − 1) − ln(x + 3) + c ft
A1
y = A(x − 1) ÷ (x + 3)
M1
x = 5, y = 2: 2 = 4A ÷ 8
M1
y = 4(x − 1) ÷ (x + 3)
A1
Part (c)
4
B1
Exam tip: Collect all the logarithms, including the constant written as ln A, into a single log before exponentiating. Then the solution comes out as one fraction in one line.
The value of a car, £V thousand, t years after a fixed date, decreases at a rate proportional to its value. Figure 1 shows the graph of V against t. The two marked points lie on grid intersections.
(a)
Write down a differential equation connecting V and t, and show that its general solution is V = Ae−kt, where A and k are positive constants.
(3)
(b)
Use the two marked points to find the exact value of k and the value of A.
(4)
(c)
Find the time at which the model predicts that the value of the car will be £5000, giving your answer to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
dV/dt = −kV, k > 0
B1
∫ 1 ÷ V dV = ∫ −k dt ⇒ ln V = −kt + c
M1
V = ec × e−kt = Ae−kt
A1
Part (b)
18 = Ae−2k and 8 = Ae−6k
M1
dividing: e4k = 9⁄4
M1
k = ½ ln 1.5 (= ¼ ln 2.25)
A1
A = 18e2k = 18 × 1.5 = 27
A1
Part (c)
27e−kt = 5 ⇒ t = ln 5.4 ÷ k = 2 ln 5.4 ÷ ln 1.5
M1
t = 8.32 years (8.318…)
A1
Exam tip: When neither marked point is at t = 0, divide one equation by the other. A cancels and leaves a single exponential equation for k; then substitute back for A.
The curve C in Figure 1 satisfies the differential equation dy/dx = −kxy2, where k is a positive constant and y > 0. C passes through the points A and B, which lie on grid intersections. The tangent to C at B meets the x-axis at the point T, which also lies on a grid intersection.
(a)
Use the positions of B and T to find the gradient of C at B, and hence show that k = ⅓.
(2)
(b)
Using k = ⅓, find the general solution of the differential equation, giving 1 ÷ y in terms of x.
Common error: Taking the reciprocal term by term. From 1 ÷ y = ⅙x2 + ⅓ it does not follow that y = 6 ÷ x2 + 3. Combine the right-hand side into one fraction first, then invert.
Figure 1 shows part of a curve C in the first quadrant. P(x, y) is any point on C, M is the foot of the perpendicular from P to the x-axis, and the tangent to C at P meets the x-axis at T, to the left of M. For every position of P on C, the length TM is equal to x2. The curve passes through the point (1, 1).
(a)
Use triangle PTM to show that dy/dx = y ÷ x2.
(2)
(b)
Find the general solution, giving y in terms of x.
(3)
(c)
Show that C has equation y = e1 − 1/x.
(1)
(d)
Find the x-coordinate of the point on C where y = 2, giving your answer to 3 significant figures.
(2)
(e)
Describe the behaviour of y as x becomes very large.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
gradient of the tangent = PM ÷ TM
M1
PM = y and TM = x2, so dy/dx = y ÷ x2
A1
Part (b)
∫ 1 ÷ y dy = ∫ x−2 dx
M1
ln y = −1 ÷ x + c
A1
y = Ae−1/x
A1
Part (c)
1 = Ae−1 ⇒ A = e, so y = e1 − 1/x
B1
Part (d)
1 − 1 ÷ x = ln 2
M1
x = 1 ÷ (1 − ln 2) = 3.26 (3.258…)
A1
Part (e)
1 ÷ x → 0, so y increases towards e (≈ 2.72) but never reaches it: y = e is an asymptote
B1
Exam tip: Turn a geometric condition into a gradient by reading lengths off the right-angled triangle: the rise PM over the run TM. Both lengths are positive here, because T lies to the left of M and the curve is increasing.
A bowl of soup is left to cool in a kitchen where the temperature is a constant 20 °C. The temperature of the soup, θ °C, t minutes after it is served, is modelled by the differential equation dθ/dt = −k(θ − 20), where k is a positive constant. Figure 1 shows the graph of the model; the two marked points lie on grid intersections. The cross shows a temperature of 35 °C measured with a thermometer at t = 20.
(a)
Using the initial temperature shown in Figure 1, solve the differential equation to show that θ = 20 + 70e−kt.
(4)
(b)
Use the second marked point to find the exact value of k.
(2)
(c)
Find the rate at which the temperature of the soup is decreasing when t = 0, giving your answer to 3 significant figures.
(2)
(d)
Find the temperature predicted by the model when t = 20, and comment on the model in the light of the measured value.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
∫ 1 ÷ (θ − 20) dθ = ∫ −k dt
M1
ln(θ − 20) = −kt + c
A1
θ − 20 = Be−kt
M1
t = 0, θ = 90 ⇒ B = 70
A1
Part (b)
(10, 50): 70e−10k = 30
M1
k = 1⁄10 ln 7⁄3
A1
Part (c)
dθ/dt = −k(90 − 20) = −7 ln 7⁄3
M1
decreasing at 5.93 °C per minute (5.931…)
A1
Part (d)
20 + 70 × (3⁄7)2 = 32.9 °C (32.857…)
B1
the measured 35 °C is higher, so by t = 20 the model cools the soup a little too quickly (the value of k may need to be smaller later on)
B1
Common error: Writing the initial rate as k × 90. The rate depends on how much hotter the soup is than the room, so it is k(90 − 20).
A lake is stocked with fish. The number of fish, N, t years later is modelled by dN/dt = 0.2N − 40, where the term 40 represents the fish removed by anglers each year. N is treated as a continuous variable. Figure 1 shows two solution curves, A and B, which start from different initial numbers of fish, and the dashed horizontal line L, on which dN/dt = 0.
(a)
Find the equation of L, and explain, using the differential equation, why curve A moves away from L.
(2)
(b)
Using the initial value read from Figure 1, show that curve A has equation N = 200 + 50e0.2t.
(4)
(c)
Find the equation of curve B, and find the time at which the model predicts that no fish remain, giving your answer to 3 significant figures.
(3)
(d)
Give one limitation of the model.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
0.2N − 40 = 0 ⇒ N = 200
B1
for N > 200, dN/dt > 0, so N increases, and the rate of increase grows as N grows
B1
Part (b)
∫ 1 ÷ (0.2N − 40) dN = ∫ 1 dt
M1
5 ln|0.2N − 40| = t + c
A1
0.2N − 40 = Be0.2t
M1
t = 0, N = 250 ⇒ B = 10, so N = 200 + 50e0.2t
A1
Part (c)
t = 0, N = 150 ⇒ B = −10: N = 200 − 50e0.2t
B1
50e0.2t = 200 ⇒ e0.2t = 4
M1
t = 5 ln 4 = 6.93 years (6.931…)
A1
Part (d)
e.g. anglers could not keep removing 40 fish a year when very few remain; or curve A grows without limit, but the lake can only support a limited number of fish
B1
Exam tip: Keep the modulus in ln|0.2N − 40| until you exponentiate. Writing 0.2N − 40 = Be0.2t, with B allowed to be negative, covers both curves at once: B > 0 above L and B < 0 below it.
The variables x and y satisfy the differential equation dy/dx = x(4 − y2), where −2 < y < 2, and y = 0 when x = 0. Figure 1 shows the solution curve, which approaches the dashed line L as |x| becomes large. The point R on the curve, with x > 0, has y-coordinate 1.
(a)
Express 4 ÷ (4 − y2) in partial fractions.
(2)
(b)
Hence show that y = 2(e2x2 − 1) ÷ (e2x2 + 1).
(5)
(c)
Find the x-coordinate of R, giving your answer to 3 significant figures.
(2)
(d)
Find the equation of L, giving a reason.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
4 ≡ a(2 + y) + b(2 − y); y = 2 and y = −2
M1
1 ÷ (2 − y) + 1 ÷ (2 + y)
A1
Part (b)
∫ 4 ÷ (4 − y2) dy = ∫ 4x dx
M1
−ln(2 − y) + ln(2 + y)
A1
= 2x2 + c, and x = 0, y = 0 gives c = 0
A1
2 + y = (2 − y)e2x2
M1
y(1 + e2x2) = 2(e2x2 − 1), giving the result
A1
Part (c)
(2 + 1) ÷ (2 − 1) = e2x2 ⇒ e2x2 = 3
M1
x = √(½ ln 3) = 0.741 (0.7411…)
A1
Part (d)
dividing by e2x2: y = 2(1 − e−2x2) ÷ (1 + e−2x2) → 2 as |x| → ∞, so L is y = 2
B1
Exam tip: To make y the subject of (2 + y) ÷ (2 − y) = E, multiply out, collect the y terms on one side and factorise: y(1 + E) = 2(E − 1). Dividing the top and bottom by E then shows the limit.
A water butt is a vertical cylinder with cross-sectional area 0.5 m2. Water flows out through a tap at the base. When the depth of water is h metres, the volume of water decreases at a rate of 0.06√h m3 per minute. At time t = 0 minutes the depth is 0.81 m.
(a)
Show that dh/dt = −0.12√h.
(2)
(b)
Solve the differential equation to show that h = (0.9 − 0.06t)2.
(4)
(c)
Find the time taken for the water butt to empty, and state the values of t for which the formula in part (b) is valid.
(2)
(d)
Find the time taken for the depth to fall to half its initial value, giving your answer to 3 significant figures.
(2)
(e)
Explain why the second half of the depth takes longer to drain than the first half.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
V = 0.5h, so dV/dt = 0.5 dh/dt
M1
0.5 dh/dt = −0.06√h ⇒ dh/dt = −0.12√h
A1
Part (b)
∫ h−½ dh = ∫ −0.12 dt
M1
2√h = −0.12t + c
A1
t = 0, h = 0.81 ⇒ c = 1.8
M1
√h = 0.9 − 0.06t, so h = (0.9 − 0.06t)2
A1
Part (c)
0.9 − 0.06t = 0 ⇒ t = 15 minutes
B1
valid for 0 ≤ t ≤ 15; after that the butt is empty, though the formula would increase again
B1
Part (d)
√0.405 = 0.9 − 0.06t
M1
t = 4.39 minutes (4.393…)
A1
Part (e)
the rate of fall is proportional to √h, so it is smaller when the water is shallower: 15 − 4.39 = 10.6 min > 4.39 min
B1
Model sketch
Figure: the depth falls more slowly as h decreases; the butt is empty at t = 15
Exam tip: In a draining-tank model the square root falls linearly, so the tank empties in a finite time. State the interval of validity: the squared bracket would start increasing after the tank is empty.
A news story spreads through a town. The proportion of the town's adults who have heard the story, t days after it is first published, is x, where 0 < x < 1. The spread is modelled by dx/dt = 0.6x(1 − x). When t = 0, x = 0.1.
(a)
Express 1 ÷ [x(1 − x)] in partial fractions.
(2)
(b)
Show that ln(9x ÷ (1 − x)) = 0.6t.
(5)
(c)
Hence show that x = 1 ÷ (1 + 9e−0.6t).
(3)
(d)
Find the proportion of adults who have heard the story after 5 days, giving your answer to 3 significant figures.
(1)
(e)
Find the time at which 90% of adults have heard the story, giving your answer in days to 3 significant figures.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
1 ≡ a(1 − x) + bx; x = 0 and x = 1
M1
1 ÷ x + 1 ÷ (1 − x)
A1
Part (b)
∫ [1 ÷ x + 1 ÷ (1 − x)] dx = ∫ 0.6 dt
M1
∫ 1 ÷ x dx = ln x
A1
∫ 1 ÷ (1 − x) dx = −ln(1 − x), so ln x − ln(1 − x) = 0.6t + c
A1
t = 0, x = 0.1: c = ln(0.1 ÷ 0.9) = ln(1⁄9)
M1
ln(x ÷ (1 − x)) + ln 9 = 0.6t, giving the result
A1
Part (c)
9x = (1 − x)e0.6t
M1
x(9 + e0.6t) = e0.6t
M1
dividing by e0.6t: x = 1 ÷ (1 + 9e−0.6t)
A1
Part (d)
1 ÷ (1 + 9e−3) = 0.691 (0.6905…)
B1
Part (e)
1 + 9e−0.6t = 1 ÷ 0.9 ⇒ e0.6t = 81
M1
t = ln 81 ÷ 0.6 = 7.32 days (7.324…)
A1
Model sketch
Figure: logistic growth of the proportion who have heard the story
Common error: Integrating 1 ÷ (1 − x) as ln(1 − x). The inner derivative is −1, so it gives −ln(1 − x); the two logs then combine into ln(x ÷ (1 − x)).
A metal casting is left to cool in a workshop whose temperature is a constant θs °C. The temperature of the casting, θ °C, t minutes after it is put down, satisfies dθ/dt = −k(θ − θs), where k is a positive constant. The temperature of the casting is 100 °C when t = 0, 70 °C when t = 10 and 50 °C when t = 20.
(a)
Show that θ = θs + (100 − θs)e−kt.
(4)
(b)
Use the temperatures at t = 10 and t = 20 to show that θs = 10.
(4)
(c)
Find the exact value of k.
(2)
(d)
Find the time at which the temperature of the casting reaches 20 °C, giving your answer to 3 significant figures.
Figure: cooling of the casting towards the workshop temperature 10 °C
Exam tip: Readings at equally spaced times are linked by the same factor e−10k. Squaring one equation and comparing with the other removes k and leaves a linear equation for the unknown room temperature.
In a chemical reaction, x grams of a product have formed t minutes after the reaction starts. The reaction is modelled by dx/dt = k(2 − x)(4 − x), 0 ≤ x < 2, where k is a positive constant. When t = 0, x = 0.
Given that x = 1 when t = 5, show that k = 1⁄10 ln 1.5.
(2)
(d)
Find the time at which 1.5 grams of product have formed, giving your answer to 3 significant figures.
(3)
(e)
Explain what happens to the mass of product as t becomes large.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
2 ≡ a(4 − x) + b(2 − x); x = 2 and x = 4
M1
1 ÷ (2 − x) − 1 ÷ (4 − x)
A1
Part (b)
∫ 2 ÷ [(2 − x)(4 − x)] dx = ∫ 2k dt
M1
using (a): −ln(2 − x) + ln(4 − x)
A1
= 2kt + c
A1
t = 0, x = 0: c = ln 2
M1
ln[(4 − x) ÷ (2 − x)] − ln 2 = 2kt, giving the result
A1
Part (c)
ln[3 ÷ (2 × 1)] = 10k
M1
k = 1⁄10 ln 1.5
A1
Part (d)
(4 − 1.5) ÷ (2 × 0.5) = 2.5
M1
t = ln 2.5 ÷ (2k) = 5 ln 2.5 ÷ ln 1.5
M1
= 11.3 minutes (11.299…)
A1
Part (e)
as t → ∞ the right-hand side 2kt → ∞, so (4 − x) ÷ (2 − x) → ∞, i.e. 2 − x → 0
M1
the mass of product approaches 2 grams but never reaches it
A1
Exam tip: Multiply both sides by 2 before integrating so that the partial fractions from part (a) can be used exactly as they stand: ∫ 2 ÷ [(2 − x)(4 − x)] dx = ∫ 2k dt.
Figure 1 shows a funnel in the shape of an inverted right circular cone, of height 30 cm and top radius 15 cm, with its axis vertical. Water leaks out through a small hole at the vertex. When the depth of water is h cm, the radius of the water surface is r cm and the volume of water is V cm3. The water leaks out at a rate of 2π√h cm3 per second. At time t = 0 seconds the depth of water is 25 cm.
(a)
Use similar triangles to show that V = πh3 ÷ 12.
(2)
(b)
Show that dh/dt = −8 ÷ h3/2.
(3)
(c)
Solve the differential equation to show that h5/2 = 3125 − 20t.
(4)
(d)
Find the time taken for the funnel to empty, and the time taken for the depth to fall from 25 cm to 16 cm, giving your answers to 3 significant figures.
(2)
(e)
Describe how the rate at which the depth falls changes as the funnel empties.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
r ÷ h = 15 ÷ 30, so r = ½h
M1
V = ⅓πr2h = ⅓π(¼h2)h = πh3 ÷ 12
A1
Part (b)
dV/dh = ¼πh2
B1
dh/dt = (dV/dt) ÷ (dV/dh) with dV/dt = −2π√h
M1
dh/dt = −2π√h × 4 ÷ (πh2) = −8 ÷ h3/2
A1
Part (c)
∫ h3/2 dh = ∫ −8 dt
M1
⅖h5/2 = −8t + c
A1
t = 0, h = 25: c = ⅖ × 3125 = 1250
M1
multiplying by 5⁄2: h5/2 = 3125 − 20t
A1
Part (d)
h = 0: t = 3125 ÷ 20 = 156 s (156.25)
B1
h = 16: 165/2 = 1024, t = 2101 ÷ 20 = 105 s (105.05)
B1
Part (e)
|dh/dt| = 8 ÷ h3/2 increases as h decreases, so the depth falls faster and faster: the last 16 cm takes only about 51 s
B1
Exam tip: Link the rates with the chain rule, dh/dt = (dV/dt) ÷ (dV/dh), and put the minus sign into dV/dt because the volume is decreasing.
The curve in Figure 1 is a solution curve of the differential equation dy/dx = y ÷ [x2(x + 1)], x > 0, y > 0. As x → ∞ the curve approaches the dashed line y = 6. The point P on the curve has x-coordinate 1.
(a)
Express 1 ÷ [x2(x + 1)] in partial fractions.
(3)
(b)
Hence show that the general solution can be written as y = K(x + 1)e−1/x ÷ x, where K is a positive constant.
(4)
(c)
Use the asymptote shown in Figure 1 to find the value of K.
(2)
(d)
Find the exact y-coordinate of P.
(1)
(e)
Explain, using the differential equation, why y increases for all x > 0.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
1 ≡ ax(x + 1) + b(x + 1) + cx2
M1
x = 0: b = 1; x = −1: c = 1
A1
x2 terms: a + c = 0, a = −1, giving −1 ÷ x + 1 ÷ x2 + 1 ÷ (x + 1)
A1
Part (b)
∫ 1 ÷ y dy = ∫ [−1 ÷ x + 1 ÷ x2 + 1 ÷ (x + 1)] dx
M1
ln y = −ln x + ln(x + 1) …
A1
… − 1 ÷ x + c
A1
y = ec × (x + 1) ÷ x × e−1/x, with K = ec
A1
Part (c)
as x → ∞, (x + 1) ÷ x → 1 and e−1/x → 1, so y → K
M1
K = 6
A1
Part (d)
6 × 2 × e−1 = 12 ÷ e
B1
Part (e)
y > 0, x2 > 0 and x + 1 > 0, so dy/dx > 0 for every x > 0
B1
Common error: Leaving out the 1 ÷ x2 term. A repeated factor x2 needs both a ÷ x and b ÷ x2; the second integrates to −b ÷ x, which is where the factor e−1/x comes from.
Two hot drinks, A and B, are left in a room kept at 20 °C. Drink A is in a thin paper cup and drink B is in an insulated mug. The temperature, θ °C, of each drink t minutes later satisfies dθ/dt = −k(θ − 20), with k = kA for drink A and k = kB for drink B, where kA and kB are positive constants. Figure 1 shows the two temperature–time graphs, which cross at the point X(10, 50).
(a)
Using the initial temperature of drink A from Figure 1, show that for drink A, θ = 20 + 60e−kAt.
(3)
(b)
Write down the corresponding equation for drink B, and use X to find the exact values of kA and kB.
(3)
(c)
Find, to 3 significant figures, the rate at which the temperature of each drink is falling at X, and state which graph is steeper there.
(2)
(d)
For t > 10 drink B is warmer than drink A. Find the time at which the difference between their temperatures is greatest, giving your answer to 3 significant figures.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
∫ 1 ÷ (θ − 20) dθ = ∫ −kA dt
M1
ln(θ − 20) = −kAt + c
A1
θ − 20 = Be−kAt, and t = 0, θ = 80 gives B = 60
A1
Part (b)
drink B starts at 60 °C: θ = 20 + 40e−kBt
B1
60e−10kA = 30 ⇒ kA = 1⁄10 ln 2
B1
40e−10kB = 30 ⇒ kB = 1⁄10 ln 4⁄3
B1
Part (c)
both are 30 °C above the room: A falls at 3 ln 2 = 2.08 and B at 3 ln 4⁄3 = 0.863 °C per minute
Exam tip: Each drink has its own constant, but both curves pass through X, so the single reading at X gives both values. At X both drinks are 30 °C above the room, so the steeper graph is simply the one with the larger k.
The number of insects, P, in a nature reserve t months after the start of a year is modelled by the differential equation dP/dt = kP cos(πt ÷ 6), where k is a positive constant. Figure 1 shows the graph of P against t for 0 ≤ t ≤ 12. It has a maximum point M and a minimum point N, both on grid intersections.
(a)
Show that P = Ae(6k/π) sin(πt/6), where A is a positive constant.
(3)
(b)
Explain, using the differential equation, why the turning points occur at t = 3 and t = 9.
(2)
(c)
Use the coordinates of M and N to find the value of A, and show that k = (π ln 2) ÷ 6.
(4)
(d)
Find the first time after t = 0 at which P = 3000, giving your answer to 3 significant figures.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
∫ 1 ÷ P dP = ∫ k cos(πt ÷ 6) dt
M1
ln P = (6k ÷ π) sin(πt ÷ 6) + c
A1
P = ec × e(6k/π) sin(πt/6), with A = ec > 0
A1
Part (b)
k > 0 and P > 0, so dP/dt = 0 only when cos(πt ÷ 6) = 0
6k ÷ π = ln 2, so P = 2000 × 2sin(πt/6) = 3000 ⇒ 2sin(πt/6) = 1.5
M1
sin(πt ÷ 6) = ln 1.5 ÷ ln 2 = 0.58496…
M1
t = (6 ÷ π) arcsin(0.58496…) = 1.19 months (1.193…)
A1
Exam tip: With one equation at a maximum and one at a minimum, the exponents are equal and opposite: multiplying the two equations removes the exponential and gives A, dividing them removes A and gives k.
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