Free worked solutions · 2026 Edition

Chapter 16: Differential equations worked solutions

Worked solutions to the practice questions in Chapter 16 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[8 marks]

Consider the differential equation

\dfrac{dy}{dx} + 3y = 9x.

(a) Verify that y = 3x - 1 is a solution of the differential equation.

[2]

(b) Show that y = Ae^{-3x} + 3x - 1 is a solution of the differential equation for every constant A.

[3]

The solution in part (b) is the general solution of the differential equation.

(c) Find the particular solution for which y = 4 when x = 0.

[2]

(d) Describe the behaviour of every solution of the differential equation for large positive values of x, giving a reason.

[1]

Show solutionHide solution

(a) \dfrac{dy}{dx} = 3, so \dfrac{dy}{dx} + 3y = 3 + 3(3x - 1)

= 9x, so the equation is satisfied

(b) \dfrac{dy}{dx} = -3Ae^{-3x} + 3

\dfrac{dy}{dx} + 3y = -3Ae^{-3x} + 3 + 3Ae^{-3x} + 9x - 3

= 9x for every value of A A1AG

(c) 4 = A + 0 - 1 \Rightarrow A = 5

y = 5e^{-3x} + 3x - 1

(d) Ae^{-3x} \to 0 as x \to \infty, so every solution approaches the line y = 3x - 1

Final answer

(c) y=5e^{-3x}+3x-1
(d) every solution approaches the line y=3x-1

Question 2

[7 marks]

Write each statement as a differential equation. Define any constant you introduce and state its sign.

(a) The number of bacteria, B, in a culture increases at a rate proportional to B, where t is the time in hours.

[2]

(b) The speed, v\ \text{m s}^{-1}, of an ice skater who stops pushing decreases at a rate proportional to the square of the speed, where t is the time in seconds.

[2]

(c) At every point (x,\ y) of a curve, the gradient is three times the product of the coordinates.

[1]

(d) The balance, A euros, of a savings account t years after it is opened grows at a rate proportional to A, while money is also withdrawn continuously at a rate of 500 euros per year.

[2]

Show solutionHide solution

(a) \dfrac{dB}{dt} = kB

with k > 0 a constant

(b) \dfrac{dv}{dt} = -kv^2

with k > 0 a constant

Note: Accept \dfrac{dv}{dt} = kv^2 with k < 0.

(c) \dfrac{dy}{dx} = 3xy

(d) \dfrac{dA}{dt} = kA - 500

with k > 0 a constant

Final answer

(a) \frac{dB}{dt}=kB, k>0
(b) \frac{dv}{dt}=-kv^2, k>0
(c) \frac{dy}{dx}=3xy
(d) \frac{dA}{dt}=kA-500, k>0

Question 3

[9 marks]

A pond holds 2000\ \text{m}^3 of water in which P kg of a chemical is dissolved, t days after monitoring starts. Clean water flows into the pond at 40\ \text{m}^3 per day, and well-mixed pond water flows out at the same rate, so the volume stays constant. In addition, a nearby factory leaks 6 kg of the chemical into the pond each day. Initially there are 100 kg of the chemical in the pond.

(a) Show that \dfrac{dP}{dt} = 6 - \dfrac{P}{50}.

[2]

(b) Find the constant solution of the differential equation, and interpret it in this context.

[2]

(c) Solve the differential equation to show that P = 300 - 200e^{-\frac{t}{50}}.

[4]

(d) Find the exact time at which there are 200 kg of the chemical in the pond.

[1]

Show solutionHide solution

(a) the concentration is \dfrac{P}{2000} kg per \text{m}^3, so the chemical leaves at 40 \times \dfrac{P}{2000} = \dfrac{P}{50} kg per day

rate of change = rate in - rate out = 6 - \dfrac{P}{50} A1AG

(b) 6 - \dfrac{P}{50} = 0 \Rightarrow P = 300

if there are 300 kg in the pond, the chemical flows out exactly as fast as it leaks in, so the amount stays at 300 kg

(c) \displaystyle\int \dfrac{1}{300 - P}\,dP = \displaystyle\int \dfrac{1}{50}\,dt

-\ln(300 - P) = \dfrac{t}{50} + C

t = 0, P = 100: C = -\ln 200

\ln\left( \dfrac{300 - P}{200} \right) = -\dfrac{t}{50}, so 300 - P = 200e^{-\frac{t}{50}} and P = 300 - 200e^{-\frac{t}{50}} A1AG

(d) 200e^{-\frac{t}{50}} = 100 \Rightarrow t = 50\ln 2 days

Final answer

(b) P=300 kg: the chemical entering balances the chemical leaving
(d) t=50\ln2 days

Question 4

[7 marks]

Consider the differential equation \dfrac{dy}{dx} = e^{2x + y}, where y = 0 when x = 0.

(a) Show that the differential equation can be written as e^{-y}\dfrac{dy}{dx} = e^{2x}.

[1]

(b) Hence find y in terms of x.

[5]

(c) Explain why the solution exists only for x < \dfrac{1}{2}\ln 3.

[1]

Show solutionHide solution

(a) e^{2x + y} = e^{2x}e^{y}; dividing by e^y gives e^{-y}\dfrac{dy}{dx} = e^{2x} A1AG

(b) \displaystyle\int e^{-y}\,dy = \displaystyle\int e^{2x}\,dx

-e^{-y} = \dfrac{1}{2}e^{2x} + C A1A1

-1 = \dfrac{1}{2} + C \Rightarrow C = -\dfrac{3}{2}

e^{-y} = \dfrac{3 - e^{2x}}{2}, so y = \ln\left( \dfrac{2}{3 - e^{2x}} \right)

Note: Accept y = -\ln\left( \dfrac{3 - e^{2x}}{2} \right) or y = \ln 2 - \ln\left( 3 - e^{2x} \right).

(c) the logarithm needs 3 - e^{2x} > 0, that is e^{2x} < 3, so x < \dfrac{1}{2}\ln 3

Final answer

(b) y=\ln\frac{2}{3-e^{2x}}
(c) the logarithm needs 3-e^{2x}>0

Question 5

[5 marks]

The differential equation \dfrac{dy}{dx} = \dfrac{2y}{x + 1} has y = 1 when x = 0. Euler's method with step length h = 0.5 is used to approximate its solution.

(a) Write down formulae for x_{n+1} and y_{n+1} in terms of x_n and y_n.

[1]

(b) Use Euler's method to find an approximate value of y when x = 1. Show your values of n, x_n and y_n in a table.

[4]

Show solutionHide solution

(a) x_{n+1} = x_n + 0.5, y_{n+1} = y_n + 0.5 \times \dfrac{2y_n}{x_n + 1}

(b) y_1 = 1 + 0.5 \times \dfrac{2}{1} = 2 (M1)A1

y_2 = 2 + 0.5 \times \dfrac{4}{1.5} = 2 + \dfrac{4}{3} = \dfrac{10}{3}

n	x_n	y_n	\dfrac{2y_n}{x_n + 1}
0	0	1	2
1	0.5	2	\dfrac{8}{3}
2	1	\dfrac{10}{3}

y \approx \dfrac{10}{3} (3.33) when x = 1

Final answer

(b) y\approx\frac{10}{3}

Question 6

[8 marks]

Consider the differential equation

x\dfrac{dy}{dx} = y^2 - 3y + 2,

where x > 0 and y = 3 when x = 1.

(a) Express \dfrac{1}{y^2 - 3y + 2} in partial fractions.

[2]

(b) Hence show that y = \dfrac{4 - x}{2 - x}, for 0 < x < 2.

[6]

Show solutionHide solution

(a) \dfrac{1}{(y - 1)(y - 2)} = \dfrac{A}{y - 2} + \dfrac{B}{y - 1}, 1 = A(y - 1) + B(y - 2)

= \dfrac{1}{y - 2} - \dfrac{1}{y - 1}

(b) \displaystyle\int \left( \dfrac{1}{y - 2} - \dfrac{1}{y - 1} \right) dy = \displaystyle\int \dfrac{1}{x}\,dx

\ln\lvert y - 2 \rvert - \ln\lvert y - 1 \rvert

= \ln x + C

x = 1, y = 3: \ln 1 - \ln 2 = C, so C = -\ln 2

\ln\left( \dfrac{y - 2}{y - 1} \right) = \ln\left( \dfrac{x}{2} \right), so \dfrac{y - 2}{y - 1} = \dfrac{x}{2}

2y - 4 = xy - x \Rightarrow y(2 - x) = 4 - x \Rightarrow y = \dfrac{4 - x}{2 - x} A1AG

Note: Near y = 3 both y - 2 and y - 1 are positive, so the moduli may be dropped.

Final answer

(a) \frac{1}{y-2}-\frac{1}{y-1}

Question 7

[10 marks]

A small steel ball is released from rest in a tall cylinder of glycerine. Its velocity, v\ \text{cm s}^{-1}, t seconds after it is released satisfies

\dfrac{dv}{dt} = 30 - 5v, \quad 0 \le v < 6.

(a) By solving the differential equation, show that v = 6\left( 1 - e^{-5t} \right).

[5]

(b) Write down the terminal velocity of the ball (the limit of v as t \to \infty).

[1]

(c) Find the time taken for the ball to reach 95\% of its terminal velocity.

[2]

(d) Find the distance that the ball falls in the first second.

[2]

Show solutionHide solution

(a) \displaystyle\int \dfrac{1}{30 - 5v}\,dv = \displaystyle\int 1\,dt

-\dfrac{1}{5}\ln(30 - 5v)

= t + C

t = 0, v = 0: C = -\dfrac{1}{5}\ln 30

\ln\left( \dfrac{30 - 5v}{30} \right) = -5t \Rightarrow 30 - 5v = 30e^{-5t} \Rightarrow v = 6\left( 1 - e^{-5t} \right) A1AG

(b) 6\ \text{cm s}^{-1}

(c) 6\left( 1 - e^{-5t} \right) = 5.7 \Rightarrow e^{-5t} = 0.05

t = \dfrac{\ln 20}{5} = 0.599 s (0.599146\ldots)

(d) distance = \displaystyle\int_{0}^{1} 6\left( 1 - e^{-5t} \right) dt (M1)

= 4.81 cm (4.80808\ldots)

Note: The exact value is 6 - \dfrac{6}{5}\left( 1 - e^{-5} \right).

Final answer

(b) 6 cm s^{-1}
(c) t=\frac{\ln20}{5}=0.599 s
(d) 4.81 cm

Question 8

[8 marks]

Consider the differential equation

\dfrac{dy}{dx} = \dfrac{2x + 3y}{x}, \quad x > 0,

where y = 0 when x = 1.

(a) Given that y = vx, where v is a function of x, show that \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}.

[1]

(b) Hence show that x\dfrac{dv}{dx} = 2(1 + v).

[2]

(c) Hence find y in terms of x.

[5]

Show solutionHide solution

(a) by the product rule, \dfrac{dy}{dx} = v \times 1 + x\dfrac{dv}{dx} A1AG

(b) \dfrac{2x + 3y}{x} = 2 + 3\dfrac{y}{x} = 2 + 3v, so v + x\dfrac{dv}{dx} = 2 + 3v

x\dfrac{dv}{dx} = 2 + 2v = 2(1 + v) A1AG

(c) \displaystyle\int \dfrac{1}{1 + v}\,dv = \displaystyle\int \dfrac{2}{x}\,dx

\ln\lvert 1 + v \rvert = 2\ln x + C A1A1

x = 1, v = 0: C = 0, so 1 + v = x^2

\dfrac{y}{x} = x^2 - 1, so y = x^3 - x

Final answer

Question 9

[7 marks]

Consider the differential equation

x\dfrac{dy}{dx} + 4y = 7x^3, \quad x > 0,

where y = 3 when x = 1.

(a) Show that x^4 is an integrating factor for this differential equation.

[3]

(b) Hence find y in terms of x.

[4]

Show solutionHide solution

(a) dividing by x: \dfrac{dy}{dx} + \dfrac{4}{x}y = 7x^2

P(x) = \dfrac{4}{x}, \displaystyle\int \dfrac{4}{x}\,dx = 4\ln x

integrating factor e^{4\ln x} = x^4 A1AG

(b) \dfrac{d}{dx}\left( x^4y \right) = x^4 \times 7x^2 = 7x^6

x^4y = x^7 + C

x = 1: 3 = 1 + C \Rightarrow C = 2

y = x^3 + \dfrac{2}{x^4}

Note: Award M0 in part (a) if the integrating factor is found from P(x) = 4 without dividing by x.

Final answer

(b) y=x^3+\frac{2}{x^4}

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 10

[5 marks]

Consider the differential equation \dfrac{dy}{dx} = 2x - y - 1.

(a) Show that y = 2x - 3 is a solution of the differential equation.

[2]

(b) A solution curve C passes through (0,\,4). Explain why C lies above the line y = 2x - 3 for all x.

[2]

(c) Find the gradient of C at (0,\,4).

[1]

Show solutionHide solution

(a) If y = 2x - 3 then \dfrac{dy}{dx} = 2.

[1]

Right-hand side: 2x - (2x - 3) - 1 = 2. The two sides agree, so the line is a solution.

[1]

(b) Two different solution curves can never cross, because the differential equation fixes a single gradient at every point.

[1]

At x = 0, C is at y = 4, above the line (y = -3); since it cannot cross the line, it stays above it for all x.

[1]

(c) \dfrac{dy}{dx} = 2(0) - 4 - 1 = -5.

[1]

Final answer

(b) solution curves cannot cross the solution y=2x-3
(c) -5

Question 11

[10 marks]

Consider \dfrac{dy}{dx} = (4 - y^2)\cos x, with y = 0 when x = 0.

(a) Write down the two constant solutions of the differential equation.

[2]

(b) Show that \dfrac{1}{4-y^2} \equiv \dfrac14\left(\dfrac{1}{2-y} + \dfrac{1}{2+y}\right).

[2]

(c) Hence solve the differential equation, giving y in terms of x.

[6]

Show solutionHide solution

(a) A constant solution has \dfrac{dy}{dx} = 0 for all x, so 4 - y^2 = 0.

y = 2 and y = -2.

[2]

(b) \dfrac14\left(\dfrac{1}{2-y} + \dfrac{1}{2+y}\right) = \dfrac14\cdot\dfrac{(2+y)+(2-y)}{4-y^2}

[1]

= \dfrac14\cdot\dfrac{4}{4-y^2} = \dfrac{1}{4-y^2}. [1];

(c) Separate and use (b): \displaystyle\int\frac14\left(\frac{1}{2-y} + \frac{1}{2+y}\right)\,dy = \int\cos x\,dx.

[1]

\dfrac14\ln\left|\dfrac{2+y}{2-y}\right| = \sin x + C.

[2]

At (0,\,0): \ln 1 = 0, so C = 0.

[1]

Since -2 < y < 2 the fraction is positive: \dfrac{2+y}{2-y} = e^{4\sin x}, so 2 + y = (2 - y)e^{4\sin x} and y\left(1 + e^{4\sin x}\right) = 2\left(e^{4\sin x} - 1\right).

y = \dfrac{2\left(e^{4\sin x} - 1\right)}{e^{4\sin x} + 1}.

[2]

Final answer

(a) y=\pm2
(c) y=\frac{2\left(e^{4\sin x}-1\right)}{e^{4\sin x}+1}

Question 12

[10 marks]

Consider \dfrac{dy}{dx} = 2 - 3y, with y = 0 when x = 0.

(a) Use Euler's method with step length 0.2 to estimate y when x = 1.

[3]

(b) Find \dfrac{d^2y}{dx^2} in terms of y and hence explain whether your estimate in part (a) is an overestimate or an underestimate.

[3]

(c) Solve the differential equation to show that y = \dfrac23\left(1 - e^{-3x}\right), and hence find the percentage error in your estimate from part (a).

[4]

Show solutionHide solution

(a) y_{n+1} = y_n + 0.2(2 - 3y_n), five steps.

[1]

y_1 = 0.4,\ y_2 = 0.56,\ y_3 = 0.624,\ y_4 = 0.6496,\ y_5 = 0.65984.

[1]

y(1) \approx 0.660 (3 s.f.).

[1]

(b) \dfrac{d^2y}{dx^2} = -3\dfrac{dy}{dx} = -3(2 - 3y).

[2]

For 0 \le y < \dfrac23, 2 - 3y > 0, so \dfrac{d^2y}{dx^2} < 0: the curve is concave down, the tangent lines lie above it, and Euler's method gives an overestimate.

[1]

(c) Separate: \displaystyle\int\frac{1}{2 - 3y}\,dy = \int\,dx, so -\dfrac13\ln(2 - 3y) = x + C.

[2]

At (0,\,0): C = -\dfrac13\ln 2, so 2 - 3y = 2e^{-3x} and y = \dfrac23\left(1 - e^{-3x}\right). [1];

Exact y(1) = \dfrac23\left(1 - e^{-3}\right) = 0.6335. Percentage error = \dfrac{0.65984 - 0.63347}{0.63347}\times 100 = 4.16\%.

[1]

Final answer

(a) 0.660
(b) \frac{d^2y}{dx^2}=-3(2-3y)<0, so an overestimate
(c) 4.16\%

Question 13

[12 marks]

A falling object experiences air resistance proportional to the square of its speed. Its speed v m s^{-1} at time t seconds satisfies \dfrac{dv}{dt} = 10 - 0.1v^2, with v = 0 when t = 0.

(a) Find the terminal speed.

[2]

(b) Show that \dfrac{1}{100 - v^2} \equiv \dfrac{1}{20}\left(\dfrac{1}{10 - v} + \dfrac{1}{10 + v}\right).

[2]

(c) Hence show that v = \dfrac{10\left(e^{2t} - 1\right)}{e^{2t} + 1}.

[5]

(d) Find the time taken to reach a speed of 9 m s^{-1}.

[2]

(e) Use your GDC to find the distance fallen in the first 3 seconds.

[1]

Show solutionHide solution

(a) Terminal speed: \dfrac{dv}{dt} = 0, so 10 - 0.1v^2 = 0.

[1]

v^2 = 100, so the terminal speed is 10 m s^{-1}.

[1]

(b) \dfrac{1}{20}\left(\dfrac{1}{10 - v} + \dfrac{1}{10 + v}\right) = \dfrac{1}{20}\cdot\dfrac{(10 + v) + (10 - v)}{100 - v^2}

[1]

= \dfrac{1}{100 - v^2}. [1];

(c) \dfrac{dv}{dt} = 0.1\left(100 - v^2\right); separate: \displaystyle\int\frac{1}{100 - v^2}\,dv = \int 0.1\,dt.

[1]

Using (b): \dfrac{1}{20}\left[-\ln(10 - v) + \ln(10 + v)\right] = 0.1t + C.

[2]

At t = 0, v = 0: C = 0, so \ln\dfrac{10 + v}{10 - v} = 2t.

[1]

10 + v = (10 - v)e^{2t}, so v\left(1 + e^{2t}\right) = 10\left(e^{2t} - 1\right) and v = \dfrac{10\left(e^{2t} - 1\right)}{e^{2t} + 1}. [1];

(d) \dfrac{10 + 9}{10 - 9} = e^{2t}, so e^{2t} = 19.

[1]

t = \dfrac12\ln 19 = 1.47 seconds.

[1]

(e) \displaystyle\int_0^{3}\frac{10\left(e^{2t} - 1\right)}{e^{2t} + 1}\,dt = 23.1 m (GDC numerical integral).

[1]

Final answer

(a) 10 m s^{-1}
(d) t=\frac12\ln19=1.47 s
(e) 23.1 m

Question 14

[12 marks]

A fungus colony grows on an agar plate. The area A cm^2 covered by the colony after t days satisfies \dfrac{dA}{dt} = 0.3A - 0.002A^2, with A = 30 when t = 0.

(a) Show that \dfrac{dA}{dt} = 0.002A(150 - A).

[1]

(b) Find the constants p and q such that \dfrac{1}{A(150 - A)} \equiv \dfrac pA + \dfrac{q}{150 - A}.

[2]

(c) Hence solve the differential equation to show that A = \dfrac{150}{1 + 4e^{-0.3t}}.

[5]

(d) Find the time at which the colony first covers 100 cm^2. Give your answer in days to 3 s.f.

[2]

(e) Find the greatest rate at which the area of the colony increases, and the area of the colony at that moment.

[2]

Show solutionHide solution

(a) 0.002A(150 - A) = 0.3A - 0.002A^2, since 0.002\times 150 = 0.3. [1];

(b) 1 \equiv p(150 - A) + qA.

[1]

Put A = 0: p = \dfrac{1}{150}. Put A = 150: q = \dfrac{1}{150}.

[1]

(c) Separate: \displaystyle\int\frac{1}{A(150 - A)}\,dA =\int 0.002\,dt.

[1]

Using (b): \dfrac{1}{150}\displaystyle\int\left(\frac1A + \frac{1}{150 - A}\right)dA =0.002t + C, so \ln A - \ln(150 - A) = 0.3t + C'.

[2]

At t = 0, A = 30: C' = \ln\dfrac{30}{120} = -\ln 4, so \dfrac{A}{150 - A} = \dfrac{e^{0.3t}}{4}.

[1]

4A = (150 - A)e^{0.3t}, so A\left(4 + e^{0.3t}\right) = 150e^{0.3t} and A = \dfrac{150}{1 + 4e^{-0.3t}}. [1];

(d) 100 = \dfrac{150}{1 + 4e^{-0.3t}} gives 1 + 4e^{-0.3t} = 1.5, so e^{-0.3t} = \dfrac18.

[1]

t = \dfrac{\ln 8}{0.3} = 6.93 days.

[1]

(e) \dfrac{dA}{dt} = 0.3A - 0.002A^2 is a quadratic in A with a maximum at A = \dfrac{0.3}{2(0.002)} = 75 cm^2.

[1]

Greatest rate = 0.002(75)(75) = 11.25 cm^2 per day.

[1]

Final answer

(b) p=q=\frac{1}{150}
(d) 6.93 days
(e) 11.25 cm^2 per day, when A=75

Question 15

[7 marks]

Consider \dfrac{dy}{dx} = \dfrac{y^2 - 2xy + 2x^2}{x^2} for 0 < x < \dfrac32, with y = 4 when x = 1.

Use the substitution y = vx to find y in terms of x.

Show solutionHide solution

Divide by x^2: \dfrac{dy}{dx} = v^2 - 2v + 2 where v = \dfrac yx.

v + x\dfrac{dv}{dx} = v^2 - 2v + 2.

[2]

x\dfrac{dv}{dx} = v^2 - 3v + 2 = (v - 1)(v - 2).

[1]

Separate and use partial fractions, \dfrac{1}{(v-1)(v-2)} = \dfrac{1}{v-2} - \dfrac{1}{v-1}: \displaystyle\int\left(\frac{1}{v-2} - \frac{1}{v-1}\right)\,dv = \int\frac1x\,dx.

[1]

\ln\left|\dfrac{v - 2}{v - 1}\right| = \ln x + C.

[1]

At x = 1, v = 4: \ln\dfrac23 = C, so \dfrac{v - 2}{v - 1} = \dfrac{2x}{3}.

[1]

3v - 6 = 2xv - 2x, so v(3 - 2x) = 6 - 2x and v = \dfrac{6 - 2x}{3 - 2x}.

y = \dfrac{x(6 - 2x)}{3 - 2x}.

[1]

Final answer

y=\frac{x(6-2x)}{3-2x}

Question 16

[8 marks]

Consider the differential equation \dfrac{dy}{dx} + \dfrac yx = x^2y^2 for x > 0, y > 0.

(a) Use the substitution z = \dfrac1y to show that \dfrac{dz}{dx} - \dfrac zx = -x^2.

[3]

(b) Given that y = 1 when x = 1, solve the differential equation, giving y in terms of x.

[5]

Show solutionHide solution

(a) y = \dfrac1z, so \dfrac{dy}{dx} = -\dfrac{1}{z^2}\dfrac{dz}{dx}.

[1]

Substituting: -\dfrac{1}{z^2}\dfrac{dz}{dx} + \dfrac{1}{xz} = \dfrac{x^2}{z^2}.

[1]

Multiply by -z^2: \dfrac{dz}{dx} - \dfrac zx = -x^2. [1];

(b) Integrating factor e^{-\ln x} = \dfrac1x.

[1]

\dfrac{d}{dx}\left(\dfrac zx\right) = -x.

[1]

\dfrac zx = -\dfrac{x^2}{2} + C.

[1]

At x = 1, y = 1, so z = 1: 1 = -\dfrac12 + C, so C = \dfrac32.

[1]

z = \dfrac{3x - x^3}{2}, so y = \dfrac{2}{3x - x^3}.

[1]

Final answer

(b) y=\frac{2}{3x-x^3}

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 17

[9 marks]

Consider the differential equation

x^2\dfrac{dy}{dx} = 9x^2 + xy - y^2, \quad x > 0,

where y = 0 when x = 1.

(a) Explain why \dfrac{dy}{dx} can be written as a function of \dfrac{y}{x} alone, and hence, by writing y = vx, show that x\dfrac{dv}{dx} = (3 - v)(3 + v).

[3]

(b) Hence show that y = \dfrac{3x\left( x^6 - 1 \right)}{x^6 + 1}.

[6]

Show solutionHide solution

(a) \dfrac{dy}{dx} = 9 + \dfrac{y}{x} - \dfrac{y^2}{x^2} = 9 + v - v^2, a function of \dfrac{y}{x} only

v + x\dfrac{dv}{dx} = 9 + v - v^2

x\dfrac{dv}{dx} = 9 - v^2 = (3 - v)(3 + v) A1AG

(b) \dfrac{1}{(3 - v)(3 + v)} = \dfrac{1}{6}\left( \dfrac{1}{3 - v} + \dfrac{1}{3 + v} \right)

\displaystyle\int \dfrac{1}{6}\left( \dfrac{1}{3 + v} + \dfrac{1}{3 - v} \right) dv = \displaystyle\int \dfrac{1}{x}\,dx \Rightarrow \dfrac{1}{6}\ln\left\lvert \dfrac{3 + v}{3 - v} \right\rvert = \ln x + C M1A1

x = 1, v = 0: C = 0

\dfrac{3 + v}{3 - v} = x^6

3 + v = 3x^6 - vx^6 \Rightarrow v = \dfrac{3\left( x^6 - 1 \right)}{x^6 + 1}, so y = vx = \dfrac{3x\left( x^6 - 1 \right)}{x^6 + 1} A1AG

Note: -3 < v < 3 on this solution, so the modulus can be removed.

Final answer

(b) y=\frac{3x\left(x^6-1\right)}{x^6+1} (shown)

Question 18

[8 marks]

Consider the differential equation

\left( x^2 + 4 \right)\dfrac{dy}{dx} + 4xy = 2,

where y = 1 when x = 0.

(a) Show that \left( x^2 + 4 \right)^2 is an integrating factor for this differential equation.

[3]

(b) Hence find y in terms of x.

[4]

(c) Find \displaystyle\lim_{x \to \infty} y.

[1]

Show solutionHide solution

(a) \dfrac{dy}{dx} + \dfrac{4x}{x^2 + 4}y = \dfrac{2}{x^2 + 4}

\displaystyle\int \dfrac{4x}{x^2 + 4}\,dx = 2\ln\left( x^2 + 4 \right)

e^{2\ln(x^2 + 4)} = \left( x^2 + 4 \right)^2 A1AG

(b) \dfrac{d}{dx}\left( \left( x^2 + 4 \right)^2 y \right) = \dfrac{2\left( x^2 + 4 \right)^2}{x^2 + 4} = 2x^2 + 8

\left( x^2 + 4 \right)^2 y = \dfrac{2x^3}{3} + 8x + C

x = 0, y = 1: 16 = C

y = \dfrac{2x^3 + 24x + 48}{3\left( x^2 + 4 \right)^2}

(c) 0 (the numerator has degree 3 and the denominator degree 4)

Final answer

(b) y=\frac{2x^3+24x+48}{3(x^2+4)^2}
(c) 0

Question 19

[14 marks]

Consider the differential equation

\dfrac{dy}{dx} = 3y - e^{x}y^2,

where y = 1 when x = 0.

(a) By using the substitution z = \dfrac{1}{y}, show that the differential equation can be written as

\dfrac{dz}{dx} + 3z = e^{x}.

[4]

(b) Hence, using an integrating factor, find z in terms of x.

[4]

(c) Hence show that y = \dfrac{4e^{3x}}{e^{4x} + 3}, and find \displaystyle\lim_{x \to \infty} y.

[2]

(d) Find the exact maximum value of y, justifying that it is a maximum.

[4]

Show solutionHide solution

(a) \dfrac{dz}{dx} = -\dfrac{1}{y^2}\dfrac{dy}{dx} M1A1

= -\dfrac{1}{y^2}\left( 3y - e^{x}y^2 \right) = -\dfrac{3}{y} + e^{x}

= -3z + e^{x}, so \dfrac{dz}{dx} + 3z = e^{x} A1AG

(b) integrating factor e^{\int 3\,dx} = e^{3x}

\dfrac{d}{dx}\left( ze^{3x} \right) = e^{4x}

ze^{3x} = \dfrac{1}{4}e^{4x} + C

x = 0, z = 1: C = \dfrac{3}{4}, so z = \dfrac{1}{4}\left( e^{x} + 3e^{-3x} \right)

(c) y = \dfrac{1}{z} = \dfrac{4}{e^{x} + 3e^{-3x}} = \dfrac{4e^{3x}}{e^{4x} + 3} A1AG

\displaystyle\lim_{x \to \infty} y = 0 (divide by e^{4x}: \dfrac{4e^{-x}}{1 + 3e^{-4x}} \to 0)

(d) at a stationary point \dfrac{dy}{dx} = y\left( 3 - e^{x}y \right) = 0 and y > 0, so e^{x}y = 3

\dfrac{4e^{4x}}{e^{4x} + 3} = 3 \Rightarrow e^{4x} = 9, so x = \dfrac{1}{2}\ln 3

y = 3e^{-x} = \dfrac{3}{\sqrt{3}} = \sqrt{3}

y > 0 for all x, y \to 0 as x \to \infty and as x \to -\infty, and there is only one stationary point, so it is a maximum

Note: Accept differentiating e^{x} + 3e^{-3x} (the denominator of y in part (c)) and showing that it has a minimum at x = \dfrac{1}{2}\ln 3.

Final answer

(b) z=\frac{e^x+3e^{-3x}}{4}
(c) limit 0
(d) maximum \sqrt3 at x=\frac12\ln3

Question 20

[13 marks]

Consider \dfrac{dy}{dx} + ay = 8e^{-x}, where a > 0 is a constant, with y = 0 when x = 0.

(a) For a \ne 1, show that y = \dfrac{8}{a - 1}\left(e^{-x} - e^{-ax}\right).

[5]

(b) For a \ne 1, show that y has a stationary point at x = \dfrac{\ln a}{a - 1}.

[3]

(c) For a = 1, solve the differential equation.

[3]

(d) Show that \displaystyle\lim_{a \to 1}\frac{\ln a}{a - 1} equals the x-coordinate of the stationary point of your solution in part (c).

[2]

Show solutionHide solution

(a) Integrating factor e^{ax}.

[2]

\dfrac{d}{dx}\left(ye^{ax}\right) = 8e^{(a-1)x}.

[1]

ye^{ax} = \dfrac{8}{a - 1}e^{(a-1)x} + C.

[1]

At (0,\,0): C = -\dfrac{8}{a - 1}, so y = \dfrac{8}{a-1}\left(e^{-x} - e^{-ax}\right). [1];

(b) \dfrac{dy}{dx} = \dfrac{8}{a-1}\left(-e^{-x} + ae^{-ax}\right) = 0.

[1]

ae^{-ax} = e^{-x}, so e^{(a-1)x} = a.

[1]

x = \dfrac{\ln a}{a - 1}. [1];

(c) With a = 1 the integrating factor is e^{x}: \dfrac{d}{dx}\left(ye^{x}\right) = 8.

[1]

ye^{x} = 8x + C.

[1]

At (0,\,0): C = 0, so y = 8xe^{-x}.

[1]

(d) For y = 8xe^{-x}: \dfrac{dy}{dx} = 8e^{-x}(1 - x) = 0 at x = 1.

[1]

\displaystyle\lim_{a \to 1}\frac{\ln a}{a - 1} is the derivative of \ln a at a = 1 (or by l'Hôpital: \dfrac{1/a}{1} \to 1), which is 1, the same value.

[1]

Final answer

(b) stationary at x=\frac{\ln a}{a-1}
(c) y=8xe^{-x}
(d) both equal 1

Question 21

[14 marks]

Two chemicals react to form a compound. The mass x grams of the compound at time t minutes satisfies \dfrac{dx}{dt} = k(4 - x)(6 - x), where k > 0 is a constant and x = 0 when t = 0. Throughout, 0 \le x < 4.

(a) Write down the value of \dfrac{dx}{dt} when t = 0, in terms of k.

[1]

(b) Show that \dfrac{1}{(4 - x)(6 - x)} \equiv \dfrac12\left(\dfrac{1}{4 - x} - \dfrac{1}{6 - x}\right).

[2]

(c) Hence show that \ln\dfrac{6 - x}{4 - x} = 2kt + \ln\dfrac32.

[5]

(d) Given that x = 2 when t = 10, show that k = \dfrac{1}{20}\ln\dfrac43.

[2]

(e) Find the mass of the compound when t = 30.

[2]

(f) Find the time at which the mass reaches 3.9 grams. Give your answer in minutes to 3 s.f.

[2]

Show solutionHide solution

(a) \dfrac{dx}{dt} = k(4)(6) = 24k.

[1]

(b) \dfrac12\left(\dfrac{1}{4 - x} - \dfrac{1}{6 - x}\right) = \dfrac12\cdot\dfrac{(6 - x) - (4 - x)}{(4 - x)(6 - x)}

[1]

= \dfrac12\cdot\dfrac{2}{(4 - x)(6 - x)} = \dfrac{1}{(4 - x)(6 - x)}. [1];

(c) Separate and use (b): \displaystyle\int\frac12\left(\frac{1}{4 - x} - \frac{1}{6 - x}\right)\,dx = \int k\,dt.

[1]

\dfrac12\left[-\ln(4 - x) + \ln(6 - x)\right] = kt + C.

[2]

At t = 0, x = 0: \dfrac12\ln\dfrac64 = C.

[1]

Multiply by 2: \ln\dfrac{6 - x}{4 - x} = 2kt + \ln\dfrac32. [1];

(d) At t = 10, x = 2: \ln\dfrac42 = 20k + \ln\dfrac32.

[1]

20k = \ln 2 - \ln\dfrac32 = \ln\dfrac43, so k = \dfrac{1}{20}\ln\dfrac43. [1];

(e) At t = 30: 2kt = 60k = 3\ln\dfrac43, so \ln\dfrac{6 - x}{4 - x} = \ln\dfrac{64}{27} + \ln\dfrac32 = \ln\dfrac{32}{9}.

[1]

9(6 - x) = 32(4 - x), so 23x = 74 and x = 3.22 g.

[1]

(f) At x = 3.9: \dfrac{6 - 3.9}{4 - 3.9} = 21 = \dfrac32e^{2kt}, so e^{2kt} = 14.

[1]

t = \dfrac{\ln 14}{2k} = \dfrac{10\ln 14}{\ln(4/3)} = 91.7 minutes.

[1]

Final answer

(a) 24k
(e) \frac{74}{23}\approx3.22 g
(f) 91.7 minutes

Question 22

[15 marks]

The number of fish, P, in a lake t years after a fish farm starts is modelled by

\dfrac{dP}{dt} = 0.5P\left( 1 - \dfrac{P}{500} \right) - 40, \quad t \ge 0,

where the term -40 represents fish that are caught.

(a) Show that \dfrac{dP}{dt} = -0.001(P - 100)(P - 400).

[2]

(b) Find the two constant solutions of the differential equation, and describe what happens to a population that starts with a value of P between them.

[2]

(c) Given that P = 200 when t = 0, show that

P = \dfrac{200 + 400e^{0.3t}}{2 + e^{0.3t}}.

[6]

(d) Find the time at which the number of fish reaches 350.

[2]

(e) The farm instead starts with 80 fish. Find the time at which, according to the model, there are no fish left.

[3]

Show solutionHide solution

(a) 0.5P - 0.001P^2 - 40 = -0.001\left( P^2 - 500P + 40\,000 \right)

= -0.001(P - 100)(P - 400) A1AG

(b) P = 100 and P = 400

for 100 < P < 400, \dfrac{dP}{dt} > 0, so the population increases and approaches 400

(c) \dfrac{dP}{dt} = 0.001(P - 100)(400 - P) and \dfrac{1}{(P - 100)(400 - P)} = \dfrac{1}{300}\left( \dfrac{1}{P - 100} + \dfrac{1}{400 - P} \right)

\displaystyle\int \dfrac{1}{300}\left( \dfrac{1}{P - 100} + \dfrac{1}{400 - P} \right) dP = \displaystyle\int 0.001\,dt

\dfrac{1}{300}\ln\left\lvert \dfrac{P - 100}{400 - P} \right\rvert = 0.001t + C, so \dfrac{P - 100}{400 - P} = Ae^{0.3t}

t = 0, P = 200: A = \dfrac{100}{200} = \dfrac{1}{2}

2(P - 100) = e^{0.3t}(400 - P)

P\left( 2 + e^{0.3t} \right) = 200 + 400e^{0.3t}, which gives the result A1AG

(d) \dfrac{350 - 100}{400 - 350} = 5 = \dfrac{1}{2}e^{0.3t} \Rightarrow e^{0.3t} = 10

t = \dfrac{10}{3}\ln 10 = 7.68 years (7.67528\ldots)

(e) t = 0, P = 80: A = \dfrac{80 - 100}{400 - 80} = -\dfrac{1}{16}

P = 0: \dfrac{-100}{400} = -\dfrac{1}{16}e^{0.3t} \Rightarrow e^{0.3t} = 4

t = \dfrac{10}{3}\ln 4 = 4.62 years (4.62098\ldots)

Note: In part (e) the constant A is negative, which is why the general solution keeps A of either sign.

Final answer

(b) P=100 and P=400; a population between them rises towards 400
(d) 7.68 years
(e) 4.62 years

Question 23

[14 marks]

Consider the differential equation

\dfrac{dy}{dx} = (2 - y)\cos x,

where y = 5 when x = 0.

(a) Use Euler's method with step length 0.25 to find an approximate value of y when x = 1.

[3]

(b) By separating the variables, show that y = 2 + 3e^{-\sin x}.

[5]

(c) Write the differential equation in the form \dfrac{dy}{dx} + P(x)y = Q(x), and solve it again using an integrating factor to confirm the result of part (b).

[4]

(d) Find the percentage error in the approximation found in part (a).

[2]

Show solutionHide solution

(a) y_{n+1} = y_n + 0.25\left( 2 - y_n \right)\cos x_n (M1)

y_1 = 4.25, y_2 = 3.70499\ldots, y_3 = 3.33092\ldots (A1)

y \approx 3.09 when x = 1 (3.08747\ldots)

(b) \displaystyle\int \dfrac{1}{2 - y}\,dy = \displaystyle\int \cos x\,dx

-\ln\lvert 2 - y \rvert

= \sin x + C

x = 0, y = 5: -\ln 3 = C

\ln\lvert y - 2 \rvert = \ln 3 - \sin x, and y - 2 > 0 near x = 0, so y - 2 = 3e^{-\sin x} and y = 2 + 3e^{-\sin x} A1AG

(c) \dfrac{dy}{dx} + y\cos x = 2\cos x, integrating factor e^{\int \cos x\,dx} = e^{\sin x}

\dfrac{d}{dx}\left( ye^{\sin x} \right) = 2\cos x\,e^{\sin x}

ye^{\sin x} = 2e^{\sin x} + C

x = 0: 5 = 2 + C, C = 3, so y = 2 + 3e^{-\sin x}, as in part (b)

(d) true value 2 + 3e^{-\sin 1} = 3.29322\ldots; percentage error = \dfrac{\lvert 3.08747\ldots - 3.29322\ldots \rvert}{3.29322\ldots} \times 100\% (M1)

= 6.25\% (6.24805\ldots)

Final answer

(a) 3.09
(c) result confirmed
(d) 6.25\%

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 A tank that empties

G1 Worked example

[11 marks]

Water drains from a cylindrical tank through a small hole in its base. The depth of the water is h metres, t minutes after the hole is opened, and \dfrac{dh}{dt}=-0.02\sqrt h. Initially h=4.

(a) Solve the differential equation to show that h=(2-0.01t)^2 for 0\le t\le200.

[4]

(b) Find the time at which the depth is 1 m, and the time at which the tank is empty.

[2]

(c) The depth y metres in a second tank satisfies \dfrac{dy}{dt}=-ky, with y=4 when t=0. At t=0 the depth in the second tank falls at the same rate as in the first. Find k, and find the depth in the second tank at the moment the first tank becomes empty.

[3]

(d) Explain why h=0 for t\ge200 is consistent with the differential equation, and why the second tank never becomes empty.

[2]

Show solutionHide solution

(a) \displaystyle\int h^{-1/2}\,dh=-\int0.02\,dt

2\sqrt h=-0.02t+c

t=0, h=4: c=4

\sqrt h=2-0.01t, so h=(2-0.01t)^2, valid while 2-0.01t\ge0, that is 0\le t\le200

A1AG

(b) h=1: 2-0.01t=1, t=100 minutes

empty: 2-0.01t=0, t=200 minutes

Note: the first half of the time removes three quarters of the depth.

(c) at t=0: -0.02\sqrt4=-0.04, so -4k=-0.04

k=0.01

y=4e^{-0.01t}; at t=200, y=4e^{-2}=0.541 m

(d) if h=0 then \dfrac{dh}{dt}=0 and -0.02\sqrt0=0, so both sides agree: the empty tank stays empty

y=4e^{-0.01t}>0 for every t, because an exponential is never zero, so the second tank is never empty

Final answer

(b) 100 min; 200 min
(c) k=0.01; 4e^{-2}=0.541 m
(d) h=0 makes both sides 0; 4e^{-0.01t}>0

G1 Investigation

[13 marks]

Water drains from a tank through a hole in its base. When the depth is h, the horizontal cross-section of the water has area A(h), and the volume leaves at the rate k\sqrt h, so A(h)\dfrac{dh}{dt}=-k\sqrt h, where k>0. The initial depth is H and the tank is empty at time T. The last half of the water is the water that remains when half of the initial volume has drained.

(a) For a cylinder, A(h)=c, a constant. Show that T=\dfrac{2c\sqrt H}{k}, and that the last half of the water takes \dfrac{1}{\sqrt2} of the time T to drain.

[3]

(b) For a cone with its vertex at the bottom, A(h)=ch^2. Find T in terms of c, H and k, and show that the last half of the water takes 2^{-5/6} of the time T to drain.

[3]

(c) For A(h)=ch^p, where p\ge0, show that the last half of the water takes 2^{-\frac{2p+1}{2p+2}} of the time T to drain.

[3]

(d) Show that this fraction is greater than \dfrac12 and at most \dfrac1{\sqrt2} for every p\ge0, and find its limit as p\to\infty.

[2]

(e) Find the value of p for which the depth falls at a constant rate. For such a tank with circular cross-sections of radius r, show that r is proportional to h^{1/4}.

[2]

Show solutionHide solution

(a) \displaystyle\int h^{-1/2}\,dh=-\frac kc\int dt, so 2\sqrt h=2\sqrt H-\dfrac kct

h=0 when t=T=\dfrac{2c\sqrt H}{k}

the volume is ch, so half the water is left at depth \dfrac H2; from there, h reaches 0 after \dfrac{2c}{k}\sqrt{\dfrac H2}=\dfrac{T}{\sqrt2}

(b) \displaystyle\int h^{3/2}\,dh=-\frac kc\int dt: \dfrac25h^{5/2}=\dfrac25H^{5/2}-\dfrac kct, so T=\dfrac{2cH^{5/2}}{5k}

the volume below depth h is \displaystyle\int_0^hcs^2\,ds=\frac{ch^3}{3}, so half is left when h^3=\dfrac{H^3}2, h=2^{-1/3}H

from depth h the tank empties in \dfrac{2ch^{5/2}}{5k}=T\left(\dfrac hH\right)^{5/2}=T\cdot2^{-5/6}

(c) \displaystyle\int h^{p-1/2}\,dh=-\frac kc\int dt: from depth h the tank empties in \dfrac{c\,h^{p+1/2}}{(p+\frac12)k}=T\left(\dfrac hH\right)^{p+1/2}

volume \dfrac{ch^{p+1}}{p+1}; half is left when \left(\dfrac hH\right)^{p+1}=\dfrac12

fraction =\left(\dfrac12\right)^{\frac{p+1/2}{p+1}}=2^{-\frac{2p+1}{2p+2}}

A1AG

(d) \dfrac{2p+1}{2p+2}=1-\dfrac{1}{2p+2} increases from \dfrac12 (at p=0) and stays below 1

so 2^{-\frac{2p+1}{2p+2}} falls from \dfrac1{\sqrt2} and stays above 2^{-1}=\dfrac12; the limit as p\to\infty is \dfrac12

(e) \dfrac{dh}{dt}=-\dfrac kc\,h^{\frac12-p} is constant when p=\dfrac12

\pi r^2=ch^{1/2}, so r=\sqrt{\dfrac c\pi}\,h^{1/4}

Final answer

(b) T=\frac{2cH^{5/2}}{5k}
(d) limit \frac12
(e) p=\frac12

G2 Up versus down

G2 Worked example

[12 marks]

A ball is thrown vertically upwards with speed 15 m s^{-1}. While it rises, its velocity v m s^{-1}, measured upwards, satisfies \dfrac{dv}{dt}=-10-0.2v, where t is the time in seconds after it is thrown.

(a) Solve the differential equation to show that v=65e^{-0.2t}-50.

[4]

(b) Find the exact time at which the ball reaches its highest point, and give this time to three significant figures.

[2]

(c) Show that the greatest height above the point of projection is 75-250\ln1.3 metres.

[3]

(d) If 0.2 is replaced by k, the ball reaches its highest point after \dfrac{\ln(1+1.5k)}{k} seconds. Find the limit of this time as k\to0, and interpret your answer.

[3]

Show solutionHide solution

(a) \displaystyle\int\frac{dv}{10+0.2v}=-\int dt

5\ln(10+0.2v)=-t+c

t=0, v=15: c=5\ln13

10+0.2v=13e^{-0.2t}, so v=65e^{-0.2t}-50

A1AG

Note: no modulus is needed, since 10+0.2v>0 while the ball rises.

(b) v=0: e^{-0.2t}=\dfrac{50}{65}, t=5\ln1.3

t=1.31 s

(c) height =\displaystyle\int_0^{5\ln1.3}\left(65e^{-0.2t}-50\right)dt

=\Bigl[-325e^{-0.2t}-50t\Bigr]_0^{5\ln1.3}

=-\dfrac{325}{1.3}-250\ln1.3+325=75-250\ln1.3

A1AG

(d) as k\to0 the limit has the form \dfrac00; by l'Hôpital's rule it equals \displaystyle\lim_{k\to0}\frac{1.5/(1+1.5k)}{1}

=1.5 s

1.5=\dfrac{15}{10} is the time to the top with no air resistance (v=15-10t): for small k the model behaves like motion without air resistance

Final answer

(b) 5\ln1.3=1.31 s
(c) 9.41 m
(d) 1.5 s, the time without air resistance

G2 Investigation

[11 marks]

A ball is thrown vertically upwards with speed u m s^{-1}. Air resistance gives it an extra deceleration of kv m s^{-2} against its motion, where k>0 and v is its speed. While it rises, its upward velocity v satisfies \dfrac{dv}{dt}=-10-kv. When it falls from rest at its highest point, its downward velocity w satisfies \dfrac{dw}{dt}=10-kw, with t now measured from the highest point.

(a) For k=1 and u=10, the ball rises for \ln2 seconds, to a height of 10-10\ln2 metres. Show that, on the way down, w=10(1-e^{-t}), and that the distance fallen after t seconds is 10t-10+10e^{-t}. Hence find the time taken to fall back to the point of projection, and compare it with the time taken to rise.

[4]

(b) Now let k>0 and u>0 be general, and let the ball rise for t_r seconds. Imagine a film of the rise played backwards. Let \tau=t_r-t and W(\tau)=v(t_r-\tau), the speed at which the ball appears to move down in the film. Show that \dfrac{dW}{d\tau}=10+kW with W(0)=0, and hence that W=\dfrac{10}{k}\left(e^{k\tau}-1\right).

[3]

(c) Show that \dfrac{10}{k}\left(e^{k\tau}-1\right)>\dfrac{10}{k}\left(1-e^{-k\tau}\right) for every \tau>0.

[2]

(d) Hence prove that, for every k>0, the ball takes longer to fall than to rise.

[2]

Show solutionHide solution

(a) \displaystyle\int\frac{dw}{10-w}=\int dt: -\ln(10-w)=t+c, and w=0 at t=0 gives c=-\ln10, so 10-w=10e^{-t}

w=10\left(1-e^{-t}\right)

distance =\displaystyle\int_0^t10\left(1-e^{-s}\right)ds=10t-10+10e^{-t}

GDC: 10t-10+10e^{-t}=10-10\ln2 gives t=0.900 s, longer than the rise, \ln2=0.693 s

(b) by the chain rule, \dfrac{dW}{d\tau}=-\dfrac{dv}{dt}=-(-10-kv)=10+kW; and W(0)=v(t_r)=0, since the ball is at rest at the top

M1A1

\displaystyle\int\frac{dW}{10+kW}=\int d\tau: \dfrac1k\ln(10+kW)=\tau+\dfrac1k\ln10, so 10+kW=10e^{k\tau} and W=\dfrac{10}{k}\left(e^{k\tau}-1\right)

(c) \left(e^{k\tau}-1\right)-\left(1-e^{-k\tau}\right)=e^{k\tau}-2+e^{-k\tau}

=\left(e^{k\tau/2}-e^{-k\tau/2}\right)^2>0 for \tau>0, and \dfrac{10}{k}>0

(d) the film and the real fall both start at rest at the top and move down through the same height; by (c) the film is faster at every moment after the start, so after any time \tau>0 it has covered more distance: \displaystyle\int_0^\tau W\,ds>\int_0^\tau w\,ds

the film reaches the point of projection at \tau=t_r; at that moment the falling ball has covered less than the full height, so it arrives later: the fall takes longer than t_r

Final answer

(a) 0.900 s, against 0.693 s for the rise
(b) W=\frac{10}{k}(e^{k\tau}-1)
(c), (d) proved in the solutions

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 16 teaches differential equations from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book