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Differential Equations · Paper 1 — Free Maths AA HL Practice Questions

What this topic covers

Calculus is where AA HL stops rewarding memorised algebra and starts rewarding process. Every mark scheme in this topic is built around specific words — show that, hence, determine, justify — and examiners are trained to withhold marks the instant your working skips a step, even when your final answer is correct. The four subtopics here build on each other in a strict chain: limits define continuity, continuity is the gatekeeper for differentiability, derivatives feed every application (tangents, optimisation, kinematics), and integration — the reverse process — feeds differential equations. Learn it in that order and the HL-only material (implicit differentiation, integration by parts, volumes of revolution, and the whole of differential equations) stops looking like extra content and starts looking like the natural next question.

Read the full Calculus lesson notes →

1FoundationSAQ-SSolving first-order differential equationss5 marksPaper 1~8 min

A population of bacteria,

P

(measured in thousands), grows according to the differential equation

\frac{\mathrm{d}P}{\mathrm{d}t} = kP

where

t

is the time in hours and

k

is a positive constant.

(a)

By separating variables, show that the differential equation can be written as

\int \frac{1}{P}\,\mathrm{d}P = \int k\,\mathrm{d}t.

[1 mark]

(b)

Hence find the general solution of the differential equation, expressing

P

in terms of

t

and an arbitrary constant

A

. [2 marks]

(c)

Given that

P = 2

when

t = 0

, and

P = 6

when

t = 3

, find the exact value of

k

. [2 marks]

Solutions

2MasterySAQ-SSolving first-order differential equationss8 marksPaper 1~12 min

A population of bacteria grows according to the differential equation

\frac{dP}{dt} = kP(1000 - P)

where

P

is the population size at time

t

days and

k

is a positive constant.

(a)

Show that

\dfrac{1}{P(1000-P)} = \dfrac{1}{1000}\left(\dfrac{1}{P} + \dfrac{1}{1000-P}\right)

. [2 marks]

(b)

Hence, given that

P(0) = 100

, solve the differential equation to show that

P = \frac{1000\,e^{1000kt}}{9 + e^{1000kt}}.

[3 marks]

(c)

Deduce the long-run behaviour of

P

t \to \infty

and explain, with reference to the differential equation, why this value is expected. [2] (Paper 1 extension — no calculator) Total: [7] > (If restricted to 5 marks, omit (c) and award (a) [2], (b) .) [3 marks]

Solutions

3ChallengeSAQ-LSolving first-order differential equationss10 marksPaper 1~15 min

A tank initially contains

100

litres of pure water. A salt solution of concentration

0.5\,\text{kg}\,\text{L}^{-1}

flows into the tank at

3\,\text{L}\,\text{min}^{-1}

, and the well-mixed solution flows out at the same rate. Let

S(t)\,\text{kg}

be the amount of salt in the tank at time

t

minutes.

(a)

Show that

S(t)

satisfies the differential equation

\frac{dS}{dt} = 1.5 - \frac{3S}{100}.

[2 marks]

(b)

Determine the general solution of this differential equation, giving

S

as a function of

t

. [3 marks]

(c)

State the particular solution satisfying the initial condition

S(0) = 0

. [1 mark]

(d)

Determine the time, to the nearest minute, for the salt in the tank to reach

40\,\text{kg}

, and comment on whether the tank's salt content could ever reach

50\,\text{kg}

. [4 marks]

Solutions

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