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Cambridge International AS & A Level Mathematics 9709 · Pure Mathematics 3

Differential equations

Writing a rate of change as a differential equation, solving first order equations by separating the variables, using an initial condition to find a particular solution, and interpreting the solution in the situation it models.

3.8Paper 3: Pure Mathematics 3Free lesson

What the syllabus says

3.8 (Paper 3)

formulate a simple statement involving a rate of change as a differential equation
find by integration a general form of solution for a first order differential equation in which the variables are separable
use an initial condition to find a particular solution
interpret the solution of a differential equation in the context of a problem being modelled by the equation.

Notes and examples

The introduction and evaluation of a constant of proportionality, where necessary, is included.
Including any of the integration techniques from topic 3.5 above.
Where a differential equation is used to model a ‘real-life’ situation, no specialised knowledge of the context will be required.

Quoted from the Cambridge International AS and A Level Mathematics 9709 syllabus (2026-2027 and 2028-2030 editions, which share this content).

Rates of change as differential equations Paper 3

A differential equation is an equation containing a derivative. A first order equation contains only first derivatives, such as $\frac{d y}{d x}$ or $\frac{d V}{d t}$ .

Many models start from a sentence about a rate. "Rate of change" with time means a derivative with respect to $t$ , and "is proportional to" means "equals a constant times". So:

"the population $P$ increases at a rate proportional to $P$ " is $\frac{d P}{d t} = k P$ , with $k > 0$ ;
"the temperature $θ$ decreases at a rate proportional to the amount by which it exceeds 20" is $\frac{d θ}{d t} = - k (θ - 20)$ , with $k > 0$ ;
"the mass $m$ increases at a rate inversely proportional to $m$ " is $\frac{d m}{d t} = \frac{k}{m}$ .

Writing a decrease with a minus sign and a positive $k$ makes the meaning clear; a model with $+ k$ and a negative $k$ is also correct if you say so. The constant of proportionality is usually found from one piece of data: if $P$ is increasing at 50 per year when $P = 2000$ , then $50 = 2000 k$ and $k = 0.025$ .

Check yourself Paper 3

The temperature $θ$ °C of a liquid decreases at a rate proportional to $(θ - 20)$ , so $\frac{d θ}{d t} = - k (θ - 20)$ , where $t$ is in minutes. When $θ = 80$ the temperature is falling at 3 °C per minute. Find $k$ .

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Separating the variables Paper 3

An equation is separable if it can be written as $\frac{d y}{d x} = f (x) g (y)$ : a function of $x$ times a function of $y$ . Then put everything involving $y$ on one side and everything involving $x$ on the other, and integrate both sides:
$\int \frac{1}{g (y)} d y = \int f (x) d x$
Why this is allowed: if $G (y)$ is an antiderivative of $\frac{1}{g (y)}$ , the chain rule gives $\frac{d}{d x} G (y) = \frac{1}{g (y)} \frac{d y}{d x} = f (x)$ . So $G (y)$ and $\int f (x) d x$ have the same derivative and differ only by a constant. Moving " $d x$ " across is shorthand for exactly this.

The steps:

separate: $y$ terms with $d y$ , $x$ terms with $d x$ ;
integrate both sides, adding one arbitrary constant (on the $x$ side);
use any given condition to find the constant;
rearrange if the question asks for $y$ in terms of $x$ .

Integrating $\frac{1}{y}$ gives $\ln | y |$ ; when $y$ is known to be positive, as in most models, write $\ln y$ .

Worked example 1 Paper 3

Find the general solution of $\frac{d y}{d x} = 6 x^{2} y$ , where $y > 0$ , giving $y$ in terms of $x$ .

Separate: $\int \frac{1}{y} d y = \int 6 x^{2} d x$ .
Integrate: $\ln y = 2 x^{3} + c$ .
Take exponentials: $y = e^{2 x^{3} + c} = e^{c} e^{2 x^{3}}$ .
$e^{c}$ is just a positive constant, so write it as $A$ : $y = A e^{2 x^{3}}$ .

Answer: $y = A e^{2 x^{3}}$

The arbitrary constant Paper 3

A general solution contains an arbitrary constant: it describes a whole family of curves, one for each value of the constant, all with the gradient the equation describes. A particular solution is the one member of the family that fits extra information, such as an initial value.

Handle the constant with care:

add it once, when you integrate; a constant on each side combines into one;
after exponentiating, $e^{x + c}$ becomes $A e^{x}$ ; the constant now multiplies, so it is no longer " $+ c$ ";
writing the constant as $\ln A$ can make the algebra neater: $\ln y = 2 \ln x + \ln A$ gives $y = A x^{2}$ .

If a question asks for the general solution in a form such as " $\sin y$ in terms of $x$ ", leave it in that form with its constant added.

Check yourself Paper 3

Find the general solution of $\frac{d y}{d x} = \frac{4 x^{3}}{\cos y}$ , giving $\sin y$ in terms of $x$ . Include the arbitrary constant.

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Particular solutions Paper 3

An initial condition (or boundary condition) such as " $y = 1$ when $x = 0$ " fixes the constant. Substitute it straight after integrating, while the equation is still in its simplest form, then rearrange. Finding the constant after rearranging is allowed but invites mistakes, because the constant may by then sit inside a power or a logarithm.

Read the question for the form of the answer: "find $y$ in terms of $x$ " needs $y = \dots$ ; "find an expression for $t$ in terms of $x$ " needs $t = \dots$ .

Worked example 2 Paper 3

Solve $\frac{d y}{d x} = \frac{e^{2 x}}{y^{2}}$ , given that $y = 1$ when $x = 0$ . Give $y$ in terms of $x$ .

Separate: $\int y^{2} d y = \int e^{2 x} d x$ .
Integrate: $\frac{1}{3} y^{3} = \frac{1}{2} e^{2 x} + c$ .
At $x = 0$ , $y = 1$ : $\frac{1}{3} = \frac{1}{2} + c$ , so $c = - \frac{1}{6}$ .
Multiply by 3: $y^{3} = \frac{3}{2} e^{2 x} - \frac{1}{2} = \frac{3 e^{2 x} - 1}{2}$ .
$y = \sqrt[3]{\frac{3 e^{2 x} - 1}{2}}$ .

Answer: $y = \sqrt[3]{\frac{3 e^{2 x} - 1}{2}}$

Integration techniques in differential equations Paper 3

After separating, either side may need any method from the integration topic of this paper:

partial fractions, for a side such as $\frac{1}{x (3 - x)}$ ;
integration by parts, for a side such as $x e^{2 x}$ or $x \cos x$ ;
substitution, or recognising $\frac{f' (x)}{f (x)}$ , for a side such as $\frac{x}{x^{2} + 1}$ ;
trigonometric identities, for example $\frac{1}{\cos^{2} y} = \sec^{2} y$ , whose integral is $\tan y$ .

Separate first, then decide the technique for each side on its own. A $\ln$ on both sides is a signal to combine logarithms before exponentiating.

Worked example 3 Paper 3

Given that $x = 1$ when $t = 0$ , solve $\frac{d x}{d t} = \frac{1}{3} x (3 - x)$ for $0 < x < 3$ , giving $x$ in terms of $t$ .

Separate: $\int \frac{3}{x (3 - x)} d x = \int 1 d t$ .
Partial fractions: $\frac{3}{x (3 - x)} = \frac{1}{x} + \frac{1}{3 - x}$ .
Integrate: $\ln x - \ln (3 - x) = t + c$ . Note the minus sign from the $- x$ in $3 - x$ .
At $t = 0$ , $x = 1$ : $\ln 1 - \ln 2 = c$ , so $c = - \ln 2$ .
$\ln \frac{2 x}{3 - x} = t$ , so $2 x = (3 - x) e^{t}$ .
$x (2 + e^{t}) = 3 e^{t}$ , so $x = \frac{3 e^{t}}{2 + e^{t}}$ .

Answer: $x = \frac{3 e^{t}}{2 + e^{t}}$

Interpreting the solution Paper 3

A model is only useful once its solution is read back into the situation. The usual questions are:

What happens in the long term? Let $t$ become large. In Worked example 3, $x = \frac{3}{2 e^{- t} + 1} \to 3$ , so $x$ approaches 3 but never reaches it. A term $e^{- k t}$ with $k > 0$ tends to 0.
When does the quantity reach a given value? Substitute it and solve for $t$ , usually with logarithms.
Is the model sensible? A tank model that predicts a negative depth after it empties is only valid until the moment it empties.

You do not need specialist knowledge of the context: everything needed is in the question.

Worked example 4 (exam style) Paper 3

Liquid is poured into a container at a constant rate of 30 cm³ per second and leaks out at a rate proportional to the volume $V$ cm³ in the container.
(a) Show that $\frac{d V}{d t} = 30 - k V$ , and given that $V$ is increasing at 10 cm³ per second when $V = 200$ , find $k$ .
(b) The container is empty at $t = 0$ . Find $V$ in terms of $t$ .
(c) State what happens to $V$ in the long term, and find the time taken for $V$ to reach 150.

(a) Rate in 30, rate out $k V$ , so $\frac{d V}{d t} = 30 - k V$ . Then $10 = 30 - 200 k$ , so $k = 0.1$ .
(b) $\int \frac{1}{30 - 0.1 V} d V = \int 1 d t$ gives $- 10 \ln (30 - 0.1 V) = t + c$ .
At $t = 0$ , $V = 0$ : $c = - 10 \ln 30$ . So $\ln \frac{30 - 0.1 V}{30} = - \frac{t}{10}$ .
$30 - 0.1 V = 30 e^{- \frac{t}{10}}$ , so $V = 300 (1 - e^{- \frac{t}{10}})$ .
(c) As $t \to \infty$ , $e^{- \frac{t}{10}} \to 0$ , so $V$ approaches 300 cm³, where the leak balances the inflow.
$150 = 300 (1 - e^{- \frac{t}{10}})$ gives $e^{- \frac{t}{10}} = \frac{1}{2}$ , so $t = 10 \ln 2 = 6.93$ seconds.

Answer: $k = 0.1$ ; $V = 300 (1 - e^{- \frac{t}{10}})$ ; $V \to 300$ ; $t = 6.93$ s

Check yourself Paper 3

For the container in Worked example 4, $V = 300 (1 - e^{- \frac{t}{10}})$ . Find the time at which $V = 270$ , correct to 3 significant figures.

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Key points

"Rate of change of $y$ " is $\frac{d y}{d t}$ ; "proportional to" introduces a constant $k$ , found from data.
Separable: $\frac{d y}{d x} = f (x) g (y)$ gives $\int \frac{1}{g (y)} d y = \int f (x) d x$ .
Add one arbitrary constant when you integrate; it may become a multiplying constant $A$ after exponentiating.
Use the initial condition straight after integrating to get the particular solution.
Any integration method may be needed: partial fractions, parts, substitution, identities.
Interpret: long-term value as $t \to \infty$ , time to reach a value, and when the model stops making sense.

Where marks are lost

Leaving out the constant of integration, or adding it after exponentiating as " $+ c$ ".
Writing $e^{x + c} = e^{x} + e^{c}$ instead of $e^{x} e^{c}$ .
Losing the minus sign in $\int \frac{1}{3 - x} d x = - \ln (3 - x)$ .
Using the wrong sign for a decreasing quantity, then a negative $k$ without comment.
Substituting the initial condition into the original differential equation instead of the solution.
Stopping at $\ln y = \dots$ when the question asks for $y$ in terms of $x$ .

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An original lesson written to the Cambridge International AS & A Level Mathematics 9709 syllabus (2026 to 2027 and 2028 to 2030). The worked examples and questions are our own, not Cambridge questions. Back to every AS & A Level Mathematics 9709 lesson.