Constructing Differential Equations

Learn to translate real-world problems into mathematical models using derivatives

SPEC 7.6 • DIFFERENTIATION

What is a Differential Equation?

Definition

A differential equation is an equation that contains one or more derivatives of an unknown function.

In A-Level Mathematics, we focus on first-order differential equations, which contain only the first derivative.

Standard Forms

Common forms you will encounter:

\frac{d y}{d x} = f (x)

— derivative depends only on x

\frac{d y}{d x} = f (y)

— derivative depends only on y

\frac{d y}{d x} = f (x, y)

— derivative depends on both

Key Insight

Differential equations describe how things change. They are essential for modelling growth, decay, and many real-world phenomena.

Key Language for Translation

The key skill is translating verbal descriptions into mathematical equations.

Verbal PhraseMathematical Translation

"The rate of change of y with respect to x"

\frac{d y}{d x}

"The rate of change of P with respect to time t"

\frac{d P}{d t}

"...is proportional to..."

\propto

then

= k \times ...

"...is inversely proportional to..."

\propto \frac{1}{...}

then

= \frac{k}{...}

Important

"Rate of decrease" means we need a negative sign. If something is decreasing,

\frac{d y}{d t} < 0

.

Proportionality: ∝ to = k

When converting proportionality statements, always show the two-step process:

Direct Proportion

"A is proportional to B"

A \propto B

→

A = k B

Inverse Proportion

"A is inversely proportional to B"

A \propto \frac{1}{B}

→

A = \frac{k}{B}

Remember

The constant

k

is always positive. Use a negative sign separately if something is decreasing.

Common Differential Equation Models

Model TypeVerbal DescriptionEquation

Exponential GrowthRate of increase ∝ amount present

\frac{d y}{d t} = k y

Exponential DecayRate of decrease ∝ amount present

\frac{d y}{d t} = - k y

Inverse ProportionRate of change ∝ 1/(variable)

\frac{d y}{d t} = \frac{k}{y}

or

\frac{k}{y ^{2}}

Logistic GrowthRate ∝ product of y and (max − y)

\frac{d y}{d t} = k y (M - y)

Watch the Sign

For decrease/decay, always include the negative sign:

\frac{d y}{d t} = - k y

Worked Example01

The rate of change of y with respect to x is proportional to y. Write down a differential equation to model this situation.

1

"Rate of change of y with respect to x" →

\frac{d y}{d x}

2

"is proportional to y" → first write

\frac{d y}{d x} \propto y

3

Convert ∝ to = k:

\frac{d y}{d x} = k y

Answer

\frac{d y}{d x} = k y

where k is a positive constant

Worked Example02

A population of bacteria, P, grows at a rate proportional to its current size. Write down a differential equation for this model.

1

Population P changes with time t →

\frac{d P}{d t}

2

"grows at a rate proportional to its current size" →

\frac{d P}{d t} \propto P

3

Convert ∝ to = k:

\frac{d P}{d t} = k P

4

"grows" means increasing, so k > 0 (no negative sign needed)

Answer

\frac{d P}{d t} = k P

where k > 0

Worked Example03

A radioactive substance decays at a rate proportional to the mass M remaining. Write down a differential equation for this process.

1

Mass M changes with time t →

\frac{d M}{d t}

2

"decays at a rate proportional to M" →

\frac{d M}{d t} \propto M

3

Convert ∝ to = k: but "decays" means decreasing

4

Add negative sign:

\frac{d M}{d t} = - k M

Note: Sometimes written as $\frac{d M}{d t} = - λ M$ where λ is the decay constant.

Answer

\frac{d M}{d t} = - k M

where k > 0

Worked Example04

A spherical mint dissolves in a cup of tea. In a simple model, the rate of decrease of the radius r is inversely proportional to the square of the radius. Write down a differential equation for r.

1

Radius r changes with time t →

\frac{d r}{d t}

2

"inversely proportional to the square of the radius" →

\frac{d r}{d t} \propto \frac{1}{r ^{2}}

3

Convert ∝ to = k:

\frac{d r}{d t} = \frac{k}{r ^{2}}

4

"rate of decrease" means add negative sign:

\frac{d r}{d t} = - \frac{k}{r ^{2}}

Answer

\frac{d r}{d t} = - \frac{k}{r ^{2}}

where k > 0

Worked Example05

Water drains from a tank. The rate of decrease of the depth h is proportional to the square root of h. Write down a differential equation for h.

1

Depth h changes with time t →

\frac{d h}{d t}

2

"proportional to the square root of h" →

\frac{d h}{d t} \propto h

3

Convert ∝ to = k:

\frac{d h}{d t} = k h

4

"rate of decrease" means add negative sign:

\frac{d h}{d t} = - k h

Answer

\frac{d h}{d t} = - k h

or

\frac{d h}{d t} = - k h^{1/2}

where k > 0

Worked Example06

In a population of 1000 people, a rumour spreads such that the rate of spread is proportional to the product of the number N who have heard it and the number who have not. Write down a differential equation for N.

1

N changes with time t →

\frac{d N}{d t}

2

Number who have NOT heard = 1000 − N

3

"proportional to the product" →

\frac{d N}{d t} \propto N (1000 - N)

4

Convert ∝ to = k:

\frac{d N}{d t} = k N (1000 - N)

5

Rumour is spreading (increasing), so positive — no negative sign

Answer

\frac{d N}{d t} = k N (1000 - N)

where k > 0

Worked Example07

The value V of a car decreases at a rate proportional to the square of its current value. Write down a differential equation for V in terms of time t.

1

Value V changes with time t →

\frac{d V}{d t}

2

"proportional to the square of its current value" →

\frac{d V}{d t} \propto V^{2}

3

Convert ∝ to = k:

\frac{d V}{d t} = k V^{2}

4

"decreases" means add negative sign:

\frac{d V}{d t} = - k V^{2}

Answer

\frac{d V}{d t} = - k V^{2}

where k > 0

Practice

Question 1: The concentration C of a drug in a patient's bloodstream decreases at a rate proportional to C. Write down a differential equation for C in terms of time t.

Step 1: C changes with time t →

\frac{d C}{d t}

Step 2: "proportional to C" →

\frac{d C}{d t} \propto C

Step 3: Convert ∝ to = k:

\frac{d C}{d t} = k C

Step 4: "decreases" means negative sign:

\frac{d C}{d t} = - k C

Answer

\frac{d C}{d t} = - k C

where k > 0

Practice

Question 2: The rate of increase of a plant's height H is inversely proportional to H. Write down a differential equation for H in terms of time t.

Step 1: H changes with time t →

\frac{d H}{d t}

Step 2: "inversely proportional to H" →

\frac{d H}{d t} \propto \frac{1}{H}

Step 3: Convert ∝ to = k:

\frac{d H}{d t} = \frac{k}{H}

Step 4: "rate of increase" means positive — no negative sign needed

Answer

\frac{d H}{d t} = \frac{k}{H}

where k > 0

Practice

Question 3: In a school of 500 students, a piece of news spreads. The rate at which students hear the news is proportional to the product of those who have heard (N) and those who haven't. Write down a differential equation for N.

Step 1: N changes with time t →

\frac{d N}{d t}

Step 2: Number who haven't heard = 500 − N

Step 3: "proportional to the product" →

\frac{d N}{d t} \propto N (500 - N)

Step 4: Convert ∝ to = k:

\frac{d N}{d t} = k N (500 - N)

Step 5: News spreading means increasing — positive, no negative sign

Answer

\frac{d N}{d t} = k N (500 - N)

where k > 0

Summary

The Two-Step Process

Write with ∝

Convert to = k

Key Translations

PhraseTranslationRate of change of y w.r.t. t

\frac{d y}{d t}

Proportional to y

\propto y

→

= k y

Inversely proportional to y

\propto \frac{1}{y}

→

= \frac{k}{y}

Rate of decreaseAdd negative sign

Exam Tip

Always check: Does "decrease" need a negative sign? Have you included the constant k? Are you using the correct variables?