1.10.6 Constructing differential equations (A-level only)

What you'll learn

How to recognise the variables and rates of change in a model.
How to translate words such as proportional to, increases and decreases into equations.
How to construct differential equations in pure mathematics and real contexts.
How to check that a proposed differential equation is sensible.

Prerequisites: derivatives as rates of change

A derivative measures the instantaneous rate at which one variable changes with respect to another.

For example, if $y$ depends on $x$ , then $dydx\frac{dy}{dx}$ is the rate of change of $y$ with respect to $x$ .

In a context involving time, the derivative often describes how quickly a quantity changes:

$dPdt\frac{dP}{dt}$ is the rate of change of a population $P$ with time $t$ .
$dvdt\frac{dv}{dt}$ is the rate of change of velocity $v$ with time $t$ .
$dDdp\frac{dD}{dp}$ is the rate of change of demand $D$ with price $p$ .

Definition

Differential equation

A differential equation is an equation containing a derivative, such as $dydx\frac{dy}{dx}$ or $d2ydx2\frac{d^2y}{dx^2}$ , together with the function or variables being modelled.

In this topic, constructing a differential equation means translating information into an equation involving a derivative. You may not be asked to solve it.

Example

Describing a rate of change

The volume $V$ of water in a tank decreases at a rate equal to $3V3\sqrt{V}$ units per minute. Construct a differential equation.

Since the volume changes with time, represent its rate of change by $dVdt\frac{dV}{dt}$ .
The word decreases tells you that the rate is negative.
The magnitude of the rate is $3V3\sqrt{V}$ , so the model is

$dVdt=−3V.\frac{dV}{dt}=-3\sqrt{V}.$

Translating proportionality statements

The most common modelling instruction is that one quantity is proportional to another.

Definition

Direct proportionality

If $A$ is proportional to $B$ , then

$A = k B,$

where $k$ is a constant of proportionality. Unless the sign is included separately, $k$ is usually taken to be positive.

Therefore, if the rate of change of $y$ is proportional to a function $f (y)$ , write

$dydx=kf(y).\frac{dy}{dx}=kf(y).$

You must include the constant $k$ . Writing only $dydx=f(y)\frac{dy}{dx}=f(y)$ assumes that the constant of proportionality is exactly one.

Example

Translating proportional growth

A quantity $N$ increases at a rate proportional to the square of its current value. Construct a differential equation.

The rate at which $N$ changes with time is $dNdt\frac{dN}{dt}$ .
The square of the current value is $N^2$ , so proportionality gives a term of the form $kN^2$ .
Because the quantity increases, the derivative is positive:

$dNdt=kN2,k>0.\frac{dN}{dt}=kN^2,\qquad k>0.$

Common Mistake

Missing the proportionality constant

“Proportional to $N^2$ ” means $kN^2$ , not just $N^2$ . Introduce a constant unless its value is given or can be calculated.

Choosing the sign

Language about the direction of change determines the sign of the derivative.

Increases, grows or rises usually gives a positive derivative.
Decreases, decays or falls usually gives a negative derivative.
“At a rate proportional to” describes the size and form of the rate; the surrounding wording determines its sign.

For example, if $M$ decreases at a rate proportional to $M$ , you can write

$dMdt=−kM,k>0.\frac{dM}{dt}=-kM,\qquad k>0.$

Alternatively, $dMdt=kM\frac{dM}{dt}=kM$ could be used with $k < 0$ , but explicitly writing $- k M$ with $k > 0$ makes the model clearer.

Key Idea

Translate the whole sentence

Identify both the form of the rate and its direction. “Proportional to $Q$ ” gives $k Q$ , while “decreases” supplies the negative sign.

Constructing equations in pure mathematics

A pure mathematics question may specify a relationship between a function and one or more of its derivatives without giving a real-world context.

Common translations include:

“The gradient is equal to twice the value of $y$ ” gives $dydx=2y\frac{dy}{dx}=2y$ .
“The rate of change of $y$ is proportional to $x^2-y$ ” gives $dydx=k(x2−y)\frac{dy}{dx}=k(x^2-y)$ .
“The second derivative is the negative of the function” gives $d2ydx2=−y\frac{d^2y}{dx^2}=-y$ .

Example

Constructing a second-order equation

A function $y$ has a second derivative proportional to $x - y$ . When $x - y$ is positive, the second derivative is positive. Construct a differential equation.

The relevant rate is the second derivative, so use $d2ydx2\frac{d^2y}{dx^2}$ .
Proportionality to $x - y$ introduces a constant $k$ .
The quantities have the same sign, so take $k > 0$ and write

$d2ydx2=k(x−y),k>0.\frac{d^2y}{dx^2}=k(x-y),\qquad k>0.$

Kinematics models

In kinematics, displacement, velocity and acceleration are connected by differentiation.

If displacement $s$ , velocity $v$ and acceleration $a$ depend on time $t$ , then

$v=dsdtanda=dvdt=d2sdt2.v=\frac{ds}{dt} \qquad\text{and}\qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.$

A question may describe acceleration in terms of displacement, velocity or time. Your task is to substitute the description into one of these derivative relationships.

Example

Modelling acceleration depending on velocity

A particle moves in a straight line. Its acceleration is the sum of a constant driving acceleration of $s−212\text{ m s}^{-2}$ and a resistance whose magnitude is proportional to its velocity $v$ . Construct a differential equation for $v$ .

Acceleration is the rate of change of velocity, so $a=dvdta=\frac{dv}{dt}$ .
The driving acceleration contributes $12$ . Resistance acts against the motion, so its contribution is $- k v$ , where $k > 0$ .
Combining the two effects gives

$dvdt=12−kv,k>0.\frac{dv}{dt}=12-kv,\qquad k>0.$

This model predicts that the acceleration decreases as the velocity increases. At the velocity satisfying $12 - k v = 0$ , the acceleration is zero, so the velocity is no longer changing.

Common Mistake

Confusing velocity and acceleration

If a statement describes how quickly velocity changes, it describes $dvdt\frac{dv}{dt}$ , not $v$ . Similarly, a rate of change of displacement is $dsdt\frac{ds}{dt}$ .

Population growth models

Let $P (t)$ be a population at time $t$ . Its rate of growth is $dPdt\frac{dP}{dt}$ .

Growth proportional to population

A simple population model assumes that the growth rate is proportional to the current population:

$dPdt=kP,k>0.\frac{dP}{dt}=kP,\qquad k>0.$

This reflects the idea that a larger population produces more new individuals per unit time.

Growth with a limiting population

Unlimited growth is often unrealistic. A model may include a maximum sustainable population, called the carrying capacity.

Definition

Carrying capacity

The carrying capacity is the population size that the environment can sustain in the long term. It is often represented by a constant such as $C$ .

If growth is proportional both to the current population and to the remaining capacity, then

$dPdt=kP(C−P),k>0.\frac{dP}{dt}=kP(C-P),\qquad k>0.$

Example

Modelling limited population growth

A population $P$ grows at a rate proportional to both its current size and the difference between 5000 and its current size. Construct a differential equation.

The rate of population change is $dPdt\frac{dP}{dt}$ .
The current size contributes a factor $P$ , while the difference from the limiting population contributes $5000 - P$ .
Introducing a positive proportionality constant gives

$dPdt=kP(5000−P),k>0.\frac{dP}{dt}=kP(5000-P),\qquad k>0.$
This is consistent with the description: for $0 < P < 5000$ , both factors are positive, so the population grows; at $P = 5000$ , the growth rate is zero.

Tip

Test important values

Substitute boundary or equilibrium values into the right-hand side. For a limited-growth model, the rate should normally be positive below the carrying capacity and zero at the carrying capacity.

Price and demand models

Demand usually decreases as price increases. Let $D$ be demand and $p$ be price. Then $dDdp\frac{dD}{dp}$ measures how demand changes as price changes.

Because greater prices usually lead to lower demand, a sensible model often has

$dDdp<0.\frac{dD}{dp}<0.$

Example

Modelling demand as price changes

The rate of change of demand $D$ with respect to price $p$ is proportional to the difference between demand and a minimum demand level $D_0$ . Demand decreases as price increases. Construct a differential equation.

The required rate is $dDdp\frac{dD}{dp}$ because demand is changing with respect to price.
The stated difference is $D-D_0$ , so proportionality gives $k(D-D_0)$ .
Demand decreases as price rises, so include a negative sign:

$dDdp=−k(D−D0),k>0.\frac{dD}{dp}=-k(D-D_0),\qquad k>0.$
When $D>D_0$ , the derivative is negative, while at $D=D_0$ the derivative is zero. The model therefore treats $D_0$ as a limiting demand level.

Using numerical information to find a constant

After constructing a model containing $k$ , you may be given a particular rate at a particular value. Substitute both pieces of information into the differential equation to find $k$ .

Example

Finding a population model constant

A population satisfies $dPdt=kP\frac{dP}{dt}=kP$ . When the population is 800, it is increasing at 40 individuals per year. Find the differential equation.

Substitute $P = 800$ and $dPdt=40\frac{dP}{dt}=40$ into the model:

$40 = 800 k .$
Solve for the constant:

$k=40800=120.k=\frac{40}{800}=\frac{1}{20}.$
Substitute this value back into the model:

$dPdt=120P.\frac{dP}{dt}=\frac{1}{20}P.$

Initial conditions and modelling assumptions

An initial condition gives the value of the dependent variable at a specified starting point, such as $P (0) = 200$ .

The initial condition is usually written alongside the differential equation:

$dPdt=kP,P(0)=200.\frac{dP}{dt}=kP, \qquad P(0)=200.$

It does not normally help you construct the relationship between the derivative and the variables, but it may later be used to find an unknown constant when solving the equation.

Every differential equation is also a model, so it is based on assumptions. For example, $dPdt=kP\frac{dP}{dt}=kP$ assumes that environmental conditions and the proportional growth rate remain constant.

Key Idea

Check the model

After constructing the equation, check its variables, sign and behaviour. Ask whether the predicted rate is positive, negative or zero in the situations described.

Exam technique

In the exam

Identify the quantity changing and the variable it changes with respect to; this determines the derivative.
Translate each proportionality statement using a constant such as $k$ , and use the wording to choose the sign.
Combine separate effects carefully, then test a simple or limiting value to check that the model behaves sensibly.
Use any numerical rate information to calculate $k$ , and state any given initial condition separately.

Self review

Check yourself

How would you model a quantity $Q$ that decreases at a rate proportional to $Q\sqrt{Q}$ ?
A particle has acceleration proportional to the difference between a fixed speed $V$ and its current velocity $v$ . What differential equation describes its velocity?
How would you construct a population model with carrying capacity $C$ , and what should happen to the growth rate when $P = C$ ?

What you'll learn

How to recognise the variables and rates of change in a model.
How to translate words such as proportional to, increases and decreases into equations.
How to construct differential equations in pure mathematics and real contexts.
How to check that a proposed differential equation is sensible.

Prerequisites: derivatives as rates of change

A derivative measures the instantaneous rate at which one variable changes with respect to another.

For example, if $y$ depends on $x$ , then $dydx\frac{dy}{dx}$ is the rate of change of $y$ with respect to $x$ .

In a context involving time, the derivative often describes how quickly a quantity changes:

$dPdt\frac{dP}{dt}$ is the rate of change of a population $P$ with time $t$ .
$dvdt\frac{dv}{dt}$ is the rate of change of velocity $v$ with time $t$ .
$dDdp\frac{dD}{dp}$ is the rate of change of demand $D$ with price $p$ .

Definition

Differential equation

A differential equation is an equation containing a derivative, such as $dydx\frac{dy}{dx}$ or $d2ydx2\frac{d^2y}{dx^2}$ , together with the function or variables being modelled.

In this topic, constructing a differential equation means translating information into an equation involving a derivative. You may not be asked to solve it.

Example

Describing a rate of change

The volume $V$ of water in a tank decreases at a rate equal to $3V3\sqrt{V}$ units per minute. Construct a differential equation.

Since the volume changes with time, represent its rate of change by $dVdt\frac{dV}{dt}$ .
The word decreases tells you that the rate is negative.
The magnitude of the rate is $3V3\sqrt{V}$ , so the model is

$dVdt=−3V.\frac{dV}{dt}=-3\sqrt{V}.$

Translating proportionality statements

The most common modelling instruction is that one quantity is proportional to another.

Definition

Direct proportionality

If $A$ is proportional to $B$ , then

$A = k B,$

where $k$ is a constant of proportionality. Unless the sign is included separately, $k$ is usually taken to be positive.

Therefore, if the rate of change of $y$ is proportional to a function $f (y)$ , write

$dydx=kf(y).\frac{dy}{dx}=kf(y).$

You must include the constant $k$ . Writing only $dydx=f(y)\frac{dy}{dx}=f(y)$ assumes that the constant of proportionality is exactly one.

Example

Translating proportional growth

A quantity $N$ increases at a rate proportional to the square of its current value. Construct a differential equation.

The rate at which $N$ changes with time is $dNdt\frac{dN}{dt}$ .
The square of the current value is $N^2$ , so proportionality gives a term of the form $kN^2$ .
Because the quantity increases, the derivative is positive:

$dNdt=kN2,k>0.\frac{dN}{dt}=kN^2,\qquad k>0.$

Common Mistake

Missing the proportionality constant

“Proportional to $N^2$ ” means $kN^2$ , not just $N^2$ . Introduce a constant unless its value is given or can be calculated.

Choosing the sign

Language about the direction of change determines the sign of the derivative.

Increases, grows or rises usually gives a positive derivative.
Decreases, decays or falls usually gives a negative derivative.
“At a rate proportional to” describes the size and form of the rate; the surrounding wording determines its sign.

For example, if $M$ decreases at a rate proportional to $M$ , you can write

$dMdt=−kM,k>0.\frac{dM}{dt}=-kM,\qquad k>0.$

Alternatively, $dMdt=kM\frac{dM}{dt}=kM$ could be used with $k < 0$ , but explicitly writing $- k M$ with $k > 0$ makes the model clearer.

Key Idea

Translate the whole sentence

Identify both the form of the rate and its direction. “Proportional to $Q$ ” gives $k Q$ , while “decreases” supplies the negative sign.

Constructing equations in pure mathematics

A pure mathematics question may specify a relationship between a function and one or more of its derivatives without giving a real-world context.

Common translations include:

“The gradient is equal to twice the value of $y$ ” gives $dydx=2y\frac{dy}{dx}=2y$ .
“The rate of change of $y$ is proportional to $x^2-y$ ” gives $dydx=k(x2−y)\frac{dy}{dx}=k(x^2-y)$ .
“The second derivative is the negative of the function” gives $d2ydx2=−y\frac{d^2y}{dx^2}=-y$ .

Example

Constructing a second-order equation

A function $y$ has a second derivative proportional to $x - y$ . When $x - y$ is positive, the second derivative is positive. Construct a differential equation.

The relevant rate is the second derivative, so use $d2ydx2\frac{d^2y}{dx^2}$ .
Proportionality to $x - y$ introduces a constant $k$ .
The quantities have the same sign, so take $k > 0$ and write

$d2ydx2=k(x−y),k>0.\frac{d^2y}{dx^2}=k(x-y),\qquad k>0.$

Kinematics models

In kinematics, displacement, velocity and acceleration are connected by differentiation.

If displacement $s$ , velocity $v$ and acceleration $a$ depend on time $t$ , then

$v=dsdtanda=dvdt=d2sdt2.v=\frac{ds}{dt} \qquad\text{and}\qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.$

A question may describe acceleration in terms of displacement, velocity or time. Your task is to substitute the description into one of these derivative relationships.

Example

Modelling acceleration depending on velocity

Acceleration is the rate of change of velocity, so $a=dvdta=\frac{dv}{dt}$ .
The driving acceleration contributes $12$ . Resistance acts against the motion, so its contribution is $- k v$ , where $k > 0$ .
Combining the two effects gives

$dvdt=12−kv,k>0.\frac{dv}{dt}=12-kv,\qquad k>0.$

This model predicts that the acceleration decreases as the velocity increases. At the velocity satisfying $12 - k v = 0$ , the acceleration is zero, so the velocity is no longer changing.

Common Mistake

Confusing velocity and acceleration

If a statement describes how quickly velocity changes, it describes $dvdt\frac{dv}{dt}$ , not $v$ . Similarly, a rate of change of displacement is $dsdt\frac{ds}{dt}$ .

Population growth models

Let $P (t)$ be a population at time $t$ . Its rate of growth is $dPdt\frac{dP}{dt}$ .

Growth proportional to population

A simple population model assumes that the growth rate is proportional to the current population:

$dPdt=kP,k>0.\frac{dP}{dt}=kP,\qquad k>0.$

This reflects the idea that a larger population produces more new individuals per unit time.

Growth with a limiting population

Unlimited growth is often unrealistic. A model may include a maximum sustainable population, called the carrying capacity.

Definition

Carrying capacity

The carrying capacity is the population size that the environment can sustain in the long term. It is often represented by a constant such as $C$ .

If growth is proportional both to the current population and to the remaining capacity, then

$dPdt=kP(C−P),k>0.\frac{dP}{dt}=kP(C-P),\qquad k>0.$

Example

Modelling limited population growth

A population $P$ grows at a rate proportional to both its current size and the difference between 5000 and its current size. Construct a differential equation.

The rate of population change is $dPdt\frac{dP}{dt}$ .
The current size contributes a factor $P$ , while the difference from the limiting population contributes $5000 - P$ .
Introducing a positive proportionality constant gives

$dPdt=kP(5000−P),k>0.\frac{dP}{dt}=kP(5000-P),\qquad k>0.$
This is consistent with the description: for $0 < P < 5000$ , both factors are positive, so the population grows; at $P = 5000$ , the growth rate is zero.

Tip

Test important values

Substitute boundary or equilibrium values into the right-hand side. For a limited-growth model, the rate should normally be positive below the carrying capacity and zero at the carrying capacity.

Price and demand models

Demand usually decreases as price increases. Let $D$ be demand and $p$ be price. Then $dDdp\frac{dD}{dp}$ measures how demand changes as price changes.

Because greater prices usually lead to lower demand, a sensible model often has

$dDdp<0.\frac{dD}{dp}<0.$

Example

Modelling demand as price changes

The required rate is $dDdp\frac{dD}{dp}$ because demand is changing with respect to price.
The stated difference is $D-D_0$ , so proportionality gives $k(D-D_0)$ .
Demand decreases as price rises, so include a negative sign:

$dDdp=−k(D−D0),k>0.\frac{dD}{dp}=-k(D-D_0),\qquad k>0.$
When $D>D_0$ , the derivative is negative, while at $D=D_0$ the derivative is zero. The model therefore treats $D_0$ as a limiting demand level.

Using numerical information to find a constant

After constructing a model containing $k$ , you may be given a particular rate at a particular value. Substitute both pieces of information into the differential equation to find $k$ .

Example

Finding a population model constant

A population satisfies $dPdt=kP\frac{dP}{dt}=kP$ . When the population is 800, it is increasing at 40 individuals per year. Find the differential equation.

Substitute $P = 800$ and $dPdt=40\frac{dP}{dt}=40$ into the model:

$40 = 800 k .$
Solve for the constant:

$k=40800=120.k=\frac{40}{800}=\frac{1}{20}.$
Substitute this value back into the model:

$dPdt=120P.\frac{dP}{dt}=\frac{1}{20}P.$

Initial conditions and modelling assumptions

An initial condition gives the value of the dependent variable at a specified starting point, such as $P (0) = 200$ .

The initial condition is usually written alongside the differential equation:

$dPdt=kP,P(0)=200.\frac{dP}{dt}=kP, \qquad P(0)=200.$

It does not normally help you construct the relationship between the derivative and the variables, but it may later be used to find an unknown constant when solving the equation.

Key Idea

Check the model

After constructing the equation, check its variables, sign and behaviour. Ask whether the predicted rate is positive, negative or zero in the situations described.

Exam technique

In the exam

Identify the quantity changing and the variable it changes with respect to; this determines the derivative.
Translate each proportionality statement using a constant such as $k$ , and use the wording to choose the sign.
Combine separate effects carefully, then test a simple or limiting value to check that the model behaves sensibly.
Use any numerical rate information to calculate $k$ , and state any given initial condition separately.

Self review

Check yourself

How would you model a quantity $Q$ that decreases at a rate proportional to $Q\sqrt{Q}$ ?
A particle has acceleration proportional to the difference between a fixed speed $V$ and its current velocity $v$ . What differential equation describes its velocity?
How would you construct a population model with carrying capacity $C$ , and what should happen to the growth rate when $P = C$ ?

What you'll learn

Prerequisites: derivatives as rates of change

Translating proportionality statements

Choosing the sign

Constructing equations in pure mathematics

Kinematics models

Population growth models

Growth proportional to population

Growth with a limiting population

Price and demand models

Using numerical information to find a constant

Initial conditions and modelling assumptions

Review 1.10.6 Constructing differential equations (A-level only) by teaching Genie

1.10.6 Constructing differential equations (A-level only)

What you'll learn

Prerequisites: derivatives as rates of change

Translating proportionality statements

Choosing the sign

Constructing equations in pure mathematics

Kinematics models

Population growth models

Growth proportional to population

Growth with a limiting population

Price and demand models

Using numerical information to find a constant

Initial conditions and modelling assumptions

Review 1.10.6 Constructing differential equations (A-level only) by teaching Genie

1.10.6 Constructing differential equations (A-level only)

What you'll learn

Prerequisites: derivatives as rates of change

Translating proportionality statements

Choosing the sign

Constructing equations in pure mathematics

Kinematics models

Population growth models

Growth proportional to population

Growth with a limiting population

Price and demand models

Using numerical information to find a constant

Initial conditions and modelling assumptions

Review 1.10.6 Constructing differential equations (A-level only) by teaching Genie

1.10.6 Constructing differential equations (A-level only) Revision Guide

1.10.6 Constructing differential equations (A-level only)

What you'll learn

Prerequisites: derivatives as rates of change

Translating proportionality statements

Choosing the sign

Constructing equations in pure mathematics

Kinematics models

Population growth models

Growth proportional to population

Growth with a limiting population

Price and demand models

Using numerical information to find a constant

Initial conditions and modelling assumptions

Review 1.10.6 Constructing differential equations (A-level only) by teaching Genie

1.10.6 Constructing differential equations (A-level only) Revision Guide