1.7.20 Constructing differential equations (A-level only)

What you'll learn

What a differential equation is and how it represents change.
How to translate words such as “proportional to” and “decreases at a constant rate” into equations.
How to construct models for kinematics, population growth, and price–demand relationships.
How to check that a proposed differential equation is sensible.

From differentiation to differential equations

You already know that a derivative represents a rate of change. For example, if $y$ depends on $x$ , then $dydx\frac{dy}{dx}$ describes how quickly $y$ changes as $x$ changes.

Usually in differentiation, you are given $y$ and asked to find $dydx\frac{dy}{dx}$ . In this topic, you work in the opposite direction: you are given information about a rate of change and use it to construct an equation involving a derivative.

Definition

Differential equation

A differential equation is an equation containing a derivative of an unknown function. For example,

$dydx=3x2−y\frac{dy}{dx}=3x^2-y$

is a differential equation because it connects the unknown function $y$ with its derivative.

At this stage, the main skill is constructing the equation. You may not be asked to solve it.

Example

Describing a rate of change

The rate of change of $y$ with respect to $x$ is equal to the square of $x$ plus twice the value of $y$ . Construct a differential equation.

“The rate of change of $y$ with respect to $x$ ” means $dydx\frac{dy}{dx}$ .
“The square of $x$ ” is $x^2$ , while “twice the value of $y$ ” is $2 y$ .
Combining these in the stated order gives

$dydx=x2+2y.\frac{dy}{dx}=x^2+2y.$

Choosing the variables

Before forming an equation, define the quantities involved and identify which is the dependent variable.

The dependent variable is the quantity being changed or measured. The independent variable is the quantity with respect to which the change is measured.

For example, if a population $P$ changes with time $t$ , then:

$P$ is the dependent variable;
$t$ is the independent variable;
$dPdt\frac{dP}{dt}$ is the rate of change of population with time.

Key Idea

Read derivative notation as a sentence

Read $dydx\frac{dy}{dx}$ as “the rate of change of $y$ with respect to $x$ ”. This helps you decide which variable belongs on the top and which belongs on the bottom.

Translating common phrases

Many questions use a small number of standard verbal descriptions.

Constant rates

If $y$ increases at a constant rate $c$ , then

$dydt=c.\frac{dy}{dt}=c.$

If it decreases at a constant rate $c$ , where $c > 0$ , then

$dydt=−c.\frac{dy}{dt}=-c.$

The sign matters: a negative derivative represents a decreasing quantity.

Proportional rates

The statement “ $A$ is proportional to $B$ ” means

$A = k B,$

where $k$ is a constant of proportionality.

Therefore, if the rate of change of $y$ is proportional to $y$ , then

$dydt=ky.\frac{dy}{dt}=ky.$

If $y$ is known to decrease, you may write

$dydt=−ky,\frac{dy}{dt}=-ky,$

where $k > 0$ .

Common Mistake

Omitting the proportionality constant

Do not translate “ $dydt\frac{dy}{dt}$ is proportional to $y$ ” as $dydt=y\frac{dy}{dt}=y$ . Introduce a constant: $dydt=ky\frac{dy}{dt}=ky$ .

Example

Constructing a proportional growth model

A quantity $M$ grows at a rate proportional to the product of its current value and the square of time. Construct a differential equation.

Since $M$ changes with time $t$ , its rate of change is $dMdt\frac{dM}{dt}$ .
The product of the current value and the square of time is $Mt^2$ .
Introducing a positive constant of proportionality $k$ gives

$dMdt=kMt2,k>0.\frac{dM}{dt}=kMt^2,\qquad k>0.$

Pure mathematics models

A differential equation can be constructed without a real-world context. You may be told how the gradient of a curve depends on its coordinates.

For a curve with equation $y = f (x)$ :

its gradient is $dydx\frac{dy}{dx}$ ;
its concavity or rate of change of gradient is described by $d2ydx2\frac{d^2y}{dx^2}$ .

Definition

Order of a differential equation

The order of a differential equation is the order of its highest derivative. An equation containing $dydx\frac{dy}{dx}$ but no higher derivative is first order; one containing $d2ydx2\frac{d^2y}{dx^2}$ is second order.

Example

Modelling the gradient of a curve

At every point $(x, y)$ on a curve, its gradient is proportional to the difference between $x$ and $y$ . At the point $(1, 3)$ , the gradient is $- 6$ . Construct the differential equation.

Translating “the gradient is proportional to the difference between $x$ and $y$ ” gives

$dydx=k(x−y).\frac{dy}{dx}=k(x-y).$
At $(1, 3)$ , substitute $x = 1$ , $y = 3$ and $dydx=−6\frac{dy}{dx}=-6$ :

$- 6 = k (1 - 3) = - 2 k .$
Hence $k = 3$ , so the required equation is

$dydx=3(x−y).\frac{dy}{dx}=3(x-y).$

Kinematics models

In kinematics, the position or displacement $s$ of a particle is usually a function of time $t$ .

Definition

Velocity and acceleration

Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity:

$v=dsdt,a=dvdt=d2sdt2.v=\frac{ds}{dt}, \qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.$

A description involving velocity produces a first derivative of displacement. A description involving acceleration may produce either $dvdt\frac{dv}{dt}$ or the second derivative $d2sdt2\frac{d^2s}{dt^2}$ .

Example

Constructing an acceleration model

A particle moves along a straight line. Its acceleration is proportional to its displacement from the origin and acts towards the origin. Construct a differential equation for its displacement $s$ .

Acceleration is the second derivative of displacement:

$a=d2sdt2.a=\frac{d^2s}{dt^2}.$
Proportionality to displacement gives a magnitude of $k ∣ s ∣$ . Because the acceleration acts towards the origin, its direction is opposite to the sign of $s$ .
Taking $k > 0$ , the model is therefore

$d2sdt2=−ks.\frac{d^2s}{dt^2}=-ks.$

If $s > 0$ , the acceleration is negative; if $s < 0$ , the acceleration is positive. Both directions point towards the origin.

Tip

Check the direction

In kinematics, test the sign of your model on each side of the origin. A force or acceleration “towards the origin” should have the opposite sign to the displacement.

Population growth models

Let $P$ be a population at time $t$ . Then $dPdt\frac{dP}{dt}$ is its population growth rate.

A common simple model assumes that the population 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 can produce more new individuals per unit time.

A more realistic model may include a limiting factor. For example, if $M$ is a maximum sustainable population, the growth rate might be proportional both to $P$ and to the remaining capacity $M - P$ .

Example

Modelling population with limited capacity

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 two factors controlling growth are the current population $P$ and the remaining capacity $5000 - P$ .
Introducing a positive constant $k$ gives

$dPdt=kP(5000−P),k>0.\frac{dP}{dt}=kP(5000-P),\qquad k>0.$
This is consistent with the description: the growth rate is zero when $P = 0$ or $P = 5000$ , and it is positive when $0 < P < 5000$ .

Common Mistake

Models have a valid range

The model $dPdt=kP(5000−P)\frac{dP}{dt}=kP(5000-P)$ predicts negative growth when $P > 5000$ . Whether that is meaningful depends on the context and the assumptions of the model.

Price and demand models

Let $D$ represent demand and $p$ represent price. Then

$dDdp\frac{dD}{dp}$

measures how demand changes as price changes.

In a typical market model, increasing the price reduces demand, so you would expect

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

Example

Modelling demand as price changes

The rate at which demand $D$ changes with price $p$ is proportional to the demand and inversely proportional to the square of the price. Demand decreases as price increases. Construct a differential equation.

“The rate at which demand changes with price” is $dDdp\frac{dD}{dp}$ .
Proportionality to demand and inverse proportionality to the square of price produce the expression $Dp2\frac{D}{p^2}$ .
Because demand decreases as price increases, the derivative must be negative. Therefore,

$dDdp=−kDp2,k>0.\frac{dD}{dp}=-\frac{kD}{p^2}, \qquad k>0.$

Common Mistake

Using time automatically

Not every derivative is with respect to time. In a price–demand model, the appropriate derivative may be $dDdp\frac{dD}{dp}$ , not $dDdt\frac{dD}{dt}$ .

Checking your model

Once you have constructed a differential equation, test whether it matches the description.

Check the variables

Confirm that the derivative represents the requested rate. For example, $dpdD\frac{dp}{dD}$ and $dDdp\frac{dD}{dp}$ are not interchangeable.

Check the sign

If a quantity is decreasing, its derivative should be negative over the relevant range. If motion is towards an equilibrium position, the direction should be consistent on both sides.

Check special values

Substitute values where the behaviour is predictable. In a limited-population model, you might expect no growth at zero population or at maximum capacity.

Check dimensions

Both sides of an equation must have compatible units. For example, if $s$ is measured in m and $t$ in s, then $d2sdt2\frac{d^2s}{dt^2}$ has units m s $^{-2}$ . In

$d2sdt2=−ks,\frac{d^2s}{dt^2}=-ks,$

the constant $k$ must therefore have units s $^{-2}$ .

Key Idea

Construct, then interpret

A differential equation is a model of how a quantity changes. After translating the words, check its sign, special values, units and predicted behaviour against the original context.

Exam technique

In the exam

Define the variables and identify exactly which rate of change is being described.
Translate phrases such as “proportional to”, “inversely proportional to” and “towards the origin” carefully, introducing a constant $k$ where needed.
Use any given point, rate or initial information to determine the constant of proportionality.
Check the sign of the derivative and test values where the model’s behaviour should be obvious.
Stop once you have constructed the required differential equation unless the question also asks you to solve it.

Self review

Check yourself

How would you model a quantity $N$ that decreases at a rate proportional to $N^2$ ?
What differential equation represents acceleration proportional to velocity but acting in the opposite direction?
If demand falls more rapidly when both demand and price are large, what variables and derivative might you use?

What you'll learn

What a differential equation is and how it represents change.
How to translate words such as “proportional to” and “decreases at a constant rate” into equations.
How to construct models for kinematics, population growth, and price–demand relationships.
How to check that a proposed differential equation is sensible.

From differentiation to differential equations

You already know that a derivative represents a rate of change. For example, if $y$ depends on $x$ , then $dydx\frac{dy}{dx}$ describes how quickly $y$ changes as $x$ changes.

Definition

Differential equation

A differential equation is an equation containing a derivative of an unknown function. For example,

$dydx=3x2−y\frac{dy}{dx}=3x^2-y$

is a differential equation because it connects the unknown function $y$ with its derivative.

At this stage, the main skill is constructing the equation. You may not be asked to solve it.

Example

Describing a rate of change

The rate of change of $y$ with respect to $x$ is equal to the square of $x$ plus twice the value of $y$ . Construct a differential equation.

“The rate of change of $y$ with respect to $x$ ” means $dydx\frac{dy}{dx}$ .
“The square of $x$ ” is $x^2$ , while “twice the value of $y$ ” is $2 y$ .
Combining these in the stated order gives

$dydx=x2+2y.\frac{dy}{dx}=x^2+2y.$

Choosing the variables

Before forming an equation, define the quantities involved and identify which is the dependent variable.

The dependent variable is the quantity being changed or measured. The independent variable is the quantity with respect to which the change is measured.

For example, if a population $P$ changes with time $t$ , then:

$P$ is the dependent variable;
$t$ is the independent variable;
$dPdt\frac{dP}{dt}$ is the rate of change of population with time.

Key Idea

Read derivative notation as a sentence

Read $dydx\frac{dy}{dx}$ as “the rate of change of $y$ with respect to $x$ ”. This helps you decide which variable belongs on the top and which belongs on the bottom.

Translating common phrases

Many questions use a small number of standard verbal descriptions.

Constant rates

If $y$ increases at a constant rate $c$ , then

$dydt=c.\frac{dy}{dt}=c.$

If it decreases at a constant rate $c$ , where $c > 0$ , then

$dydt=−c.\frac{dy}{dt}=-c.$

The sign matters: a negative derivative represents a decreasing quantity.

Proportional rates

The statement “ $A$ is proportional to $B$ ” means

$A = k B,$

where $k$ is a constant of proportionality.

Therefore, if the rate of change of $y$ is proportional to $y$ , then

$dydt=ky.\frac{dy}{dt}=ky.$

If $y$ is known to decrease, you may write

$dydt=−ky,\frac{dy}{dt}=-ky,$

where $k > 0$ .

Common Mistake

Omitting the proportionality constant

Do not translate “ $dydt\frac{dy}{dt}$ is proportional to $y$ ” as $dydt=y\frac{dy}{dt}=y$ . Introduce a constant: $dydt=ky\frac{dy}{dt}=ky$ .

Example

Constructing a proportional growth model

A quantity $M$ grows at a rate proportional to the product of its current value and the square of time. Construct a differential equation.

Since $M$ changes with time $t$ , its rate of change is $dMdt\frac{dM}{dt}$ .
The product of the current value and the square of time is $Mt^2$ .
Introducing a positive constant of proportionality $k$ gives

$dMdt=kMt2,k>0.\frac{dM}{dt}=kMt^2,\qquad k>0.$

Pure mathematics models

A differential equation can be constructed without a real-world context. You may be told how the gradient of a curve depends on its coordinates.

For a curve with equation $y = f (x)$ :

its gradient is $dydx\frac{dy}{dx}$ ;
its concavity or rate of change of gradient is described by $d2ydx2\frac{d^2y}{dx^2}$ .

Definition

Order of a differential equation

Example

Modelling the gradient of a curve

At every point $(x, y)$ on a curve, its gradient is proportional to the difference between $x$ and $y$ . At the point $(1, 3)$ , the gradient is $- 6$ . Construct the differential equation.

Translating “the gradient is proportional to the difference between $x$ and $y$ ” gives

$dydx=k(x−y).\frac{dy}{dx}=k(x-y).$
At $(1, 3)$ , substitute $x = 1$ , $y = 3$ and $dydx=−6\frac{dy}{dx}=-6$ :

$- 6 = k (1 - 3) = - 2 k .$
Hence $k = 3$ , so the required equation is

$dydx=3(x−y).\frac{dy}{dx}=3(x-y).$

Kinematics models

In kinematics, the position or displacement $s$ of a particle is usually a function of time $t$ .

Definition

Velocity and acceleration

Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity:

$v=dsdt,a=dvdt=d2sdt2.v=\frac{ds}{dt}, \qquad a=\frac{dv}{dt}=\frac{d^2s}{dt^2}.$

Example

Constructing an acceleration model

A particle moves along a straight line. Its acceleration is proportional to its displacement from the origin and acts towards the origin. Construct a differential equation for its displacement $s$ .

Acceleration is the second derivative of displacement:

$a=d2sdt2.a=\frac{d^2s}{dt^2}.$
Proportionality to displacement gives a magnitude of $k ∣ s ∣$ . Because the acceleration acts towards the origin, its direction is opposite to the sign of $s$ .
Taking $k > 0$ , the model is therefore

$d2sdt2=−ks.\frac{d^2s}{dt^2}=-ks.$

If $s > 0$ , the acceleration is negative; if $s < 0$ , the acceleration is positive. Both directions point towards the origin.

Tip

Check the direction

In kinematics, test the sign of your model on each side of the origin. A force or acceleration “towards the origin” should have the opposite sign to the displacement.

Population growth models

Let $P$ be a population at time $t$ . Then $dPdt\frac{dP}{dt}$ is its population growth rate.

A common simple model assumes that the population 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 can produce more new individuals per unit time.

Example

Modelling population with limited capacity

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 two factors controlling growth are the current population $P$ and the remaining capacity $5000 - P$ .
Introducing a positive constant $k$ gives

$dPdt=kP(5000−P),k>0.\frac{dP}{dt}=kP(5000-P),\qquad k>0.$
This is consistent with the description: the growth rate is zero when $P = 0$ or $P = 5000$ , and it is positive when $0 < P < 5000$ .

Common Mistake

Models have a valid range

The model $dPdt=kP(5000−P)\frac{dP}{dt}=kP(5000-P)$ predicts negative growth when $P > 5000$ . Whether that is meaningful depends on the context and the assumptions of the model.

Price and demand models

Let $D$ represent demand and $p$ represent price. Then

$dDdp\frac{dD}{dp}$

measures how demand changes as price changes.

In a typical market model, increasing the price reduces demand, so you would expect

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

Example

Modelling demand as price changes

“The rate at which demand changes with price” is $dDdp\frac{dD}{dp}$ .
Proportionality to demand and inverse proportionality to the square of price produce the expression $Dp2\frac{D}{p^2}$ .
Because demand decreases as price increases, the derivative must be negative. Therefore,

$dDdp=−kDp2,k>0.\frac{dD}{dp}=-\frac{kD}{p^2}, \qquad k>0.$

Common Mistake

Using time automatically

Not every derivative is with respect to time. In a price–demand model, the appropriate derivative may be $dDdp\frac{dD}{dp}$ , not $dDdt\frac{dD}{dt}$ .

Checking your model

Once you have constructed a differential equation, test whether it matches the description.

Check the variables

Confirm that the derivative represents the requested rate. For example, $dpdD\frac{dp}{dD}$ and $dDdp\frac{dD}{dp}$ are not interchangeable.

Check the sign

If a quantity is decreasing, its derivative should be negative over the relevant range. If motion is towards an equilibrium position, the direction should be consistent on both sides.

Check special values

Substitute values where the behaviour is predictable. In a limited-population model, you might expect no growth at zero population or at maximum capacity.

Check dimensions

Both sides of an equation must have compatible units. For example, if $s$ is measured in m and $t$ in s, then $d2sdt2\frac{d^2s}{dt^2}$ has units m s $^{-2}$ . In

$d2sdt2=−ks,\frac{d^2s}{dt^2}=-ks,$

the constant $k$ must therefore have units s $^{-2}$ .

Key Idea

Construct, then interpret

A differential equation is a model of how a quantity changes. After translating the words, check its sign, special values, units and predicted behaviour against the original context.

Exam technique

In the exam

Define the variables and identify exactly which rate of change is being described.
Translate phrases such as “proportional to”, “inversely proportional to” and “towards the origin” carefully, introducing a constant $k$ where needed.
Use any given point, rate or initial information to determine the constant of proportionality.
Check the sign of the derivative and test values where the model’s behaviour should be obvious.
Stop once you have constructed the required differential equation unless the question also asks you to solve it.

Self review

Check yourself

How would you model a quantity $N$ that decreases at a rate proportional to $N^2$ ?
What differential equation represents acceleration proportional to velocity but acting in the opposite direction?
If demand falls more rapidly when both demand and price are large, what variables and derivative might you use?

What you'll learn

From differentiation to differential equations

Choosing the variables

Translating common phrases

Constant rates

Proportional rates

Pure mathematics models

Kinematics models

Population growth models

Price and demand models

Checking your model

Check the variables

Check the sign

Check special values

Check dimensions

Review 1.7.20 Constructing differential equations (A-level only) by teaching Genie

1.7.20 Constructing differential equations (A-level only)

What you'll learn

From differentiation to differential equations

Choosing the variables

Translating common phrases

Constant rates

Proportional rates

Pure mathematics models

Kinematics models

Population growth models

Price and demand models

Checking your model

Check the variables

Check the sign

Check special values

Check dimensions

Review 1.7.20 Constructing differential equations (A-level only) by teaching Genie

1.7.20 Constructing differential equations (A-level only)

What you'll learn

From differentiation to differential equations

Choosing the variables

Translating common phrases

Constant rates

Proportional rates

Pure mathematics models

Kinematics models

Population growth models

Price and demand models

Checking your model

Check the variables

Check the sign

Check special values

Check dimensions

Review 1.7.20 Constructing differential equations (A-level only) by teaching Genie

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

1.7.20 Constructing differential equations (A-level only)

What you'll learn

From differentiation to differential equations

Choosing the variables

Translating common phrases

Constant rates

Proportional rates

Pure mathematics models

Kinematics models

Population growth models

Price and demand models

Checking your model

Check the variables

Check the sign

Check special values

Check dimensions

Review 1.7.20 Constructing differential equations (A-level only) by teaching Genie

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