1.11.8 Interpreting solutions of differential equations (A-level only)

What you'll learn

How to translate a differential equation and its solution into the language of a real problem.
How to interpret gradients, constants, limiting values and long-term behaviour.
How differential equations connect displacement, velocity and acceleration in kinematics.
How to identify assumptions and limitations in a mathematical model.

Prerequisites: rates of change

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

If a quantity $y$ depends on time $t$ , then $dydt\frac{dy}{dt}$ is its rate of change at time $t$ . On a graph of $y$ against $t$ , this derivative is the gradient of the curve.

If $dydt>0\frac{dy}{dt}>0$ , then $y$ is increasing.
If $dydt<0\frac{dy}{dt}<0$ , then $y$ is decreasing.
If $dydt=0\frac{dy}{dt}=0$ , then the graph has a stationary point or the quantity is at an equilibrium value.

Definition

Differential equation

A differential equation is an equation containing a derivative, such as $dydt\frac{dy}{dt}$ , together with the quantity being modelled.

For example,

$dydt=3y\frac{dy}{dt}=3y$

says that the rate of change of $y$ is three times its current value. It describes a relationship between a quantity and the way that quantity changes.

Key Idea

Read the derivative in words

Before doing any algebra, translate $dydt=f(y,t)\frac{dy}{dt}=f(y,t)$ as “the rate of change of $y$ is given by $f (y, t)$ ”.

Example

Interpreting a population model

A population $P$ , measured in thousands, is modelled by

$dPdt=0.2P,\frac{dP}{dt}=0.2P,$

where $t$ is measured in years.

Since $P$ is a population and $t$ is time, $dPdt\frac{dP}{dt}$ represents the population’s instantaneous rate of change, in thousands per year.
For any positive population, $0.2 P > 0$ . Therefore $dPdt>0\frac{dP}{dt}>0$ , so the model predicts that the population increases.
The rate is proportional to $P$ . A larger population therefore produces a larger absolute increase per year; for example, when $P = 10$ , the rate is $2$ thousand people per year.

What a solution represents

A solution of a differential equation is a function that satisfies the differential equation over a stated interval.

For example, a solution might be

$y=Ce^{3t},$

where $C$ is an arbitrary constant. Differentiating gives

$dydt=3Ce3t=3y,\frac{dy}{dt}=3Ce^{3t}=3y,$

so the function satisfies $dydt=3y\frac{dy}{dt}=3y$ .

A differential equation often has a whole family of solutions. An initial condition gives the value of the modelled quantity at a particular time and allows you to determine the arbitrary constant.

Definition

Initial condition

An initial condition is a known value such as $y (0) = 5$ , specifying the state of the system at a chosen starting time.

Example

Using an initial population

The population model above has the general solution

$P=Ae^{0.2t}.$

Initially, the population is 10 thousand.

“Initially” means $t = 0$ , so the initial condition is $P (0) = 10$ .
Substitute this into the general solution:

$10=Ae^{0.2(0)}=A.$

Therefore $A = 10$ .
The particular solution is

$P=10e^{0.2t}.$

At time $t$ , this gives the model’s predicted population in thousands.

Common Mistake

Ignoring the meaning of the constant

An arbitrary constant is not just an algebraic extra. Once an initial condition is used, it records information about the system’s starting state.

Interpreting the shape of a solution

Once you have a solution, ask what it says about:

the initial value;
whether the quantity increases or decreases;
how quickly it changes;
any maximum, minimum or limiting value;
its behaviour as time becomes large.

Suppose a velocity is modelled by

$v=20(1−e−0.4t).v=20\left(1-e^{-0.4t}\right).$

At $t = 0$ , the velocity is zero. As $t$ increases, $e^{-0.4t}$ decreases towards zero, so $v$ increases towards $s−120\text{ m s}^{-1}$ .

Velocity increasing towards a limiting value of 20 metres per second

The line $v = 20$ is a horizontal asymptote: the graph approaches this line as $t→∞t\to\infty$ .

Definition

Limiting value

A limiting value is the value approached by a modelled quantity as the independent variable tends towards a specified value, often as $t→∞t\to\infty$ .

Example

Interpreting a limiting velocity

A particle’s velocity is

$s−1,v=20\left(1-e^{-0.4t}\right)\text{ m s}^{-1},$

where $t≥0t\geq 0$ .

At the start,

$v (0) = 20 (1 - 1) = 0,$

so the particle starts from rest.
Differentiate to find the acceleration:

$a=dvdt=8e−0.4t.a=\frac{dv}{dt}=8e^{-0.4t}.$

Since this is positive for $t≥0t\geq 0$ , the velocity is always increasing. However, the exponential factor decreases, so the acceleration becomes smaller.
As $t→∞t\to\infty$ , $e−0.4t→0e^{-0.4t}\to 0$ , giving

$v→20.v\to 20.$

The model predicts that the particle approaches a limiting velocity of $s−120\text{ m s}^{-1}$ .
The velocity never equals 20 at a finite time in this model. It becomes arbitrarily close to 20 as time increases.

Tip

Check the endpoints

Evaluate the solution at the initial time and examine its limit as $t→∞t\to\infty$ . These two checks often reveal most of the model’s physical meaning.

Equilibrium solutions

An equilibrium value is a value at which the modelled quantity has zero rate of change.

For an autonomous differential equation

$dydt=f(y),\frac{dy}{dt}=f(y),$

equilibria are found by solving $f (y) = 0$ . If the quantity begins exactly at an equilibrium, the model predicts that it remains there.

Consider

$dPdt=0.1P(1−P500).\frac{dP}{dt}=0.1P\left(1-\frac{P}{500}\right).$

The equilibria are $P = 0$ and $P = 500$ .

Example

Predicting behaviour near an equilibrium

Interpret the model for a positive population $P$ .

For $0 < P < 500$ , both $P$ and $1−P5001-\frac{P}{500}$ are positive. Hence $dPdt>0\frac{dP}{dt}>0$ , so the population increases.
For $P > 500$ , the factor $1−P5001-\frac{P}{500}$ is negative. Hence $dPdt<0\frac{dP}{dt}<0$ , so the population decreases.
On either side of $P = 500$ , the population moves towards 500. Therefore 500 is a stable equilibrium and represents the model’s long-term limiting population.
The value $P = 0$ is also an equilibrium, but a small positive population moves away from zero. It is therefore an unstable equilibrium for positive populations.

Common Mistake

Confusing zero rate with zero quantity

The condition $dydt=0\frac{dy}{dt}=0$ means the quantity is not changing at that instant. It does not usually mean that $y = 0$ .

Links to kinematics

In one-dimensional motion, let $s$ be displacement, $v$ be velocity and $a$ be acceleration. These are connected by

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

A differential equation involving $v$ and $dvdt\frac{dv}{dt}$ therefore describes how the acceleration depends on the particle’s velocity or on time.

Remember that velocity includes direction. A negative velocity means motion in the negative direction, while a negative acceleration does not necessarily mean that the particle is slowing down.

Example

Interpreting velocity and acceleration

A particle has velocity

$s−1,v=6-10e^{-2t}\text{ m s}^{-1},$

for $t≥0t\geq 0$ .

Initially,

$s−1,v(0)=6-10=-4\text{ m s}^{-1},$

so the particle initially moves in the negative direction.
Its acceleration is

$s−2,a=\frac{dv}{dt}=20e^{-2t}\text{ m s}^{-2},$

which is positive. While $v < 0$ , velocity and acceleration have opposite signs, so the particle slows down.
The particle changes direction when $v = 0$ :

$6−10e−2t=0e−2t=35t=12ln⁡(53).\begin{aligned} 6-10e^{-2t}&=0\ e^{-2t}&=\frac{3}{5}\ t&=\frac{1}{2}\ln\left(\frac{5}{3}\right). \end{aligned}$
After this time, both velocity and acceleration are positive, so the particle speeds up in the positive direction. As $t→∞t\to\infty$ , $s−1v\to 6\text{ m s}^{-1}$ and $s−2a\to 0\text{ m s}^{-2}$ .

Common Mistake

Assuming positive acceleration means speeding up

A particle speeds up when velocity and acceleration have the same sign. It slows down when they have opposite signs.

Limitations of a solution

A differential equation is a mathematical model, meaning a simplified mathematical description of a real situation. Its solution is only as reliable as the assumptions used to create the model.

Possible limitations include:

parameters such as resistance or growth rate being treated as constant;
external influences being ignored;
the model only applying over a stated time interval;
impossible predictions, such as a negative population;
measurements or initial conditions being approximate;
a continuous model being used for a discrete quantity such as the number of people;
extreme long-term behaviour being unrealistic.

Example

Evaluating a cooling model

The temperature of a drink is modelled by

$T=20+70e^{-0.08t},$

where $T$ is in degrees Celsius and $t$ is in minutes.

The initial temperature is $T (0) = 90$ , and the limiting temperature is 20 degrees Celsius. The model therefore treats the surrounding temperature as a constant 20 degrees Celsius.
In reality, the room temperature may vary, and evaporation or a change of container could alter the cooling rate.
The model predicts that the drink approaches 20 degrees Celsius but never reaches it exactly. For practical purposes, however, the difference may eventually be too small to measure.
The equation should not be used before the chosen starting time, or after the drink is moved to different surroundings, because the assumptions would no longer match the situation.

Common Mistake

Mathematical validity is not physical validity

A formula may be defined for many values of $t$ , but the context may restrict its valid domain. Always apply physical constraints such as $t≥0t\geq 0$ and non-negative population.

Exam technique

In the exam

Identify what the variable and its derivative represent, including units where appropriate.
Use the initial condition to interpret or determine the constant in the solution.
Check signs, stationary values and the limit as $t→∞t\to\infty$ , then state what each result means in context.
In kinematics, compare the signs of velocity and acceleration before deciding whether the particle speeds up or slows down.
Give a specific limitation linked to an assumption of the model, rather than only saying that the model is “not realistic”.

Self review

Check yourself

What does $dPdt<0\frac{dP}{dt}<0$ tell you about a population $P$ ?
How would you identify an equilibrium value from $dydt=f(y)\frac{dy}{dt}=f(y)$ ?
If a particle has negative velocity and positive acceleration, what is happening to its speed?

What you'll learn

How to translate a differential equation and its solution into the language of a real problem.
How to interpret gradients, constants, limiting values and long-term behaviour.
How differential equations connect displacement, velocity and acceleration in kinematics.
How to identify assumptions and limitations in a mathematical model.

Prerequisites: rates of change

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

If a quantity $y$ depends on time $t$ , then $dydt\frac{dy}{dt}$ is its rate of change at time $t$ . On a graph of $y$ against $t$ , this derivative is the gradient of the curve.

If $dydt>0\frac{dy}{dt}>0$ , then $y$ is increasing.
If $dydt<0\frac{dy}{dt}<0$ , then $y$ is decreasing.
If $dydt=0\frac{dy}{dt}=0$ , then the graph has a stationary point or the quantity is at an equilibrium value.

Definition

Differential equation

A differential equation is an equation containing a derivative, such as $dydt\frac{dy}{dt}$ , together with the quantity being modelled.

For example,

$dydt=3y\frac{dy}{dt}=3y$

says that the rate of change of $y$ is three times its current value. It describes a relationship between a quantity and the way that quantity changes.

Key Idea

Read the derivative in words

Before doing any algebra, translate $dydt=f(y,t)\frac{dy}{dt}=f(y,t)$ as “the rate of change of $y$ is given by $f (y, t)$ ”.

Example

Interpreting a population model

A population $P$ , measured in thousands, is modelled by

$dPdt=0.2P,\frac{dP}{dt}=0.2P,$

where $t$ is measured in years.

Since $P$ is a population and $t$ is time, $dPdt\frac{dP}{dt}$ represents the population’s instantaneous rate of change, in thousands per year.
For any positive population, $0.2 P > 0$ . Therefore $dPdt>0\frac{dP}{dt}>0$ , so the model predicts that the population increases.
The rate is proportional to $P$ . A larger population therefore produces a larger absolute increase per year; for example, when $P = 10$ , the rate is $2$ thousand people per year.

What a solution represents

A solution of a differential equation is a function that satisfies the differential equation over a stated interval.

For example, a solution might be

$y=Ce^{3t},$

where $C$ is an arbitrary constant. Differentiating gives

$dydt=3Ce3t=3y,\frac{dy}{dt}=3Ce^{3t}=3y,$

so the function satisfies $dydt=3y\frac{dy}{dt}=3y$ .

Definition

Initial condition

An initial condition is a known value such as $y (0) = 5$ , specifying the state of the system at a chosen starting time.

Example

Using an initial population

The population model above has the general solution

$P=Ae^{0.2t}.$

Initially, the population is 10 thousand.

“Initially” means $t = 0$ , so the initial condition is $P (0) = 10$ .
Substitute this into the general solution:

$10=Ae^{0.2(0)}=A.$

Therefore $A = 10$ .
The particular solution is

$P=10e^{0.2t}.$

At time $t$ , this gives the model’s predicted population in thousands.

Common Mistake

Ignoring the meaning of the constant

An arbitrary constant is not just an algebraic extra. Once an initial condition is used, it records information about the system’s starting state.

Interpreting the shape of a solution

Once you have a solution, ask what it says about:

the initial value;
whether the quantity increases or decreases;
how quickly it changes;
any maximum, minimum or limiting value;
its behaviour as time becomes large.

Suppose a velocity is modelled by

$v=20(1−e−0.4t).v=20\left(1-e^{-0.4t}\right).$

At $t = 0$ , the velocity is zero. As $t$ increases, $e^{-0.4t}$ decreases towards zero, so $v$ increases towards $s−120\text{ m s}^{-1}$ .

Velocity increasing towards a limiting value of 20 metres per second

The line $v = 20$ is a horizontal asymptote: the graph approaches this line as $t→∞t\to\infty$ .

Definition

Limiting value

A limiting value is the value approached by a modelled quantity as the independent variable tends towards a specified value, often as $t→∞t\to\infty$ .

Example

Interpreting a limiting velocity

A particle’s velocity is

$s−1,v=20\left(1-e^{-0.4t}\right)\text{ m s}^{-1},$

where $t≥0t\geq 0$ .

At the start,

$v (0) = 20 (1 - 1) = 0,$

so the particle starts from rest.
Differentiate to find the acceleration:

$a=dvdt=8e−0.4t.a=\frac{dv}{dt}=8e^{-0.4t}.$

Since this is positive for $t≥0t\geq 0$ , the velocity is always increasing. However, the exponential factor decreases, so the acceleration becomes smaller.
As $t→∞t\to\infty$ , $e−0.4t→0e^{-0.4t}\to 0$ , giving

$v→20.v\to 20.$

The model predicts that the particle approaches a limiting velocity of $s−120\text{ m s}^{-1}$ .
The velocity never equals 20 at a finite time in this model. It becomes arbitrarily close to 20 as time increases.

Tip

Check the endpoints

Evaluate the solution at the initial time and examine its limit as $t→∞t\to\infty$ . These two checks often reveal most of the model’s physical meaning.

Equilibrium solutions

An equilibrium value is a value at which the modelled quantity has zero rate of change.

For an autonomous differential equation

$dydt=f(y),\frac{dy}{dt}=f(y),$

equilibria are found by solving $f (y) = 0$ . If the quantity begins exactly at an equilibrium, the model predicts that it remains there.

Consider

$dPdt=0.1P(1−P500).\frac{dP}{dt}=0.1P\left(1-\frac{P}{500}\right).$

The equilibria are $P = 0$ and $P = 500$ .

Example

Predicting behaviour near an equilibrium

Interpret the model for a positive population $P$ .

For $0 < P < 500$ , both $P$ and $1−P5001-\frac{P}{500}$ are positive. Hence $dPdt>0\frac{dP}{dt}>0$ , so the population increases.
For $P > 500$ , the factor $1−P5001-\frac{P}{500}$ is negative. Hence $dPdt<0\frac{dP}{dt}<0$ , so the population decreases.
On either side of $P = 500$ , the population moves towards 500. Therefore 500 is a stable equilibrium and represents the model’s long-term limiting population.
The value $P = 0$ is also an equilibrium, but a small positive population moves away from zero. It is therefore an unstable equilibrium for positive populations.

Common Mistake

Confusing zero rate with zero quantity

The condition $dydt=0\frac{dy}{dt}=0$ means the quantity is not changing at that instant. It does not usually mean that $y = 0$ .

Links to kinematics

In one-dimensional motion, let $s$ be displacement, $v$ be velocity and $a$ be acceleration. These are connected by

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

A differential equation involving $v$ and $dvdt\frac{dv}{dt}$ therefore describes how the acceleration depends on the particle’s velocity or on time.

Remember that velocity includes direction. A negative velocity means motion in the negative direction, while a negative acceleration does not necessarily mean that the particle is slowing down.

Example

Interpreting velocity and acceleration

A particle has velocity

$s−1,v=6-10e^{-2t}\text{ m s}^{-1},$

for $t≥0t\geq 0$ .

Initially,

$s−1,v(0)=6-10=-4\text{ m s}^{-1},$

so the particle initially moves in the negative direction.
Its acceleration is

$s−2,a=\frac{dv}{dt}=20e^{-2t}\text{ m s}^{-2},$

which is positive. While $v < 0$ , velocity and acceleration have opposite signs, so the particle slows down.
The particle changes direction when $v = 0$ :

$6−10e−2t=0e−2t=35t=12ln⁡(53).\begin{aligned} 6-10e^{-2t}&=0\ e^{-2t}&=\frac{3}{5}\ t&=\frac{1}{2}\ln\left(\frac{5}{3}\right). \end{aligned}$
After this time, both velocity and acceleration are positive, so the particle speeds up in the positive direction. As $t→∞t\to\infty$ , $s−1v\to 6\text{ m s}^{-1}$ and $s−2a\to 0\text{ m s}^{-2}$ .

Common Mistake

Assuming positive acceleration means speeding up

A particle speeds up when velocity and acceleration have the same sign. It slows down when they have opposite signs.

Limitations of a solution

A differential equation is a mathematical model, meaning a simplified mathematical description of a real situation. Its solution is only as reliable as the assumptions used to create the model.

Possible limitations include:

parameters such as resistance or growth rate being treated as constant;
external influences being ignored;
the model only applying over a stated time interval;
impossible predictions, such as a negative population;
measurements or initial conditions being approximate;
a continuous model being used for a discrete quantity such as the number of people;
extreme long-term behaviour being unrealistic.

Example

Evaluating a cooling model

The temperature of a drink is modelled by

$T=20+70e^{-0.08t},$

where $T$ is in degrees Celsius and $t$ is in minutes.

The initial temperature is $T (0) = 90$ , and the limiting temperature is 20 degrees Celsius. The model therefore treats the surrounding temperature as a constant 20 degrees Celsius.
In reality, the room temperature may vary, and evaporation or a change of container could alter the cooling rate.
The model predicts that the drink approaches 20 degrees Celsius but never reaches it exactly. For practical purposes, however, the difference may eventually be too small to measure.
The equation should not be used before the chosen starting time, or after the drink is moved to different surroundings, because the assumptions would no longer match the situation.

Common Mistake

Mathematical validity is not physical validity

A formula may be defined for many values of $t$ , but the context may restrict its valid domain. Always apply physical constraints such as $t≥0t\geq 0$ and non-negative population.

Exam technique

In the exam

Identify what the variable and its derivative represent, including units where appropriate.
Use the initial condition to interpret or determine the constant in the solution.
Check signs, stationary values and the limit as $t→∞t\to\infty$ , then state what each result means in context.
In kinematics, compare the signs of velocity and acceleration before deciding whether the particle speeds up or slows down.
Give a specific limitation linked to an assumption of the model, rather than only saying that the model is “not realistic”.

Self review

Check yourself

What does $dPdt<0\frac{dP}{dt}<0$ tell you about a population $P$ ?
How would you identify an equilibrium value from $dydt=f(y)\frac{dy}{dt}=f(y)$ ?
If a particle has negative velocity and positive acceleration, what is happening to its speed?

What you'll learn

Prerequisites: rates of change

What a solution represents

Interpreting the shape of a solution

Equilibrium solutions

Links to kinematics

Limitations of a solution

Review 1.11.8 Interpreting solutions of differential equations (A-level only) by teaching Genie

1.11.8 Interpreting solutions of differential equations (A-level only)

What you'll learn

Prerequisites: rates of change

What a solution represents

Interpreting the shape of a solution

Equilibrium solutions

Links to kinematics

Limitations of a solution

Review 1.11.8 Interpreting solutions of differential equations (A-level only) by teaching Genie

1.11.8 Interpreting solutions of differential equations (A-level only)

What you'll learn

Prerequisites: rates of change

What a solution represents

Interpreting the shape of a solution

Equilibrium solutions

Links to kinematics

Limitations of a solution

Review 1.11.8 Interpreting solutions of differential equations (A-level only) by teaching Genie

1.11.8 Interpreting solutions of differential equations (A-level only) Revision Guide

1.11.8 Interpreting solutions of differential equations (A-level only)

What you'll learn

Prerequisites: rates of change

What a solution represents

Interpreting the shape of a solution

Equilibrium solutions

Links to kinematics

Limitations of a solution

Review 1.11.8 Interpreting solutions of differential equations (A-level only) by teaching Genie

1.11.8 Interpreting solutions of differential equations (A-level only) Revision Guide