Constructing differential equation models

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A differential equation connects an unknown quantity to one or more of its rates of change. At A level, constructing a model usually means translating a sentence such as “the population grows at a rate proportional to its size” into an equation such as

$$\frac{dP}{dt}=kP.$$

This lesson is about building and interpreting such equations. Methods for finding their solutions are covered in separable differential equations.

Prerequisites

You should be able to:

• interpret $\dfrac{dy}{dx}$ as the rate at which $y$ changes with $x$;
• distinguish positive, negative and zero gradients;
• rearrange formulae;
• use units to check an equation;
• understand direct proportion.

The central translation

If $Q$ depends on time $t$, then

$$\frac{dQ}{dt}=\text{instantaneous rate of change of }Q.$$

The words after “rate” determine the right-hand side.

Statement	Differential equation
$Q$ increases at a constant rate $a$	$\dfrac{dQ}{dt}=a$
$Q$ decreases at a constant rate $a$	$\dfrac{dQ}{dt}=-a$
The rate of increase of $Q$ is proportional to $Q$	$\dfrac{dQ}{dt}=kQ$, $k>0$
The rate of decrease of $Q$ is proportional to $Q$	$\dfrac{dQ}{dt}=-kQ$, $k>0$
The rate of change of $Q$ is proportional to $M-Q$	$\dfrac{dQ}{dt}=k(M-Q)$
The rate of change of $Q$ is proportional to $Q(M-Q)$	$\dfrac{dQ}{dt}=kQ(M-Q)$

The constant of proportionality is not optional. “Proportional to $Q$” means “equal to $kQ$ for some constant $k$”.

1. Define every variable, including its units. 2. Identify the quantity whose rate is described. 3. Write its derivative on the left. 4. Translate the dependence into an expression on the right. 5. choose the sign from the direction of change. 6. Add any given initial condition. 7. Check units, signs and limiting behaviour.

Example 1: constant filling rate

Water enters a tank at $12$ litres per minute. Let $V$ be the volume of water, in litres, after $t$ minutes.

The described rate is the rate of change of volume with time, so

$$\frac{dV}{dt}=12.$$

If the tank initially contains $35$ litres, the complete initial value model is

$$\frac{dV}{dt}=12, \qquad V(0)=35.$$

The units agree:

$$\left[\frac{dV}{dt}\right] =\frac{\text{litres}}{\text{minute}} =[12].$$

Notice that $V(0)=35$ is not another differential equation. It selects the particular situation that begins with $35$ litres.

Proportional growth and decay

Suppose the rate of change of $P$ is proportional to $P$. Direct proportion gives

$$\frac{dP}{dt}=kP.$$

If $P$ is increasing and $P>0$, then $k>0$. For decay, either write $k<0$, or more clearly write

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

Both conventions are valid, but the meaning of $k$ must be stated.

Example 2: bacterial population

A bacterial population grows at a rate proportional to its current population. Initially there are $800$ bacteria. After two hours, the instantaneous growth rate is $240$ bacteria per hour.

Let $P(t)$ be the number of bacteria after $t$ hours. The verbal statement gives

$$\frac{dP}{dt}=kP, \qquad P(0)=800.$$

At $t=2$, the extra information says

$$P'(2)=240.$$

It does not yet determine $k$, because the value of $P(2)$ is unknown. If the question also stated that $P(2)=1200$, then

$$240=k(1200) \quad\Longrightarrow\quad k=0.2\text{ hour}^{-1}.$$

Thus the model would be

$$\frac{dP}{dt}=0.2P, \qquad P(0)=800.$$

The unit of $k$ follows from dimensional consistency:

$$\frac{\text{bacteria}}{\text{hour}} =[k]\times\text{bacteria} \quad\Longrightarrow\quad [k]=\text{hour}^{-1}.$$

Example 3: radioactive decay

The mass $m$ of a radioactive substance decreases at a rate proportional to the mass remaining. At time $t=0$, the mass is $50$ mg.

Using $\lambda>0$ for the decay constant,

$$\frac{dm}{dt}=-\lambda m, \qquad m(0)=50.$$

The negative sign matters. When $m>0$, the equation guarantees that $\dfrac{dm}{dt}<0$, so the mass decreases.

Growth towards a limiting value

Not all growth remains proportional to the amount already present. Sometimes the rate is controlled by the gap between the current value and a limiting value.

If $T$ approaches a surrounding temperature $A$, a simple model is

$$\frac{dT}{dt}=k(A-T), \qquad k>0.$$

• If $T<A$, then $A-T>0$, so $T$ increases.
• If $T>A$, then $A-T<0$, so $T$ decreases.
• If $T=A$, then $\dfrac{dT}{dt}=0$, so there is no further change.

This sign analysis is a powerful way to test a proposed model without solving it.

Example 4: charging a battery

A simplified battery model assumes that the rate of increase of charge $Q$ is proportional to the unused capacity. The maximum charge is $100$ units and the initial charge is $20$ units.

Unused capacity is $100-Q$, so

$$\frac{dQ}{dt}=k(100-Q), \qquad Q(0)=20, \qquad k>0.$$

A common incorrect model is $\dfrac{dQ}{dt}=kQ$. That predicts faster charging as the battery becomes fuller. The stated context says the opposite: charging slows because $100-Q$ becomes smaller.

Population models with limited resources

Unrestricted proportional growth, $P'=kP$, assumes that resources never become scarce. A simple model with a carrying capacity $K$ is

$$\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right), \qquad r>0.$$

Equivalently,

$$\frac{dP}{dt}=\frac{r}{K}P(K-P).$$

The factors have clear roles:

• $P$ makes growth slow when the population is very small;
• $K-P$ represents the remaining capacity;
• $P'=0$ at $P=0$ and $P=K$;
• $P'>0$ for $0<P<K$;
• $P'<0$ for $P>K$.

Example 5: fish population

A lake supports at most $5000$ fish. A model assumes that the growth rate is proportional both to the current population and to the unused capacity. Initially there are $1200$ fish.

Let $P$ be the population after $t$ years. Then

$$\frac{dP}{dt}=kP(5000-P), \qquad P(0)=1200, \qquad k>0.$$

If $P=4500$, then $P'>0$, but the factor $5000-P$ is relatively small. The population is still growing, but resource pressure has slowed its growth.

Kinematics as differential equations

In one-dimensional motion,

$$v=\frac{dx}{dt}, \qquad a=\frac{dv}{dt}=\frac{d^2x}{dt^2}.$$

A rule for acceleration is therefore a differential equation for velocity, position, or both.

Example 6: acceleration depending on velocity

A cyclist’s acceleration is modelled as proportional to the difference between a limiting speed of $12\text{ m s}^{-1}$ and the current speed $v$. The cyclist starts from rest.

The difference is $12-v$, so

$$\frac{dv}{dt}=k(12-v), \qquad v(0)=0, \qquad k>0.$$

Check the behaviour:

• at $v=0$, acceleration is positive;
• as $v$ approaches $12$, acceleration approaches $0$;
• if the model were used with $v>12$, acceleration would be negative.

The model therefore makes $12\text{ m s}^{-1}$ an equilibrium speed.

Example 7: resistance proportional to velocity

A particle of mass $m$ falls vertically. Take downward as positive. Its weight is $mg$ downward, while resistance of magnitude $kv$ acts upward when $v>0$.

Newton’s second law gives

$$m\frac{dv}{dt}=mg-kv.$$

Hence

$$\frac{dv}{dt}=g-\frac{k}{m}v.$$

At terminal speed, acceleration is zero:

$$0=g-\frac{k}{m}v \quad\Longrightarrow\quad v=\frac{mg}{k}.$$

This conclusion comes from interpreting the differential equation. No full solution is needed.

If upward were chosen as positive, both the velocity convention and the force signs would change. A consistent alternative model is correct. Mixing conventions halfway through is not.

Price-demand and market adjustment

Differential equations can model how a price changes in response to excess demand. Let

• $p(t)$ be price;
• $D(p)$ be demand at price $p$;
• $S(p)$ be supply at price $p$.

A basic adjustment assumption is

$$\frac{dp}{dt}=k\bigl(D(p)-S(p)\bigr), \qquad k>0.$$

If demand exceeds supply, then $p'>0$ and price rises. If supply exceeds demand, then $p'<0$ and price falls.

Example 8: linear supply and demand

Suppose

$$D(p)=140-4p, \qquad S(p)=20+2p,$$

and price changes at a rate proportional to excess demand. Then

$$\frac{dp}{dt} =k\bigl((140-4p)-(20+2p)\bigr) =k(120-6p).$$

The equilibrium price occurs when $p'=0$:

$$120-6p=0 \quad\Longrightarrow\quad p=20.$$

For $p<20$, $p'>0$. For $p>20$, $p'<0$. The model therefore predicts adjustment towards a price of $20$.

Initial conditions and parameters

A differential equation describes a family of possible histories. An initial condition provides the value of the dependent variable at a specified starting point:

$$y(0)=y_0.$$

Do not confuse these three kinds of information:

$$\underbrace{y'=ky}_{\text{model structure}}, \qquad \underbrace{y(0)=200}_{\text{initial condition}}, \qquad \underbrace{k=0.04}_{\text{parameter value}}.$$

Parameters such as $k$, $r$ and $K$ describe how the system behaves. Variables such as $t$, $P$ and $v$ change within the model.

State the assumptions

A mathematical model is a deliberate simplification. Strong answers make its assumptions visible. Depending on context, assumptions might include:

• time is treated as continuous;
• a population is large enough to be represented by a continuous variable;
• environmental conditions remain unchanged;
• the proportionality constant does not vary with time;
• no migration, harvesting or sudden external event occurs;
• forces included in the equation are the only significant forces;
• supply and demand respond instantly to price;
• measurements and initial values are exact enough for the purpose.

An equation can be manipulated perfectly and still be a poor model if its assumptions are unreasonable.

Testing a model before solving it

For an equation of the form

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

ask the following questions.

1. Does the sign make sense?

If the context says $y$ is increasing, the model should give $f(y)>0$ in the relevant range.

2. Where are the equilibria?

Equilibrium values satisfy

$$f(y)=0.$$

At such a value, the model predicts no instantaneous change.

3. What happens near a boundary?

If $y$ represents mass or population, can the model predict $y<0$? If it does, the model may only be valid for a restricted interval.

4. Are the units consistent?

Every term added or subtracted must have the same units. In

$$\frac{dP}{dt}=rP\left(1-\frac{P}{K}\right),$$

$P/K$ is dimensionless, so $r$ must have units of inverse time.

Common misconceptions

Treating proportional as equal

Incorrect:

$$\frac{dP}{dt}=P.$$

Correct general model:

$$\frac{dP}{dt}=kP.$$

The incorrect equation silently fixes $k=1$ in the chosen units.

Attaching the derivative to the wrong variable

If population changes with time, write $\dfrac{dP}{dt}$, not $\dfrac{dt}{dP}$ and not $\dfrac{dP}{dx}$.

Losing the direction of change

“Decreases at a rate proportional to $m$” requires either

$$\frac{dm}{dt}=-km, \qquad k>0,$$

or $m'=km$ with an explicit statement that $k<0$.

Using a total change as an instantaneous rate

$P(3)-P(0)$ is a change over three time units. $P'(3)$ is the instantaneous rate at $t=3$. They are different quantities and usually have different units.

Solving when only a model is requested

If a question asks you to formulate a differential equation, stop once the variables, equation, conditions and assumptions have been stated. Solving an unnecessary equation wastes time and can introduce errors.

Self-check

Question 1

The rate at which a medicine leaves the bloodstream is proportional to the amount $M$ present. Initially, $80$ mg is present. Construct a model using a positive constant $k$.

Answer

Because the amount decreases,

$$\frac{dM}{dt}=-kM, \qquad M(0)=80, \qquad k>0.$$

Question 2

A rumour spreads through a school of $900$ students. The rate of spread is proportional both to the number $R$ who know the rumour and the number who do not. Initially $15$ students know it. Construct a model.

Answer

The number who do not know the rumour is $900-R$. Therefore

$$\frac{dR}{dt}=kR(900-R), \qquad R(0)=15, \qquad k>0.$$

Question 3

A car’s acceleration is $0.3$ times the difference between its current speed $v$ and a target speed of $25\text{ m s}^{-1}$. The car begins at $10\text{ m s}^{-1}$. Correct the wording if necessary, then give the physically sensible model.

Answer

For the car to accelerate towards the target from below, the relevant difference must be target minus current speed. Thus

$$\frac{dv}{dt}=0.3(25-v), \qquad v(0)=10.$$

Writing $0.3(v-25)$ would make the initial acceleration negative and move the car away from the target.

Question 4

Demand and supply are $D(p)=90-2p$ and $S(p)=15+p$. Price changes at twice the excess demand per week. Construct the equation and find the equilibrium price without solving the differential equation.

Answer

$$\frac{dp}{dt} =2(D-S) =2\bigl((90-2p)-(15+p)\bigr) =150-6p.$$

At equilibrium, $p'=0$, so

$$150-6p=0 \quad\Longrightarrow\quad p=25.$$

What to learn next

• Learn how to solve first-order models in separable differential equations.
• Learn how to extract information from a solution in interpreting differential equation models.
• Review physical rate relationships in connected rates of change.

The essential skill is translation: define the quantities, identify what controls the rate, select the sign, add the initial condition, then test whether the model behaves sensibly.