Interpreting differential equation models

A differential equation describes a quantity through its rate of change. In a population model, for example, the equation does not state the population directly. It states how quickly the population changes. Solving the equation recovers possible population functions; an initial condition selects the one relevant to the situation.

This lesson focuses on interpreting models. For the integration technique itself, first study separable differential equations.

Prerequisites

You should be able to:

• interpret $\dfrac{dy}{dx}$ as an instantaneous rate of change
• separate variables and integrate
• use $\ln$ and exponential functions
• apply an initial condition to find a constant
• sketch exponential graphs and solve exponential equations

Review integration basics, growth and decay and A-level calculus and differentiation if needed.

From words to a differential equation

Let $Q(t)$ be a quantity depending on time. Translate the wording one piece at a time.

Words	Mathematical statement
the rate of change of $Q$	$\dfrac{dQ}{dt}$
proportional to $Q$	$\dfrac{dQ}{dt}=kQ$
proportional to the square of $Q$	$\dfrac{dQ}{dt}=kQ^2$
proportional to $A-Q$	$\dfrac{dQ}{dt}=k(A-Q)$
decreases at a constant rate $r$	$\dfrac{dQ}{dt}=-r$

The constant of proportionality carries the sign and the units required by the model. If $Q>0$ and $\dfrac{dQ}{dt}=kQ$, then $k>0$ gives growth and $k<0$ gives decay.

Worked example 1: form the equation before solving

A culture contains $P(t)$ bacteria, where $t$ is measured in hours. Its rate of growth is proportional to the number present.

“Rate of growth” means $\dfrac{dP}{dt}$ and “proportional to the number present” means $kP$. Therefore

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

If the culture were dying at a rate proportional to its population, one could write either

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

or $\dfrac{dP}{dt}=kP$ with $k<0$. Both conventions work, but the meaning of $k$ must be stated consistently.

Misconception: confusing an amount with its rate

“The population is proportional to time” would mean $P=kt$. It does not mean $\dfrac{dP}{dt}=kP$. Always identify the quantity on the left before translating the rest of the sentence.

A solution family and a particular solution

Solving

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

gives a family of solutions. Separating variables,

$$\frac1P\frac{dP}{dt}=k,$$

so

$$\int\frac1P\,dP=\int k\,dt.$$

Hence

$$\ln|P|=kt+C.$$

Exponentiating gives

$$|P|=e^Ce^{kt}.$$

The arbitrary positive factor $e^C$ can be absorbed into a new constant. Allowing that constant to have either sign gives

$$P=Ae^{kt}.$$

For a population $P>0$, so $A>0$. At $t=0$,

$$P(0)=Ae^0=A.$$

Thus $A$ is the initial population. The differential equation determines the shape of the family; the initial condition determines which member of the family models the culture.

Worked example 2: use data and interpret the constants

A population satisfies

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

Initially there are $800$ organisms, and after $3$ hours there are $1200$. Find the model.

The general positive solution is

$$P=Ae^{kt}.$$

Using $P(0)=800$ gives $A=800$. Now use $P(3)=1200$:

$$1200=800e^{3k}.$$

Therefore

$$e^{3k}=\frac32, \qquad 3k=\ln\frac32, \qquad k=\frac13\ln\frac32.$$

The exact model is

$$\boxed{P=800e^{(\ln(3/2))t/3}}.$$

Here $800$ is the population at $t=0$. The parameter

$$k=\frac13\ln\frac32\approx0.135\text{ h}^{-1}$$

is the continuous proportional growth constant. Its unit is per hour because the exponent $kt$ must have no units.

It is tempting to call $k$ a $13.5\%$ hourly increase. That is only an approximation. The exact multiplication factor over one hour is

$$e^k=\left(\frac32\right)^{1/3}\approx1.1447,$$

so the population increases by about $14.5\%$ in one hour under this model.

Read information directly from the equation

Before solving, inspect the sign and zeros of the rate.

• If $\dfrac{dy}{dt}>0$, then $y$ is increasing.
• If $\dfrac{dy}{dt}<0$, then $y$ is decreasing.
• If $\dfrac{dy}{dt}=0$, the model has an equilibrium value, provided a constant solution at that value is valid.
• A large $\left|\dfrac{dy}{dt}\right|$ means a steep solution curve, not necessarily a large value of $y$.

Worked example 3: approach to an equilibrium

The temperature $T$ in degrees Celsius of an object is modelled by

$$\frac{dT}{dt}=-0.2(T-20),$$

where $t$ is in minutes and $T(0)=80$.

The rate is zero when $T=20$, so $20^\circ\text{C}$ is an equilibrium temperature. If $T>20$, then $T-20>0$ and $\dfrac{dT}{dt}<0$, so the object cools. If $T<20$, then $\dfrac{dT}{dt}>0$, so it warms. In both cases the rate points towards $20$.

Now solve. Write $u=T-20$. Since $\dfrac{du}{dt}=\dfrac{dT}{dt}$,

$$\frac{du}{dt}=-0.2u.$$

Therefore

$$u=Ae^{-0.2t},$$

and hence

$$T=20+Ae^{-0.2t}.$$

Using $T(0)=80$ gives $80=20+A$, so $A=60$. Thus

$$\boxed{T=20+60e^{-0.2t}}.$$

The parts of the answer have distinct meanings:

• $20$ is the surrounding temperature and the long term equilibrium
• $60=80-20$ is the initial temperature difference
• $0.2\text{ min}^{-1}$ controls how rapidly that difference decays

As $t\to\infty$, $e^{-0.2t}\to0$, so $T\to20$. The model predicts that $T$ approaches $20^\circ\text{C}$ but does not equal it at any finite time, because $60e^{-0.2t}>0$.

Self-check 1

For the model

$$\frac{dN}{dt}=0.4N\left(1-\frac{N}{500}\right),$$

find the equilibrium values and determine whether $N$ increases or decreases when (a) $0<N<500$ and (b) $N>500$.

Answer

The rate is zero when

$$N=0\quad\text{or}\quad1-\frac{N}{500}=0,$$

so the equilibria are $N=0$ and $N=500$.

For $0<N<500$, both factors are positive, so $\dfrac{dN}{dt}>0$ and $N$ increases. For $N>500$, the second factor is negative, so $\dfrac{dN}{dt}<0$ and $N$ decreases. The rate therefore points towards $N=500$ from either side.

You can extract all of this without solving the differential equation.

Initial conditions are not optional decoration

An initial condition usually has the form

$$y=y_0\quad\text{when}\quad x=x_0,$$

or equivalently $y(x_0)=y_0$. Substitute both coordinates into the general solution.

Worked example 4: a reciprocal model and its valid domain

The mass $M$ of a crystal satisfies

$$\frac{dM}{dt}=0.02M^2, \qquad M(0)=10,$$

where $M$ is in grams and $t$ is in days.

Separate and integrate:

$$M^{-2}\,dM=0.02\,dt,$$ $$-M^{-1}=0.02t+C.$$

It is convenient to rearrange before using the condition:

$$\frac1M=C_1-0.02t.$$

Since $M(0)=10$,

$$\frac1{10}=C_1.$$

Therefore

$$\boxed{M=\frac1{0.1-0.02t}=\frac{50}{5-t}}.$$

The algebraic expression is defined for $t\ne5$, but the physical solution beginning at $t=0$ is meaningful only for

$$0\leq t<5.$$

The model predicts $M\to\infty$ as $t\to5^-$, which is evidence that the model must fail before day $5$. It does not predict a physically meaningful negative mass after day $5$.

This is an important distinction: the domain of a formula and the domain in which a model is credible need not be the same.

Check a proposed solution

To verify a model solution, check both requirements:

1. Differentiate it and substitute into the differential equation.
2. Substitute the initial condition.

Worked example 5: verification catches a sign error

Claim: $y=3+2e^{-4t}$ solves

$$\frac{dy}{dt}=4(3-y), \qquad y(0)=5.$$

Differentiate:

$$\frac{dy}{dt}=-8e^{-4t}.$$

The right hand side is

$$4(3-y)=4(3-3-2e^{-4t})=-8e^{-4t}.$$

The two sides agree. Also,

$$y(0)=3+2=5.$$

Therefore the claim is correct. Notice that checking only $y(0)=5$ would not establish that the function satisfies the differential equation.

Self-check 2

Does $Q=12e^{-0.3t}$ satisfy

$$\frac{dQ}{dt}=-0.3Q, \qquad Q(0)=10?$$

Answer

Differentiation gives

$$\frac{dQ}{dt}=-3.6e^{-0.3t}=-0.3Q,$$

so the differential equation is satisfied. However,

$$Q(0)=12\ne10.$$

It is a member of the correct solution family, but it is not the particular solution satisfying the stated initial condition. That solution is $Q=10e^{-0.3t}$.

Units expose modelling errors

The units on both sides of a differential equation must agree. If $V$ is measured in litres and $t$ in seconds, then

$$\frac{dV}{dt}$$

has units $\text{L s}^{-1}$.

In

$$\frac{dV}{dt}=kV,$$

$k$ must have units $\text{s}^{-1}$. In

$$\frac{dV}{dt}=kV^2,$$

$k$ must have units

$$\frac{\text{L s}^{-1}}{\text{L}^2}=\text{L}^{-1}\text{s}^{-1}.$$

The exponent in an exponential must be dimensionless. Thus $e^{-0.4t}$ is meaningful only if $0.4t$ has no units. When $t$ is in minutes, $0.4$ has units $\text{min}^{-1}$.

Self-check 3

The radius $r$ of a drop is measured in millimetres and time $t$ in seconds. In the model

$$\frac{dr}{dt}=-kr^2,$$

what are the units of $k$?

Answer

The left side has units $\text{mm s}^{-1}$, so

$$[k]=\frac{\text{mm s}^{-1}}{\text{mm}^2} =\boxed{\text{mm}^{-1}\text{s}^{-1}}.$$

Interpret predictions with care

A model answer should name the quantity, include units, and respect the model’s domain.

For the cooling model

$$T=20+60e^{-0.2t},$$

the time at which the object reaches $30^\circ\text{C}$ satisfies

$$30=20+60e^{-0.2t}.$$

Therefore

$$e^{-0.2t}=\frac16, \qquad -0.2t=\ln\frac16=-\ln6,$$

so

$$t=5\ln6\approx8.96.$$

A complete conclusion is:

$$\boxed{\text{The model predicts that the object reaches }30^\circ\text{C after about }8.96\text{ minutes}.}$$

The phrase “the model predicts” matters. A mathematical model is a simplified description, not a guarantee.

Assumptions and limitations

The form of a differential equation encodes assumptions. For example,

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

with constant $k>0$ assumes that:

• every individual has the same average contribution to growth
• the proportional growth rate stays constant
• resources, space and other limiting effects do not change the rate
• migration or harvesting is absent, or already included in the net rate
• treating a population as continuous is adequate

Its solution $P=P_0e^{kt}$ grows without bound, so it may be useful over a short interval but unrealistic over a long one.

Similarly, Newton’s cooling model

$$\frac{dT}{dt}=-k(T-T_s)$$

usually assumes constant surrounding temperature $T_s$, constant $k$, and conditions in which heat transfer is approximately proportional to the temperature difference. Changing airflow, evaporation or a varying room temperature can invalidate those assumptions.

When criticising a model, connect the limitation to the equation. “The model is unrealistic” is too vague. “The constant proportional growth rate ignores resource limitations, so long term exponential growth is unlikely” is precise.

Common misconceptions

Dropping the constant too early

An indefinite integration produces a family. Keep the arbitrary constant until an initial condition determines it.

Treating $k$ as unitless

A proportionality constant generally has units. Check that both sides of the differential equation match.

Assuming every algebraic branch is physical

Reject values only for a stated reason, such as $t<0$, negative population, a singularity, or a restriction in the question.

Assuming an equilibrium must be crossed

In $T=20+60e^{-0.2t}$, the temperature approaches $20$ from above but never crosses it. A horizontal asymptote can represent an equilibrium.

Giving a percentage interpretation for $k$ without checking

In $Q=Q_0e^{kt}$, the exact factor over one time unit is $e^k$, so the exact percentage change is $100(e^k-1)\%$. The approximation $100k\%$ is reliable only when $|k|$ is small.

Mixed exam style check

A chemical mass $M$ grams satisfies

$$\frac{dM}{dt}=-kM,$$

where $t$ is in hours and $k>0$. Initially $M=50$, and after $4$ hours $M=30$.

1. Find $M$ in terms of $t$, giving $k$ exactly.
2. Find the half-life.
3. State the units of $k$.
4. Give one modelling assumption.

Answer

The general solution is

$$M=Ae^{-kt}.$$

Since $M(0)=50$, $A=50$. Using $M(4)=30$,

$$30=50e^{-4k}, \qquad e^{-4k}=\frac35.$$

Taking logarithms,

$$-4k=\ln\frac35,$$

so

$$k=\frac14\ln\frac53.$$

Therefore

$$\boxed{M=50e^{-(\ln(5/3))t/4}}.$$

For the half-life, set $M=25$:

$$25=50e^{-kt} \quad\Rightarrow\quad e^{-kt}=\frac12 \quad\Rightarrow\quad t=\frac{\ln2}{k}.$$

Hence

$$\boxed{t=\frac{4\ln2}{\ln(5/3)}\approx5.43\text{ hours}}.$$

The units of $k$ are $\text{h}^{-1}$. A possible assumption is that the proportional decay constant remains unchanged throughout the modelled period. Equivalently, external conditions affecting the decay are assumed constant.

Final checklist

For a differential equation model, ask:

1. What does each variable represent, and what are its units?
2. Has the wording been translated into a rate equation correctly?
3. What does the sign of the derivative say before any solving?
4. Are there equilibrium values where the derivative is zero?
5. Has the general solution kept its arbitrary constant?
6. Has every initial condition been applied?
7. Does the answer satisfy both the equation and the conditions?
8. What domain is mathematically valid and physically credible?
9. What assumptions are encoded in the equation?

Next, consolidate the solving technique in separable differential equations, revisit growth and decay for exponential models, and practise translating conclusions through interpreting answers in context.