5. Applications & Modelling

Video coming soon3 worked examples

Theory

Differential equations earn their keep in modelling. Exam questions rarely hand you the equation — they describe a situation in words and expect you to form it first. The key is that the phrase “rate of change” means a derivative with respect to time, and “proportional to” introduces a constant:

“the rate of increase of $P$ is proportional to $P$ ” $\implies \dfrac{dP}{dt} = kP$
“the rate of decrease of $N$ is proportional to $N$ ” $\implies \dfrac{dN}{dt} = -kN$
“the rate of cooling is proportional to the excess temperature above the surroundings $\theta_0$ ” $\implies \dfrac{d\theta}{dt} = -k(\theta - \theta_0)$

That last one is Newton's law of cooling. The substitution $u = \theta - \theta_0$ turns it into the same separable equation as the others.

The workflow is always the same: form the equation, separate the variables and integrate, then use the given data to pin down the constants. There are usually two unknowns — the constant of integration and the constant of proportionality $k$ — so expect two pieces of information.

The Golden Rule: translate the words into an equation before doing any calculus, and count your unknowns against the data you have been given. If you have two unknowns and only one condition, you have mis-formed the equation.

⚠️ Common Examiner Traps

Sign of $k$ : for decay or cooling, write the minus sign into the equation and keep $k$ positive. Doing both — a minus sign and a negative $k$ — reverses the model.
Using conditions in the wrong order: the initial condition (at $t=0$ ) gives the constant of integration most easily, because the exponential term becomes $1$ . Find that first, then use the second condition for $k$ .
Leaving the answer in log form: convert $\ln P = kt + C$ to $P = Ae^{kt}$ before substituting — it is far less error-prone.
Forgetting the surrounding temperature: in a cooling problem the variable that decays exponentially is the excess $\theta - \theta_0$ , not $\theta$ itself. Add $\theta_0$ back at the end.

Worked examples

Example 1

A population grows at a rate proportional to its current size. Initially it is $500$ , and after 3 hours it has reached $2000$ . Find the population after 5 hours.

Step 1: Translate the description into a differential equation:

\frac{dP}{dt} = kP

Step 2: Separate the variables and integrate:

\int \frac{1}{P}\,dP = \int k\,dt \implies \ln P = kt + C

Step 3: Convert to exponential form immediately, writing $A = e^{C}$ :

P = Ae^{kt}

Step 4: Use the initial condition $P = 500$ when $t = 0$ . Since $e^0 = 1$ , this gives $A$ at once:

500 = A \implies P = 500e^{kt}

Step 5: Use the second condition, $P = 2000$ when $t = 3$ , to find $k$ :

2000 = 500e^{3k} \implies e^{3k} = 4 \implies k = \frac{\ln 4}{3}

Step 6: Evaluate at $t = 5$ :

P = 500e^{5k} = 500 \times 4^{5/3} = 500 \times 10.079 \approx 5040

Example 2

A body at $90^\circ\text{C}$ is left to cool in a room kept at $20^\circ\text{C}$ . After 10 minutes its temperature is $60^\circ\text{C}$ . Using Newton's law of cooling, find its temperature after 25 minutes.

Step 1: Newton's law of cooling gives, with room temperature $20$ :

\frac{d\theta}{dt} = -k(\theta - 20)

Step 2: Substitute $u = \theta - 20$ . Since $20$ is constant, $\frac{du}{dt} = \frac{d\theta}{dt}$ , so the equation becomes separable in $u$ :

\frac{du}{dt} = -ku \implies u = Ae^{-kt}

Step 3: At $t = 0$ the temperature is $90$ , so the excess is $u = 70$ :

A = 70 \implies u = 70e^{-kt}

Step 4: At $t = 10$ the temperature is $60$ , so the excess is $40$ :

40 = 70e^{-10k} \implies e^{-10k} = \frac{4}{7} \implies k = \frac{1}{10}\ln\!\left(\frac{7}{4}\right) = 0.05596

Step 5: Find the excess at $t = 25$ :

u = 70e^{-25(0.05596)} = 70e^{-1.399} = 70 \times 0.2468 = 17.28

Step 6: Add the room temperature back to get the actual temperature:

\theta = 20 + 17.28 = 37.3^\circ\text{C} \ \text{(1 d.p.)}

Example 3

A tank leaks so that the rate of decrease of its volume is proportional to the square root of the volume remaining. Initially it holds $100$ litres, and after 5 minutes it holds $64$ litres. Find the volume after 10 minutes.

Step 1: The rate is decreasing and proportional to $\sqrt{V}$ , so:

\frac{dV}{dt} = -k\sqrt{V}

Step 2: Separate the variables, writing the left side with a negative index:

\int V^{-1/2}\,dV = \int -k\,dt \implies 2V^{1/2} = -kt + C

Step 3: Apply the initial condition $V = 100$ at $t = 0$ :

2\sqrt{100} = C \implies C = 20

Step 4: Apply the second condition, $V = 64$ at $t = 5$ , to find $k$ :

2\sqrt{64} = -5k + 20 \implies 16 = -5k + 20 \implies k = 0.8

Step 5: The model is therefore $2\sqrt{V} = 20 - 0.8t$ . Substitute $t = 10$ :

2\sqrt{V} = 20 - 8 = 12 \implies \sqrt{V} = 6 \implies V = 36

The tank holds 36 litres after 10 minutes.