G6, H8PureStretch

Modelling with Differential Equations

Differential equations model situations where a rate of change depends on the current amount — growth, decay, cooling and mixing. The skill is translating words into a differential equation, then solving it.

30 min Integration

The big picture

“Proportional to the difference from the surroundings” is a model for cooling, unlike “proportional to the temperature” itself. Define the variable and sign convention before solving. A cooling law must predict a negative derivative while the object is hotter than its surroundings. Use an initial condition to determine the integration constant, then use additional information to determine the rate constant. The limiting temperature is a property of the solution and an opportunity to critique the physical assumptions.

Edexcel A Level Maths: 11.11 Modelling with Differential EquationsCurated by Kepler, not produced by us. Watch it, then work through the notes below.

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What you'll be able to do

Translate a worded rate into a differential equation
Use proportionality (∝) correctly
Solve and apply boundary conditions
Interpret the solution in context

Setting up the equation

“Rate of change of $N$ is proportional to $N$ ” becomes $\dfrac{dN}{dt} = kN$ . A decreasing quantity uses a negative constant, $\dfrac{dN}{dt} = -kN$ .

\frac{dN}{dt} = kN \ \Rightarrow\ N = N_0 e^{kt}

Exponential growth/decay model.

Example 1A population grows so that

\dfrac{dP}{dt} = 0.1P

P_0 = 200

. Find

P(t)

1Separate and integrate:

P = A e^{0.1t}

2At

t = 0

P = 200

, so

A = 200

Answer

P = 200 e^{0.1t}

Tip — Translate “proportional to” as “= k ×”, and decide the sign from growth vs decay.

Translate a verbal rate law, then interpret the solution

Use an initial condition to determine the integration constant, then use additional information to determine the rate constant. The limiting temperature is a property of the solution and an opportunity to critique the physical assumptions.

Example 1An object cools in a room at

20^\circ\text{C}

according to

dT/dt=-k(T-20)

, k > 0. Initially T = 80; after 10 min T = 50. Find T(t).

1Separate and integrate:

\ln|T-20|=-kt+C

2The initial condition gives

T-20=60e^{-kt}

3At t = 10,

30=60e^{-10k}

, so

k=\ln2/10

4Thus

T=20+60\,2^{-t/10}

. As

t\to\infty

, T approaches 20 from above.

Answer

T(t)=20+60\,2^{-t/10}

degrees Celsius, with t in minutes.

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

Using a positive k for a decaying quantity.

Decay needs −k (so N decreases).

Omitting the constant of proportionality.

“Proportional to” means “= k ×”.

Key results to remember

\frac{dN}{dt} = kN

Growth.

\frac{dN}{dt} = -kN

Decay.

N = N_0 e^{\pm kt}

Solution.

Key takeaways

“Rate ∝ amount” ⇒ dN/dt = ±kN.
Solution is exponential: N = N₀e^{±kt}.
Use the initial condition for N₀ and interpret in context.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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