11.11PureStretch

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.

26 min Video by Zeeshan Zamurred Integration

Edexcel A Level Maths: 11.11 Modelling with Differential EquationsWatch the full walkthrough before 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 $\frac{d N}{d t} = k N$ . A decreasing quantity uses a negative constant, $\frac{d N}{d t} = - k N$ .

\frac{d N}{d t} = k N \Rightarrow N = N_{0} e^{k t}

Exponential growth/decay model.

1Separate and integrate:

P = A e^{0.1 t}

2At

t = 0

P = 200

, so

A = 200

Answer

P = 200 e^{0.1 t}

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

Formula recap

\frac{d N}{d t} = k N

Growth.

\frac{d N}{d t} = - k N

Decay.

N = N_{0} e^{\pm k t}

Solution.

Common mistakes to avoid

Using a positive k for a decaying quantity.

Decay needs −k (so N decreases).

Omitting the constant of proportionality.

“Proportional to” means “= k ×”.

Key takeaways

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

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