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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 Integration
Edexcel A Level Maths: 11.11 Modelling with Differential EquationsCurated by Kepler, not produced by us. Watch it 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
1

Setting up the equation

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

dNdt=kN ⇒ N=N0ekt\frac{dN}{dt} = kN \ \Rightarrow\ N = N_0 e^{kt}
Exponential growth/decay model.
Example 1A population grows so that dPdt=0.1P\dfrac{dP}{dt} = 0.1P, P0=200P_0 = 200. Find P(t)P(t).
1Separate and integrate: P=Ae0.1tP = A e^{0.1t}.
2At t=0t = 0, P=200P = 200, so A=200A = 200.
Answer
P=200e0.1tP = 200 e^{0.1t}

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

Formula recap

dNdt=kN\frac{dN}{dt} = kN
Growth.
dNdt=−kN\frac{dN}{dt} = -kN
Decay.
N=N0e±ktN = N_0 e^{\pm kt}
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.

Test yourself

Answer a few questions, then review anything you missed.

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Answer each question before opening its answer. This self-check does not record a quiz score.

  1. “Rate of change of N is proportional to N” is:

    • dNdt=kN\frac{dN}{dt} = kN
    • dNdt=k\frac{dN}{dt} = k
    • N=ktN = kt
    • dNdt=N2\frac{dN}{dt} = N^2
    Show answer and explanation

    Answer: dNdt=kN\frac{dN}{dt} = kN

    Proportionality.

  2. A decaying quantity is modelled with:

    • dNdt=−kN\frac{dN}{dt} = -kN
    • dNdt=kN\frac{dN}{dt} = kN
    • dNdt=k\frac{dN}{dt} = k
    • N=ktN = kt
    Show answer and explanation

    Answer: dNdt=−kN\frac{dN}{dt} = -kN

    Negative constant.

  3. The solution of dNdt=kN\frac{dN}{dt}=kN is:

    • N=N0ektN = N_0 e^{kt}
    • N=N0+ktN = N_0 + kt
    • N=kt2N = kt^2
    • N=ln⁡(kt)N = \ln(kt)
    Show answer and explanation

    Answer: N=N0ektN = N_0 e^{kt}

    Exponential.

  4. For dPdt=0.1P\frac{dP}{dt}=0.1P, P0=200P_0=200, P(t)=P(t)=

    • 200e0.1t200 e^{0.1t}
    • 200+0.1t200 + 0.1t
    • e0.1te^{0.1t}
    • 200et200 e^{t}
    Show answer and explanation

    Answer: 200e0.1t200 e^{0.1t}

    A = 200.

  5. “Proportional to” is translated as:

    • = k ×
    • + k
    • × by t
    • squared
    Show answer and explanation

    Answer: = k ×

    Constant of proportionality.

Next step

Check your understanding

Work through questions on Modelling with Differential Equations. Use the feedback to decide what needs another look.

5 past-paper questions · 14 marks

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