Differential Equations in Modelling

This lesson covers first-order differential equations with separable variables and their application to mathematical modelling. Differential equations are used to describe how quantities change in real-world situations and are a key part of the A-Level Mathematics specification.

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What Is a Differential Equation?

A differential equation is an equation involving derivatives. A first-order ordinary differential equation (ODE) involves dy/dx (or equivalent notation):

dy/dx = f(x, y)

The solution to a differential equation is a function y = g(x) that satisfies the equation.

• A general solution contains an arbitrary constant C.
• A particular solution has a specific value of C determined by an initial condition or boundary condition.

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Separable Variables

A first-order ODE is separable if it can be written in the form:

dy/dx = f(x) × g(y)

To solve, separate the variables — put all y terms on one side and all x terms on the other:

(1/g(y)) dy = f(x) dx

Then integrate both sides:

∫ (1/g(y)) dy = ∫ f(x) dx

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Example 1: Basic Separable Equation

Solve dy/dx = 3x²y, given that y = 2 when x = 0.

Separate:

(1/y) dy = 3x² dx

Integrate:

∫ (1/y) dy = ∫ 3x² dx
ln|y| = x³ + C

Apply initial condition: When x = 0, y = 2:

ln 2 = 0 + C
C = ln 2

So:

ln|y| = x³ + ln 2
ln y = x³ + ln 2     (since y > 0 from context)
y = e^(x³ + ln 2)
y = e^(ln 2) × e^(x³)
y = 2e^(x³)

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Example 2: Another Separable Equation

Solve dy/dx = (x + 1)/y², given y = 3 when x = 0.

Separate:

y² dy = (x + 1) dx

Integrate:

y³/3 = x²/2 + x + C

Apply initial condition: y = 3, x = 0:

27/3 = 0 + 0 + C
C = 9

Particular solution:

y³/3 = x²/2 + x + 9
y³ = 3x²/2 + 3x + 27
y = (3x²/2 + 3x + 27)^(1/3)

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Exponential Growth

Many real-world quantities grow at a rate proportional to their current size. This is modelled by:

dP/dt = kP

where P is the population (or quantity), t is time, and k > 0 is the growth rate.

Solving:

(1/P) dP = k dt
ln P = kt + C
P = e^(kt + C) = Ae^(kt)    where A = e^C

If P = P₀ when t = 0, then A = P₀, giving:

P = P₀e^(kt)

Example 3

A population of bacteria doubles every 3 hours. Initially there are 500 bacteria. Find: (a) the value of k, (b) the population after 8 hours.

Part (a): P = 500e^(kt). When t = 3, P = 1000:

1000 = 500e^(3k)
2 = e^(3k)
3k = ln 2
k = ln 2/3 ≈ 0.2310

Part (b):

P = 500e^(8 ln 2/3)
= 500 × 2^(8/3)
= 500 × 6.3496...
≈ 3175 bacteria

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Exponential Decay

When k < 0 (or we write dP/dt = −kP with k > 0), the quantity decreases exponentially:

P = P₀e^(−kt)

Example 4: Radioactive Decay

A radioactive substance has a half-life of 10 years. If the initial mass is 200 g, find the mass after 25 years.

Model: M = 200e^(−kt).

Find k from the half-life: When t = 10, M = 100:

100 = 200e^(−10k)
1/2 = e^(−10k)
−10k = ln(1/2) = −ln 2
k = ln 2/10 ≈ 0.0693

At t = 25:

M = 200e^(−25 × ln 2/10)
= 200e^(−2.5 ln 2)
= 200 × 2^(−2.5)
= 200/(4√2)
= 200/5.6569
≈ 35.4 g

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Newton's Law of Cooling

Newton's law of cooling states that the rate of cooling of an object is proportional to the difference between its temperature and the surrounding (ambient) temperature:

dT/dt = −k(T − Tₐ)

where T is the temperature of the object, Tₐ is the ambient temperature, and k > 0.

Solving (let θ = T − Tₐ, so dθ/dt = dT/dt):

dθ/dt = −kθ
θ = θ₀e^(−kt)
T − Tₐ = (T₀ − Tₐ)e^(−kt)
T = Tₐ + (T₀ − Tₐ)e^(−kt)

Example 5

A cup of coffee at 90°C is placed in a room at 20°C. After 5 minutes, the coffee has cooled to 70°C. Find its temperature after 15 minutes.

Model: T = 20 + 70e^(−kt) (since T₀ − Tₐ = 90 − 20 = 70).

Find k: When t = 5, T = 70:

70 = 20 + 70e^(−5k)
50 = 70e^(−5k)
5/7 = e^(−5k)
−5k = ln(5/7)
k = −ln(5/7)/5 = ln(7/5)/5 ≈ 0.0673

At t = 15:

T = 20 + 70e^(−15 × 0.0673)
= 20 + 70e^(−1.0096)
= 20 + 70 × 0.3644
= 20 + 25.5
≈ 45.5°C

Alternatively, using exact values:

T = 20 + 70(5/7)³ = 20 + 70 × 125/343 = 20 + 8750/343 ≈ 20 + 25.51 ≈ 45.5°C

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Example 6: Logistic-Style Equation (Limited Growth)

A population P satisfies dP/dt = 0.1P(100 − P)/100 with P(0) = 10. This models growth that slows as P approaches a carrying capacity of 100.

This is separable but requires partial fractions to integrate (which is at the boundary of A-Level). At A-Level you are more likely to be given the solution and asked to interpret it or verify it.

The solution is:

P = 100/(1 + 9e^(−0.1t))

Verify: As t → ∞, e^(−0.1t) → 0, so P → 100/(1 + 0) = 100 (the carrying capacity). As t = 0, P = 100/(1 + 9) = 10 ✓.

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Forming Differential Equations from Context

At A-Level, you may be asked to set up a differential equation from a description:

• "The rate of increase of P is proportional to P" → dP/dt = kP.
• "The rate of decrease of T is proportional to (T − 20)" → dT/dt = −k(T − 20).
• "The rate of change of y with respect to x is equal to 3x/y" → dy/dx = 3x/y.
• "The volume decreases at a rate proportional to the square root of the volume" → dV/dt = −k√V.

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General Solution vs Particular Solution

• The general solution has an arbitrary constant C (or A):

  y = Ae^(3x)
  

• The particular solution uses an initial/boundary condition to fix C:

  If y = 5 when x = 0: 5 = Ae⁰, so A = 5, giving y = 5e^(3x).
  

Exam Tip: Always state your general solution clearly before applying the initial condition. The general solution earns marks even if you make an error with the boundary condition.

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Example 7: Full Worked Problem

A tank contains 100 litres of water with 10 kg of salt dissolved in it. Pure water flows into the tank at 5 litres per minute, and the well-mixed solution flows out at 5 litres per minute (so the volume remains constant at 100 litres). Find the amount of salt in the tank after t minutes.

Let S be the mass of salt (in kg) at time t.

The concentration of salt is S/100 kg per litre.

The rate of salt leaving = concentration × flow rate = (S/100) × 5 = S/20 kg per minute.

The rate of salt entering = 0 (pure water).

dS/dt = 0 − S/20 = −S/20

Separate and integrate:

(1/S) dS = −(1/20) dt
ln S = −t/20 + C
S = Ae^(−t/20)

Initial condition: S = 10 when t = 0: A = 10.

S = 10e^(−t/20)

After 40 minutes: S = 10e^(−2) ≈ 10 × 0.1353 ≈ 1.35 kg.

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Summary

• A differential equation relates a function to its derivatives.
• Separable equations can be solved by separating variables and integrating both sides.
• Exponential growth: dP/dt = kP gives P = P₀e^(kt).
• Exponential decay: dP/dt = −kP gives P = P₀e^(−kt).
• Newton's law of cooling: dT/dt = −k(T − Tₐ) gives T = Tₐ + (T₀ − Tₐ)e^(−kt).
• The general solution contains an arbitrary constant; the particular solution is found by applying initial/boundary conditions.
• When modelling, translate the English description into a differential equation using phrases like "rate of change is proportional to...".

Exam Tip: Differential equation questions often follow a sequence: (a) set up the equation from a description, (b) solve it using separation of variables, (c) use an initial condition to find the particular solution, (d) use the solution to answer a contextual question (e.g. "Find the temperature after 10 minutes"). Show all steps: the separation, both integrals, the general solution, the substitution of the initial condition, and the particular solution. This is a multi-mark question and partial credit is available at every stage.

A-Level Deep Dive: Differential Equations and Modelling

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section H (Year 2): "Construct simple differential equations in pure mathematics and in context, including the cases where variables are separable. Solve such equations by separation of variables, including those involving an exponential growth or decay model. Interpret the solution of a differential equation in the context of solving a problem, including identifying limitations of the solution; includes links to kinematics." This sits inside the broader Section H (Integration) and is examinable on Paper 2 alongside techniques such as integration by substitution, by parts, and via partial fractions — all of which can appear inside the integration step of separating variables. The AQA formula booklet does not list a worked separable-ODE template; you are expected to write $\int \dfrac{1}{f(y)}\,dy = \int g(x)\,dx$∫f(y)1​dy=∫g(x)dx from scratch.

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Worked example with full mark scheme

Question (8 marks):

A cup of coffee at $80°\text{C}$80°C is left in a room at $20°\text{C}$20°C. Newton's law of cooling states that the rate of change of temperature is proportional to the difference between the body's temperature and the ambient temperature, so

$\dfrac{dT}{dt} = -k(T - 20)$dtdT​=−k(T−20)

where $T$T is the temperature in $°\text{C}$°C, $t$t is time in minutes, and $k > 0$k>0 is a constant. After $5$5 minutes the coffee has cooled to $60°\text{C}$60°C.

(a) Solve the differential equation, expressing $T$T as a function of $t$t. (5)

(b) Find the time at which the coffee reaches $30°\text{C}$30°C, giving your answer to $3$3 significant figures. (3)

Solution with mark scheme:

(a) Step 1 — separate variables.

$\dfrac{1}{T - 20}\,dT = -k\,dt$T−201​dT=−kdt

M1 — variables correctly separated, with $T$T-terms on the left and $t$t-terms on the right. Common error: dividing by $T$T rather than $(T - 20)$(T−20), which produces a wrong (and unsolvable-in-closed-form) integrand.

Step 2 — integrate both sides.

$\int \dfrac{1}{T - 20}\,dT = \int -k\,dt \implies \ln|T - 20| = -kt + C$∫T−201​dT=∫−kdt⟹ln∣T−20∣=−kt+C

M1 — correct integration of both sides. The left integrates to $\ln|T - 20|$ln∣T−20∣ via the standard $\int \tfrac{1}{u}\,du = \ln|u|$∫u1​du=ln∣u∣. Mark-scheme expectation: a single constant of integration on one side only (combining the two constants into $C$C).

A1 — correct antiderivatives with constant $C$C.

Step 3 — apply the initial condition $T(0) = 80$T(0)=80.

$\ln|80 - 20| = -k(0) + C \implies C = \ln 60$ln∣80−20∣=−k(0)+C⟹C=ln60

So $\ln|T - 20| = -kt + \ln 60$ln∣T−20∣=−kt+ln60, giving

$\ln\left|\dfrac{T - 20}{60}\right| = -kt \implies T - 20 = 60e^{-kt}$ln​60T−20​​=−kt⟹T−20=60e−kt

(positive branch since $T > 20$T>20 throughout cooling).

M1 — correct application of the initial condition to determine $C$C. Candidates who skip this and write $T = Ae^{-kt} + 20$T=Ae−kt+20 directly should still earn the M1, but must justify $A = 60$A=60 from $T(0) = 80$T(0)=80.

Step 4 — apply the second condition $T(5) = 60$T(5)=60 to find $k$k.

$60 - 20 = 60e^{-5k} \implies \dfrac{40}{60} = e^{-5k} \implies -5k = \ln\dfrac{2}{3}$60−20=60e−5k⟹6040​=e−5k⟹−5k=ln32​

So $k = -\dfrac{1}{5}\ln\dfrac{2}{3} = \dfrac{1}{5}\ln\dfrac{3}{2}$k=−51​ln32​=51​ln23​.

A1 — final form $T = 20 + 60e^{-kt}$T=20+60e−kt with $k = \tfrac{1}{5}\ln(3/2)$k=51​ln(3/2), or equivalently $T = 20 + 60(2/3)^{t/5}$T=20+60(2/3)t/5.

(b) Step 1 — set $T = 30$T=30 and solve for $t$t.

$30 = 20 + 60e^{-kt} \implies e^{-kt} = \dfrac{1}{6} \implies -kt = \ln\dfrac{1}{6} = -\ln 6$30=20+60e−kt⟹e−kt=61​⟹−kt=ln61​=−ln6

M1 — correctly rearranging to isolate the exponential and taking logs.

Step 2 — substitute $k$k and evaluate.

$t = \dfrac{\ln 6}{k} = \dfrac{\ln 6}{(1/5)\ln(3/2)} = \dfrac{5\ln 6}{\ln(3/2)}$t=kln6​=(1/5)ln(3/2)ln6​=ln(3/2)5ln6​

M1 — correct substitution of $k$k.

Numerically: $\ln 6 \approx 1.7918$ln6≈1.7918, $\ln(3/2) \approx 0.4055$ln(3/2)≈0.4055, so $t \approx 5 \cdot 1.7918 / 0.4055 \approx 22.1$t≈5⋅1.7918/0.4055≈22.1 minutes.

A1 — final answer $t \approx 22.1$t≈22.1 minutes ($3$3 s.f.).

Total: 8 marks (M5 A3).

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Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A water tank is leaking. The volume $V$V (in litres) of water in the tank at time $t$t (in hours) satisfies

$\dfrac{dV}{dt} = -\dfrac{\sqrt{V}}{10}, \quad V(0) = 100$dtdV​=−10V​​,V(0)=100

(a) By separating variables, show that $V = \left(10 - \dfrac{t}{20}\right)^2$V=(10−20t​)2. (4)

(b) State, with reasoning, one limitation of this model in describing a real leaking tank. (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — separating variables: $V^{-1/2}\,dV = -\tfrac{1}{10}\,dt$V−1/2dV=−101​dt.
• M1 (AO1.1b) — integrating both sides: $2V^{1/2} = -\tfrac{t}{10} + C$2V1/2=−10t​+C.
• M1 (AO3.1a) — applying $V(0) = 100$V(0)=100: $2\sqrt{100} = C$2100​=C, so $C = 20$C=20, hence $2\sqrt{V} = 20 - \tfrac{t}{10}$2V​=20−10t​.
• A1 (AO2.1) — rearranging to $\sqrt{V} = 10 - \tfrac{t}{20}$V​=10−20t​, then squaring to obtain the printed form $V = (10 - t/20)^2$V=(10−t/20)2.

(b)

• B1 (AO3.5a) — identifying a specific limitation, e.g. "the model predicts $V = 0$V=0 at $t = 200$t=200, after which $V$V would increase again, which is unphysical".
• B1 (AO3.5b) — connecting the limitation to a model assumption, e.g. "the equation assumes outflow rate depends only on $\sqrt{V}$V​, ignoring effects such as the leak hole's geometry, viscosity, or the tank running dry".

Total: 6 marks split AO1 = 2, AO2 = 1, AO3 = 3. This is an AO3-heavy item — 50% of the marks reward problem-solving and modelling judgement, which is characteristic of AQA's Section H specimen items. Students who solve the ODE algorithmically but skip the modelling-evaluation step lose half the available marks.

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Synoptic links

Connects to:

1. Section H — Integration: every separable-ODE solution requires an integration step. Harder ODEs require partial fractions ($\int \tfrac{1}{N(K - N)}\,dN$∫N(K−N)1​dN for the logistic equation), substitution ($\int \tfrac{1}{1 + x^2}\,dx = \arctan x + C$∫1+x21​dx=arctanx+C), or by-parts. The ODE topic is, in effect, applied integration in disguise.

2. Exponential growth and decay (Section G): the equation $\tfrac{dy}{dt} = ky$dtdy​=ky has solution $y = Ae^{kt}$y=Aekt, the foundation for population growth, radioactive decay, compound interest, and continuously compounded growth. AQA examiners regularly ask students to derive $y = Ae^{kt}$y=Aekt from $\tfrac{dy}{dt} = ky$dtdy​=ky via separation, rather than quoting it.

3. Newton's law of cooling: $\tfrac{dT}{dt} = -k(T - T_{\text{room}})$dtdT​=−k(T−Troom​) is the canonical "shifted exponential" — the substitution $u = T - T_{\text{room}}$u=T−Troom​ reduces it to the basic decay ODE. This same trick appears in mortgage repayment, drug clearance from the bloodstream, and RC-circuit discharge.

4. The logistic equation: $\tfrac{dN}{dt} = rN(1 - N/K)$dtdN​=rN(1−N/K) models population growth with carrying capacity. Solving requires partial fractions: $\int \tfrac{1}{N(1 - N/K)}\,dN$∫N(1−N/K)1​dN. Although the closed-form solution $N(t) = \tfrac{K}{1 + Ae^{-rt}}$N(t)=1+Ae−rtK​ is just beyond the AQA syllabus, the separation step is fully on-spec and frequently appears.

5. Partial fractions in ODEs: Section D partial-fraction decomposition is the bridge between Section H integration and ODE solving when the integrand has factored denominators — a synoptic link AQA specifically rewards on Paper 2.

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Mark-scheme literacy

Differential-equation questions on AQA 7357 Paper 2 split AO marks more evenly than typical pure-maths topics: