Modelling with Differential Equations

Using differential equations to model real-world problems: population growth, decay, and flow models

SPEC 8.7 • INTEGRATION

Setting Up Differential Equations

Many real-world problems involve rates of change. We translate word problems into differential equations using the proportional sign.

Key Process

Write "proportional to" using ∝ first, then convert to = k∝→= k ×

The Modelling Process

1. Translate the word problem using ∝, then replace with = k

2. Solve the differential equation using separation of variables

3. Use initial conditions to find constants

4. Interpret the solution in context

Key Phrase Translations

Convert word phrases to mathematics using ∝ first, then replace with = k:

Word Problem PhraseUsing ∝As Equation

"Rate of change of P is proportional to P"

\frac{d P}{d t} \propto P

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

"Rate of change of h is proportional to √h"

\frac{d h}{d t} \propto h

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

"Rate of decrease of R is proportional to R"

\frac{d R}{d t} \propto - R

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

Key Point

"Rate of decrease" or "rate of decay" means a negative rate. Write the negative sign with the proportional:

\frac{d R}{d t} \propto - R

, then

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

where k > 0.

Exponential Growth Model

When "the rate of increase is proportional to the current amount":

Setting Up

\frac{d P}{d t} \propto P

→

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

Solving

Separate variables:

\frac{1}{P} d P = k d t

Integrate:

ln ∣ P ∣ = k t + c

Rearrange:

P = e^{k t + c} = e^{c} \cdot e^{k t} = A e^{k t}

Solution: P = Ae^(kt)

Limitation

Exponential growth cannot continue indefinitely — populations are limited by resources, space, etc. Always state this limitation when asked.

01Worked Example

The rate of increase of a population P of microorganisms at time t hours is proportional to P. Initially the population was of size 8.

(a) Given that the constant of proportionality is 3, find a model for P in the form P = Ae³ᵗ.

(b) Find, to the nearest hundred, the size of the population at time t = 2.

(c) Find the time at which the population will be 1000 times its starting value.

(a) "Rate of increase proportional to P":

\frac{d P}{d t} \propto P

→

\frac{d P}{d t} = 3 P

Separate:

\int \frac{1}{P} d P = \int 3 d t

ln ∣ P ∣ = 3 t + c

P = e^{3 t + c} = A e^{3 t}

When t = 0, P = 8:

8 = A e^{0} = A

P = 8e³ᵗ, where A = 8

(b) When t = 2:

P = 8 e^{6} = 8 \times 403.4... = 3227...

P ≈ 3200

(c) For 1000 times starting value: P = 8000

8000 = 8 e^{3 t}

→

1000 = e^{3 t}

→

ln 1000 = 3 t

t = (ln 1000)/3 = 2.30 hours

Exponential Decay Model

When "the rate of decrease is proportional to the current amount":

Setting Up

\frac{d R}{d t} \propto - R

→

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

(k > 0, negative sign shows decrease)

Solving

Separate variables:

\frac{1}{R} d R = - k d t

Integrate:

ln ∣ R ∣ = - k t + c

Rearrange:

R = A e^{- k t}

With initial condition R = R₀ when t = 0:

Solution: R = R₀e^(−kt)

Applications

Radioactive decay, carbon dating, drug elimination, depreciation

02Worked Example

Fossils are aged using carbon dating. The amount of carbon R remaining decreases over time t years. The rate of decrease of carbon is proportional to the remaining carbon.

(a) Given that initially the amount of carbon is R₀, show that R = R₀e⁻ᵏᵗ.

(b) Given that k = 1.21 × 10⁻⁴, find the age of a fossil with 10% of its expected carbon remaining.

(a) "Rate of decrease proportional to R":

\frac{d R}{d t} \propto - R

→

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

Separate:

\int \frac{1}{R} d R = \int - k d t

ln ∣ R ∣ = - k t + c

R = A e^{- k t}

When t = 0, R = R₀: A = R₀

R = R₀e⁻ᵏᵗ ✓

(b) When R = 0.1R₀ (10% remaining):

0.1 R_{0} = R_{0} e^{- k t}

0.1 = e^{- k t}

ln 0.1 = - k t

t = \frac{- l n 0.1}{k} = \frac{l n 10}{1.21 \times 1 0 ^{- 4}}

t = \frac{2.303}{1.21 \times 1 0 ^{- 4}} = 19030...

Age ≈ 19000 years

Rate In − Rate Out Models

When liquid flows into and out of a container:

Setting Up

\frac{d V}{d t} = (rate in) - (rate out)

Common Setup

Rate in: Often constant (e.g., tap flowing at fixed rate)

Rate out: Often proportional to volume or depth

Result: equations like

\frac{d V}{d t} = a - bV

or

\frac{d h}{d t} = c - d h

Solution Form

Equations like

\frac{d V}{d t} = a - bV

give solutions of the form:

V = A + B e^{- c t}

As t → ∞, the exponential term → 0, giving a limiting value

03Worked Example

A bath tub has base area 6000 cm². Water flows in at 12000 cm³/min. At time t, the depth is h cm. Water leaves through a plughole at rate 500h cm³/min.

(a) Show that

60 \frac{d h}{d t} = 120 - 5 h

.

(b) Given h = 6 when t = 0, find the value of t when h = 10.

(a) Volume V = Base area × height = 6000h

\frac{d V}{d t} = 6000 \frac{d h}{d t}

Rate in − Rate out:

\frac{d V}{d t} = 12000 - 500 h

So:

6000 \frac{d h}{d t} = 12000 - 500 h

Dividing by 100: $60 \frac{d h}{d t} = 120 - 5 h$ ✓

(b) Rearrange:

\frac{60}{120 - 5 h} d h = d t

\frac{60}{- 5} ln ∣120 - 5 h ∣ = t + c

- 12 ln ∣120 - 5 h ∣ = t + c

When t = 0, h = 6:

- 12 ln 90 = c

t = - 12 ln ∣120 - 5 h ∣ + 12 ln 90 = 12 ln \frac{90}{120 - 5 h}

When h = 10:

t = 12 ln \frac{90}{70} = 12 ln \frac{9}{7}

t = 3.03 minutes

04Worked Example

Liquid pours into a container at 40 cm³s⁻¹ and leaks out at rate ¼V cm³s⁻¹, where V cm³ is the volume.

(a) Show that

- 4 \frac{d V}{d t} = V - 160

.

(b) Given V = 5000 when t = 0, find V in the form

V = a + b e^{- \frac{t}{4}}

.

(c) Write down the limiting value of V as t → ∞.

(a) Rate in − Rate out:

\frac{d V}{d t} = 40 - \frac{V}{4}

Multiply by −4:

- 4 \frac{d V}{d t} = - 160 + V

$- 4 \frac{d V}{d t} = V - 160$ ✓

(b) Rearrange:

\frac{d V}{d t} = \frac{160 - V}{4}

Separate:

\frac{1}{160 - V} d V = \frac{1}{4} d t

- ln ∣160 - V ∣ = \frac{t}{4} + c

160 - V = A e^{- t /4}

V = 160 - A e^{- t /4}

When t = 0, V = 5000:

5000 = 160 - A

, so A = −4840

$V = 160 + 4840 e^{- t /4}$ (a = 160, b = 4840)

(c) As t → ∞,

e^{- t /4} \to 0

Limiting value: V = 160 cm³

More Complex Models

Some differential equations require more advanced integration techniques to solve.

Techniques You May Need

Partial fractions: For expressions like

\frac{1}{P ( M - P )}

Substitution: For expressions with roots or powers

Standard integrals: Recognising forms like

\int \frac{1}{a - b x} d x

Reverse chain rule: For composite functions

Approach

The method is always the same:

1. Set up the DE from the context (using ∝ then = k)

2. Separate variables

3. Integrate both sides (using appropriate technique)

4. Apply initial conditions

Remember

Always check your answer satisfies the original differential equation and initial conditions!

Practice

Question 1: The rate of increase of a population P of rabbits at time t years is proportional to P. Initially the population was of size 200.

(a) Solve the differential equation giving P in terms of k and t.

(b) Given k = 3, find the time taken for the population to reach 4000.

(c) State a limitation of this model for large values of t.

(a)

\frac{d P}{d t} \propto P

→

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

\int \frac{1}{P} d P = \int k d t

→

ln P = k t + c

→

P = A e^{k t}

t = 0, P = 200: A = 200. P = 200e^(kt)

(b) 4000 = 200e^(3t) → 20 = e^(3t) → t = (ln 20)/3 = 1.00 years

(c) The model predicts unlimited growth, but populations are limited by food, space, predators, etc.

Practice

Question 2: The thickness x cm of ice on a pond increases at a rate that is inversely proportional to the square of the existing thickness. Initially x = 1 cm. After 20 days, x = 2 cm.

(a) Show that the thickness can be modelled by

x = 3 \frac{7 t}{20} + 1

.

(b) Find the time taken for the ice to increase from 2 cm to 3 cm.

(a) "Rate inversely proportional to x²":

\frac{d x}{d t} \propto \frac{1}{x ^{2}}

→

\frac{d x}{d t} = \frac{k}{x ^{2}}

\int x^{2} d x = \int k d t

→

\frac{x ^{3}}{3} = k t + c

t = 0, x = 1: c = 1/3

t = 20, x = 2: 8/3 = 20k + 1/3, so k = 7/60

\frac{x ^{3}}{3} = \frac{7 t}{60} + \frac{1}{3}

→

x^{3} = \frac{7 t}{20} + 1

→ $x = 3 \frac{7 t}{20} + 1$ ✓

(b) x = 2: t = 20. x = 3: 27 = 7t/20 + 1, t = 520/7 ≈ 74.3

Time = 74.3 − 20 = 54.3 days

Practice

Question 3: Oil spreads on water. The surface area A mm² at time t minutes satisfies

\frac{d A}{d t} = \frac{A ^{3/2}}{10 t ^{2}}

.

Given that A = 1 when t = 1:

(a) Find an expression for A in terms of t.

(b) Show that the surface area of the drop cannot exceed

\frac{400}{361}

mm².

(a)

\int A^{- 3/2} d A = \int \frac{1}{10 t ^{2}} d t

\frac{A ^{- 1/2}}{- 1/2} = \frac{- 1}{10 t} + c

→

- 2 A^{- 1/2} = - \frac{1}{10 t} + c

t = 1, A = 1: −2 = −0.1 + c, so c = −1.9

- 2 A^{- 1/2} = - \frac{1}{10 t} - 1.9 = - \frac{1 + 19 t}{10 t}

A^{- 1/2} = \frac{1 + 19 t}{20 t}

→ $A = \frac{400 t ^{2}}{( 1 + 19 t ) ^{2}}$

(b) As t → ∞:

A \to \frac{400 t ^{2}}{361 t ^{2}} = \frac{400}{361}

Maximum A = 400/361 mm² ✓

Summary

Key Models

ContextUsing ∝As Equation

Growth proportional to P

\frac{d P}{d t} \propto P

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

→

P = A e^{k t}

Decay proportional to R

\frac{d R}{d t} \propto - R

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

→

R = R_{0} e^{- k t}

Rate in − rate out—

\frac{d V}{d t} = a - bV

→

V = A + B e^{- c t}

Key Skills

Use ∝ first, then = k

Separate and integrate

Apply initial conditions

Find limiting values (t → ∞)

Exam Tips

"Rate of decrease" → negative

State model limitations

Use integration techniques as needed