Coupled Equations Revision Notes for AQA A-Level Further Maths

Coupled Equations

Introduction to coupled differential equations

Coupled differential equations are systems where the rates of change of multiple variables depend on each other. Unlike the simpler differential equations you've studied before (which typically involve just two variables), coupled systems involve three or more variables that interact simultaneously.

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These equations are particularly useful for modelling real-world situations where multiple quantities influence each other's behaviour. Common applications include:

Population dynamics (predator-prey relationships)
Chemical reactions (decay chains and equilibria)
Physical systems with multiple interacting components
Economic models with interdependent variables

Predator-prey problems

One of the most important applications of coupled equations is in predator-prey models, which describe populations of two species where one species (the predator) hunts the other species (the prey).

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Key definitions:

Predator: A species that hunts and eats another species
Prey: A species that is eaten by the predator

Understanding these roles is essential for setting up the correct equations with appropriate signs (positive for growth, negative for predation).

Consider an island populated by foxes and rabbits. Let $F$ represent the number of foxes and $R$ represent the number of rabbits. The foxes hunt and eat the rabbits (making foxes the predators and rabbits the prey). Meanwhile, rabbits eat grass but are hunted by foxes.

The rate of change of each population depends on both populations:

$\frac{dR}{dt}$ depends on both $R$ and $F$ . More rabbits lead to more breeding and deaths, but more foxes lead to more rabbits being eaten.
$\frac{dF}{dt}$ depends on both $R$ and $F$ . More rabbits provide more food for foxes, but more foxes lead to increased competition and breeding.

This gives rise to a system of coupled equations of the form:

$\frac{dR}{dt} = \alpha R - \beta F \quad (1)$

$\frac{dF}{dt} = \lambda R + \mu F \quad (2)$

where $\alpha$ , $\beta$ , $\lambda$ , and $\mu$ are constants determined by the specific situation.

Method for solving coupled systems

To solve a system of coupled differential equations, you need to convert it into a single second-order differential equation. This is done through a systematic process of differentiation and substitution.

Step-by-step method:

Choose one equation to differentiate (usually the simpler one, or the one that makes substitution easier)
Differentiate with respect to $t$ to obtain a second derivative
Substitute from the original equations to eliminate the other variable, leaving you with a single second-order differential equation in one variable
Solve the second-order equation using standard techniques:
- Find the characteristic equation
- Solve for the roots
- Write the complementary function
- Find the particular integral if needed
- Form the general solution
Use initial conditions to find unknown coefficients
Find the other variable by substituting back into one of the original equations

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Demonstration using predator-prey equations:

Starting with equations (1) and (2) above, differentiate equation (1):

$\frac{d^2R}{dt^2} = \alpha \frac{dR}{dt} - \beta \frac{dF}{dt}$

Now substitute for $\frac{dF}{dt}$ from equation (2):

$\frac{d^2R}{dt^2} = \alpha \frac{dR}{dt} - \beta(\lambda R + \mu F)$

$\frac{d^2R}{dt^2} = \alpha \frac{dR}{dt} - \beta\lambda R - \mu\beta F$

Now substitute $\beta F = \alpha R - \frac{dR}{dt}$ from equation (1):

$\frac{d^2R}{dt^2} = \alpha \frac{dR}{dt} - \beta\lambda R - \mu\left(\alpha R - \frac{dR}{dt}\right)$

Rearranging gives:

$\frac{d^2R}{dt^2} - (\alpha + \mu)\frac{dR}{dt} + (\beta\lambda + \alpha\mu)R = 0$

This is now a second-order differential equation in $R$ and $t$ that you can solve using standard techniques.

Worked example 1: Basic coupled system

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Question: A system of differential equations is given by:

$\frac{dx}{dt} = x + 4y \quad (1)$

$\frac{dy}{dt} = 2x + 3y - 10 \quad (2)$

where $(x, y) = (3, 2)$ when $t = 0$ .

Find expressions for $x$ and $y$ in terms of $t$ .

Solution:

We'll differentiate equation (2) and use equation (1) to substitute for $\frac{dx}{dt}$ .

Differentiating equation (2) gives:

$\frac{d^2y}{dt^2} = 2\frac{dx}{dt} + 3\frac{dy}{dt}$

Substitute dx/dt = x + 4y from equation (1):

$\frac{d^2y}{dt^2} = 2(x + 4y) + 3\frac{dy}{dt}$

$\frac{d^2y}{dt^2} = 2x + 8y + 3\frac{dy}{dt}$

Now substitute 2x = dy/dt - 3y + 10 from equation (2):

$\frac{d^2y}{dt^2} = \frac{dy}{dt} - 3y + 10 + 8y + 3\frac{dy}{dt}$

$\frac{d^2y}{dt^2} - 4\frac{dy}{dt} - 5y = 10$

The characteristic equation is:

$m^2 - 4m - 5 = 0$

$(m - 5)(m + 1) = 0$

So $m = 5$ or $m = -1$ .

The complementary function is:

$y = Ae^{5t} + Be^{-t}$

For the particular integral, try $y = c$ (a constant):

$0 - 0 - 5c = 10$

$c = -2$

The general solution is:

$y = Ae^{5t} + Be^{-t} - 2 \quad (3)$

Differentiating gives:

$\dot{y} = 5Ae^{5t} - Be^{-t} \quad (4)$

Finding the constants using initial conditions:

From equation (2) at $t = 0$ :

$\dot{y}(0) = 2x(0) + 3y(0) - 10 = 2(3) + 3(2) - 10 = 2$

Substituting $t = 0$ into equation (3):

$2 = A + B - 2$

$A + B = 4$

Substituting $t = 0, \dot{y} = 2$ into equation (4):

$2 = 5A - B$

Solving simultaneously: $A + B = 4$ and $5A - B = 2$

Adding: $6A = 6$ , so $A = 1$

Therefore $B = 3$ .

So:

$y = e^{5t} + 3e^{-t} - 2$

Finding $x$ :

From equation (2): $2x = \dot{y} - 3y + 10$

Using $\dot{y} = 5e^{5t} - 3e^{-t}$ :

$2x = 5e^{5t} - 3e^{-t} - 3(e^{5t} + 3e^{-t} - 2) + 10$

$2x = 5e^{5t} - 3e^{-t} - 3e^{5t} - 9e^{-t} + 6 + 10$

$2x = 2e^{5t} - 12e^{-t} + 16$

$x = e^{5t} - 6e^{-t} + 8$

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Exam tip: Always verify your solutions by substituting back into the original equations to check they satisfy both differential equations and the initial conditions.

Worked example 2: Chemical reaction problem

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Question: In a chemical reaction, substance $X$ decays into substance $Y$ , which itself decays.

The rate of decay of $X$ , in grams per hour, is given by twice the amount of substance $Y$ , in grams.
The rate of change of $Y$ , in grams per hour, is given by the amount of substance $X$ , in grams, minus three times the amount of substance $Y$ , in grams.

a) Set up two differential equations for $x$ and $y$ , the amounts in grams, of substances $X$ and $Y$ respectively.

b) Given that initially $x = 20$ and $y = 0$ , solve for $x$ and $y$ at time $t$ hours.

c) Prove that there can never be equal amounts of $X$ and $Y$ .

Solution:

Part a) Setting up the equations

Let x = amount of substance $X$ (in grams) and y = amount of substance $Y$ (in grams).

Since $X$ decays, the rate must be negative:

$\frac{dx}{dt} = -2y$

The rate of change of $Y$ is:

$\frac{dy}{dt} = x - 3y$

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Exam tip: Remember to include the negative sign when a quantity is decaying. The rate of decrease is negative.

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Part b) Solving the system

Differentiate the second equation:

$\frac{d^2y}{dt^2} = \frac{dx}{dt} - 3\frac{dy}{dt}$

Substitute $\frac{dx}{dt} = -2y$ :

$\frac{d^2y}{dt^2} = -2y - 3\frac{dy}{dt}$

$\frac{d^2y}{dt^2} + 3\frac{dy}{dt} + 2y = 0$

The characteristic equation is:

$m^2 + 3m + 2 = 0$

$(m + 1)(m + 2) = 0$

So $m = -1$ or $m = -2$ .

The general solution is:

$y(t) = Ae^{-t} + Be^{-2t}$

Using initial conditions:

At $t = 0$ : $y(0) = 0$

$A + B = 0 \Rightarrow A = -B$

From the second equation: $x = \frac{dy}{dt} + 3y$

Differentiating $y$ :

$\dot{y} = -Ae^{-t} - 2Be^{-2t}$

So:

$x(t) = -Ae^{-t} - 2Be^{-2t} + 3(Ae^{-t} + Be^{-2t})$

$x(t) = 2Ae^{-t} + Be^{-2t}$

Since $A = -B$ , we can write:

$x(t) = 2Ae^{-t} + Ae^{-2t}$

At $t = 0$ : $x(0) = 20$

$2A + A = 20$

$A = \frac{20}{3}$

Therefore $B = -\frac{20}{3}$ .

The solutions are:

$y(t) = \frac{20}{3}e^{-t} - \frac{20}{3}e^{-2t} = \frac{20}{3}(e^{-t} - e^{-2t})$

$x(t) = \frac{40}{3}e^{-t} + \frac{20}{3}e^{-2t}$

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Exam tip: You could have chosen to differentiate the first equation instead, but it's usually easier to eliminate $\frac{dx}{dt}$ from this equation as it appears without other terms.

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Part c) Proving $X$ and $Y$ can never be equal

Suppose at some time $t$ there are equal amounts: $x(t) = y(t)$ .

$\frac{40}{3}e^{-t} + \frac{20}{3}e^{-2t} = \frac{20}{3}e^{-t} - \frac{20}{3}e^{-2t}$

$\frac{40}{3}e^{-t} + \frac{40}{3}e^{-2t} = \frac{20}{3}e^{-t}$

$40e^{-t} + 40e^{-2t} = 20e^{-t}$

$20e^{-t} + 40e^{-2t} = 0$

$20e^{-t}(1 + 2e^{-t}) = 0$

But $e^{-t} > 0$ for all values of $t$ (exponential functions are always positive), and $(1 + 2e^{-t}) > 0$ for all $t$ .

Therefore this equation is impossible, which means $x$ and $y$ can never be equal.

Answer in context: There can never be equal amounts of substances $X$ and $Y$ at any time during the reaction.

Worked example 3: Predator-prey population dynamics

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Question:
An isolated island supports populations of sparrowhawks and finches.

The number of sparrowhawks increases at a rate proportional to the number of finches. When there are 64 finches present, the rate of increase of sparrowhawks is 16 per year.
If there are no sparrowhawks present, then the finch population would increase by $120\%$ per year.
If there are sparrowhawks present, then, on average, each sparrowhawk kills $1.44$ finches per year.

a) Set up two differential equations to model the population of sparrowhawks and finches.
Initially there are $75$ sparrowhawks and $120$ finches.

b) Find the number of sparrowhawks and finches $t$ years later.

c) What happens to the populations of finches and sparrowhawks in the distant future?

Solution

Part (a): Setting up the model

Let
$s(t)$ = number of sparrowhawks
$f(t)$ = number of finches
$t$ = time in years

Sparrowhawks

The rate of increase of sparrowhawks is proportional to the number of finches:

\frac{ds}{dt} = kf

When $f = 64$ , $\frac{ds}{dt} = 16$ :

16 = 64k \quad \Rightarrow \quad k = 0.25

So,

\frac{ds}{dt} = 0.25f

Finches

With no sparrowhawks present:

\frac{df}{dt} = 1.2f

With sparrowhawks present:

\frac{df}{dt} = 1.2f - 1.44s

Part (b): Solving the system

The system is:

\frac{ds}{dt} = 0.25f \tag{1}

\frac{df}{dt} = 1.2f - 1.44s \tag{2}

Differentiate equation (2):

\frac{d^2 f}{dt^2} = 1.2\frac{df}{dt} - 1.44\frac{ds}{dt}

Substitute $\frac{ds}{dt} = 0.25f$ from (1):

\frac{d^2 f}{dt^2} = 1.2\frac{df}{dt} - 0.36f

\frac{d^2 f}{dt^2} - 1.2\frac{df}{dt} + 0.36f = 0

The characteristic equation is:

m^2 - 1.2m + 0.36 = 0

m = \frac{1.2 \pm \sqrt{1.44 - 1.44}}{2} = 0.6

This is a repeated root, so:

f(t) = (A + Bt)e^{0.6t}

From equation (2):

s = \frac{1.2f - \frac{df}{dt}}{1.44}

Differentiate $f(t)$ :

\frac{df}{dt} = [0.6A + B + 0.6Bt]e^{0.6t}

Substitute:

s(t) = \frac{[0.6A - B + 0.6Bt]e^{0.6t}}{1.44}

Applying initial conditions

At $t = 0$ :

f(0) = 120 \Rightarrow A = 120

At $t = 0$ :

s(0) = 75

75 = \frac{0.6(120) - B}{1.44}

75 = \frac{72 - B}{1.44}

108 = 72 - B

B = -36

Therefore,

f(t) = (120 - 36t)e^{0.6t}

s(t) = \frac{[72 + 36 - 21.6t]e^{0.6t}}{1.44}

s(t) = (75 - 15t)e^{0.6t}

Part (c): Long-term behaviour

The finch population becomes zero when:

120 - 36t = 0 \quad \Rightarrow \quad t = \frac{120}{36} = \frac{10}{3}

The sparrowhawk population becomes zero when:

75 - 15t = 0 \quad \Rightarrow \quad t = 5

Conclusion:
The finch population dies out after $\frac{10}{3}$ years. After this point, sparrowhawks have no food source, so their population would also decline. In reality, the model would no longer apply once the finch population reaches zero, but it predicts the eventual extinction of both populations.

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Exam tip: After finding your solution, always interpret it in context. If a population becomes negative, explain that the model breaks down and describe what would actually happen in the real situation.

Problem-solving strategy for coupled systems

When solving problems involving coupled differential equations, follow this systematic approach:

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Five-Step Strategy:

Define any variables that you need. Be clear about what each variable represents, including units where appropriate.
Use the information in the question to set up the coupled differential equations. Translate word descriptions into mathematical relationships carefully, paying attention to signs (growth vs decay, increase vs decrease).
Differentiate one of the equations and use the original equations to eliminate the other variable to obtain a second-order differential equation. Choose the equation that's easier to differentiate or that makes substitution simpler.
Solve the second-order differential equation and hence solve for both variables. Use the characteristic equation method, find the complementary function and particular integral, then apply initial conditions.
Interpret your solution in the context of the question. Explain what your mathematical answer means in real-world terms. Check if the solution makes physical sense (e.g., populations shouldn't be negative).

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Exam tips:

Always include negative signs for decay or decrease
Choose the easier differentiation path (usually the one with simpler terms)
Use initial conditions to find all unknown coefficients
Check your solutions satisfy both original equations
Answer questions in context, explaining the meaning of your results

Remember!

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Key Points to Remember:

Coupled differential equations involve multiple variables whose rates of change depend on each other. You solve them by converting to a single second-order differential equation.
The solution method has four key steps: choose an equation to differentiate, substitute to eliminate one variable, solve the resulting second-order equation, then find the other variable.
Predator-prey problems model populations where one species hunts another. The prey equation typically has a growth term and a predation term (negative), while the predator equation has terms depending on food availability.
Always interpret solutions in context. If populations become negative, explain that the model breaks down. Consider what happens in the long term and what this means in reality.
Use initial conditions carefully to find all unknown coefficients. You'll often need to use both original equations and the differentiated form to generate enough simultaneous equations.

Coupled Equations (AQA A-Level Further Maths): Revision Notes