Coupled First Order Linear Equations (Edexcel A-Level Further Mathematics): Revision Notes

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8.2.2 Coupled First Order Linear Equations

Introduction to Coupled Differential Equations

Coupled first-order linear equations involve two or more dependent variables, such as f(t)f(t) and r(t)r(t), that interact and influence each other.

These equations often model real-world situations such as predator-prey dynamics, where the populations of two species depend on one another over time.

The general form for a pair of coupled equations is:

dfdt=af+br,drdt=cf+dr\frac{df}{dt} = af + br, \quad \frac{dr}{dt} = cf + dr

where f(t)f(t) and r(t)r(t) are the dependent variables, and a,b,c,da, b, c, d are constants.

To solve these equations:

  1. Rewrite one equation as a second-order equation for one variable.
  2. Solve the second-order equation.
  3. Back-substitute to find the other variable.

Worked Example: Predator-Prey Dynamics

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Problem

In a population of foxes (ff) and rabbits (rr), the dynamics are governed by:

dfdt=3r−6f,drdt=4r−8f\frac{df}{dt} = 3r - 6f, \quad \frac{dr}{dt} = 4r - 8f

The initial populations are f(0)=2f(0) = 2 and dfdt∣t=0=−2\frac{df}{dt}\big|_{t=0} = -2

Find f(t)f(t), r(t)r(t), and the long-term populations of foxes and rabbits.

Step 1: Rewrite as a Second-Order Equation

Start with dfdt=3r−6f\frac{df}{dt} = 3r - 6f

Differentiate both sides with respect to tt:

d2fdt2=3drdt−6dfdt\frac{d^2f}{dt^2} = 3\frac{dr}{dt} - 6\frac{df}{dt}

Substitute drdt=4r−8f\frac{dr}{dt} = 4r - 8f

d2fdt2=3(4r−8f)−6dfdt\frac{d^2f}{dt^2} = 3(4r - 8f) - 6\frac{df}{dt}

Substitute r=dfdt+6fr = \frac{df}{dt} + 6f from the first equation:

d2fdt2=3(4(dfdt+6f)−8f)−6dfdt\frac{d^2f}{dt^2} = 3(4(\frac{df}{dt} + 6f) - 8f) - 6\frac{df}{dt}

Simplify:

d2fdt2=12dfdt+72f−24f−6dfdt\frac{d^2f}{dt^2} = 12\frac{df}{dt} + 72f - 24f - 6\frac{df}{dt}d2fdt2=6dfdt+48f\frac{d^2f}{dt^2} = 6\frac{df}{dt} + 48f

Step 2: Solve the Second-Order Equation

The second-order equation is:

d2fdt2−6dfdt−48f=0\frac{d^2f}{dt^2} - 6\frac{df}{dt} - 48f = 0

Solve the characteristic equation:

m2−6m−48=0m^2 - 6m - 48 = 0

Using the quadratic formula:

m=−(−6)±(−6)2−4(1)(−48)2(1)m = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-48)}}{2(1)} =6±36+1922=6±152= \frac{6 \pm \sqrt{36 + 192}}{2} = \frac{6 \pm 15}{2}m=10.5, m=−4.5m = 10.5, \, m = -4.5

General solution for f(t)f(t)

f(t)=Ae10.5t+Be−4.5tf(t) = A e^{10.5t} + B e^{-4.5t}

Step 3: Use Initial Conditions to Solve for Constants

At t=0t = 0, f(0)=2f(0) = 2

2=Ae0+Be0⇒A+B=22 = A e^{0} + B e^{0} \quad \Rightarrow \quad A + B = 2

At t=0,dfdt=−2t=0, \frac{df}{dt} = -2

Differentiate f(t)f(t)

dfdt=10.5Ae10.5t−4.5Be−4.5t\frac{df}{dt} = 10.5A e^{10.5t} - 4.5B e^{-4.5t}

Substituting t=0t=0

−2=10.5A−4.5B-2 = 10.5A - 4.5B

Solve the simultaneous equations:

A+B=2,10.5A−4.5B=−2.A + B = 2, \quad 10.5A - 4.5B = -2.

Solve for AA and BB:

A=0.5, B=1.5A = 0.5, \, B = 1.5

Final solution for f(t)f(t)

f(t)=0.5e10.5t+1.5e−4.5tf(t) = 0.5e^{10.5t} + 1.5e^{-4.5t}

Step 4: Find r(t)r(t)

Use r=dfdt+6fr = \frac{df}{dt} + 6f

dfdt=10.5(0.5)e10.5t−4.5(1.5)e−4.5t\frac{df}{dt} = 10.5(0.5)e^{10.5t} - 4.5(1.5)e^{-4.5t}r(t)=[5.25e10.5t−6.75e−4.5t]+6[0.5e10.5t+1.5e−4.5t]r(t) = \left[5.25e^{10.5t} - 6.75e^{-4.5t}\right] + 6\left[0.5e^{10.5t} + 1.5e^{-4.5t}\right]

Simplify:

r(t)=8.25e10.5t+1.25e−4.5tr(t) = 8.25e^{10.5t} + 1.25e^{-4.5t}

Step 5: Long-Term Behavior

As t→∞t \to \infty

e10.5t→∞e^{10.5t} \to \infty, suggesting unbounded growth of f(t)f(t) and r(t)r(t) if no damping or external factors are applied.

Note Summary

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Common Mistakes

  1. Confusing roles of variables: Be clear about dependent and independent variables.
  2. Errors in substitution: Check carefully when replacing one variable in terms of another.
  3. Mismanaging constants in solutions: Use initial conditions consistently to find AA and BB.
  4. Skipping interpretation: Always analyse the physical or long-term meaning of the solution.
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Key Formulas

  1. Coupled Differential Equations:
dfdt=af+br,drdt=cf+dr\frac{df}{dt} = af + br, \quad \frac{dr}{dt} = cf + dr
  1. General Solution for Second-Order Linear Equations:
y=Aem1t+Bem2ty = Ae^{m_1t} + Be^{m_2t}
  1. Long-Term Behaviour: Analyse terms emte^{mt}, where mm determines exponential growth, decay, or oscillation.

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