Pc18 - Coupled first order equations | OCR A-Level Further Mathematics B (MEI)

When two changing quantities affect one another, their differential equations must be solved together. Pc18 turns a coupled pair of first-order linear equations into one second-order equation, recovers the other variable and applies consistent initial data. You will also interpret population-rate models and explain the long-term behaviour of a coupled system.

Two quantities sharing two rate equations

Two processing compartments contain amounts $x$ and $y$ of a chemical. Their model is $\dot{x}=-2x+y+6$ , $\dot{y}=x-2y+3$ . If they start at $x=8$ , $y=5$ , can the second amount initially rise even though both amounts eventually settle below their starting values? We will solve the model and find out.

The first rate is not determined by $x$ alone; it also depends on $y$ . The same is true in reverse. Population models work in a similar way: a species' own growth may be reduced by the presence of a competitor.

Coupled first-order linear equations

A coupled system uses the same independent variable, here time $t$ , with dependent variables $x$ and $y$ influencing the rate equations together. A constant-coefficient first-order linear system has the form

\dot{x}=ax+by+f(t),\qquad \dot{y}=cx+dy+g(t)

where $a,b,c,d$ are constants and $f,g$ are specified functions of time. Terms such as $xy$ or $x^2$ would make the system non-linear. A dot means differentiation with respect to $t$ .

Translate each rate statement separately. Growth proportional to the same population gives a positive self-term; a competitor reducing that growth gives a negative term involving the other population. Do not confuse a rate with the population itself.

Worked example 1: competition and initial rates

Question. In a model, each of two species would grow at a rate $0.1$ times its population per year if alone. Competition reduces the growth rate of species X by $0.2y$ and that of species Y by $0.05x$ . Form the rate equations and find the initial rates when $x=100$ , $y=40$ .

Plan. Write each independent growth term, subtract its competition term and substitute the starting populations.

\begin{aligned} \dot{x}&=0.1x-0.2y && \text{X growth minus Y effect} \\ \dot{y}&=0.1y-0.05x && \text{Y growth minus X effect} \\ \dot{x}(0)&=0.1(100)-0.2(40)=2 && \text{initial X rate} \\ \dot{y}(0)&=0.1(40)-0.05(100)=-1 && \text{initial Y rate} \end{aligned}

Why this step? The competition term involves the other species, not the species whose rate is being calculated.

Answer. The displayed equations model the competition. Initially X increases by $2$ animals per year and Y decreases by $1$ animal per year.

Check. Without competition the rates would be $10$ and $4$ . The reductions are $8$ and $5$ , giving the calculated net rates.

Worked example 2 changes only the initial Y population, from $40$ to $60$ . Both rate signs reverse even though the model's coefficients are unchanged.

Worked example 2: a different starting balance

Question. For the same model, find the initial rates when $x=100$ , $y=60$ .

Plan. Retain both rate equations and substitute the new initial state.

\begin{aligned} \dot{x}(0)&=0.1(100)-0.2(60)=10-12=-2 && \text{new competition on X} \\ \dot{y}(0)&=0.1(60)-0.05(100)=6-5=1 && \text{new Y growth} \end{aligned}

Why this step? A system's behaviour depends on its starting state as well as its coefficients. A change in one population can change both rates.

Answer. X initially decreases by $2$ animals per year and Y increases by $1$ animal per year.

Check. Raising $y$ by $20$ changes $\dot{x}$ by $-0.2(20)=-4$ and $\dot{y}$ by $0.1(20)=2$ , matching the changes from the first example.

These equations are mathematical approximations to competing populations. If a solution later predicts a negative population, it has left the physically meaningful range; do not interpret negative values as actual animals.

Watch out

Wrong: from $\dot{x}<0$ , conclude that $x<0$ . Right: a negative derivative means the population is decreasing at that instant; the population itself can still be positive.

The next model uses different growth and competition coefficients.

Eliminate one variable, then recover it

Differentiating one first-order equation introduces a second derivative; the other equation then lets you eliminate the unwanted variable.

You already know

A constant-coefficient second-order equation is solved using its auxiliary equation, with a particular integral when the right side is non-zero. The resulting two constants must describe the whole coupled system, not two unrelated solutions.

If the first equation is $\dot{x}=ax+by+f(t)$ , with $b\ne0$ , it also gives a way back to $y$ . Keep that rearrangement before differentiating: you will need it both for elimination and for recovery.

Elimination and recovery

From

\dot{x}=ax+by+f(t),\qquad b\ne0,

write

y=\frac{\dot{x}-ax-f(t)}b

Differentiate the first equation, substitute the given equation for $\dot{y}$ , then substitute this expression for $y$ . Solve the resulting second-order equation for $x$ , and recover $y$ from the original first-order equation. If $b=0$ , use the other equation for elimination when possible, or solve the resulting one-way system in sequence.

Not in the formula booklet: learn it.

Recovering the second variable from an original equation ensures the constants remain linked. It also gives a useful check against the other original equation.

Worked example 3: two distinct exponential modes

Question. Find the general solution of

\dot{x}=2x+y,\qquad \dot{y}=x+2y.

Plan. Eliminate $y$ using $y=\dot{x}-2x$ , solve for $x$ , then recover $y$ .

\begin{aligned} \ddot{x}&=2\dot{x}+\dot{y} && \text{differentiate the first equation} \\ &=2\dot{x}+x+2y && \text{substitute the second} \\ &=2\dot{x}+x+2(\dot{x}-2x) && \text{eliminate }y \\ \ddot{x}-4\dot{x}+3x&=0 && \text{collect terms} \\ r^2-4r+3&=(r-1)(r-3)=0 && \text{auxiliary equation} \\ x&=Ae^t+Be^{3t} && \text{general solution} \\ y&=\dot{x}-2x=-Ae^t+Be^{3t} && \text{recover with the same constants} \end{aligned}

Why this step? Differentiation alone does not eliminate $y$ ; both original equations are needed before the second-order equation contains only $x$ .

Answer. $x=Ae^t+Be^{3t}$ , $y=-Ae^t+Be^{3t}$ .

Check. Differentiating $y$ gives $-Ae^t+3Be^{3t}$ , while $x+2y$ gives exactly the same expression.

Worked example 4 changes only the coefficient of $x$ in the second equation from $1$ to $4$ . One auxiliary root becomes zero, so the corresponding term is a constant.

Worked example 4: a zero auxiliary root

Question. Find the general solution of $\dot{x}=2x+y$ , $\dot{y}=4x+2y$ .

Plan. Use the same recovery relation but repeat the elimination with the changed coefficient.

\begin{aligned} \ddot{x}&=2\dot{x}+4x+2y && \text{differentiate and substitute} \\ &=2\dot{x}+4x+2(\dot{x}-2x) && \text{eliminate }y \\ \ddot{x}-4\dot{x}&=0 && \text{the }x\text{ terms cancel} \\ r(r-4)&=0\quad\Longrightarrow\quad r=0,4 && \text{auxiliary roots} \\ x&=A+Be^{4t} && e^{0t}=1 \\ y&=\dot{x}-2x=-2A+2Be^{4t} && \text{recover the second variable} \end{aligned}

Why this step? A zero root supplies a constant solution mode. Omitting it would leave only one arbitrary constant for a system needing two.

Answer. $x=A+Be^{4t}$ , $y=-2A+2Be^{4t}$ .

Check. $\dot{y}=8Be^{4t}=4x+2y$ ; the constant terms cancel.

Under “Show that”, the eliminated differential equation and its auxiliary equation must be visible. Jumping directly to a supplied exponential expression does not establish it.

Exam tip

Show the differentiation, both substitutions and the second-order equation before solving it. After recovery, simplify the expression for the second variable rather than leaving an avoidable unsimplified derivative.

The next system changes the diagonal coefficients as well as the coupling, so repeat the elimination.

Watch out

Wrong: after $x=A+Be^{4t}$ , take $y=C+De^{4t}$ with independent $C,D$ . The original equations link the constants. The correct recovery is $y=\dot x-2x=-2A+2Be^{4t}$ , so $C=-2A$ and $D=2B$ .

Use an original equation to recover the second variable with the same constants.

Initial data must describe one shared state

Giving $x(0)$ and $y(0)$ already determines both initial derivatives through the original equations.

For $\dot{x}=2x+y$ , $\dot{y}=x+2y$ , the general solution found above gives

x(0)=A+B,\qquad y(0)=-A+B.

These are two equations for the same two constants. Alternatively, use the original system to calculate $\dot{x}(0)$ , then apply $x(0)$ and $\dot{x}(0)$ to the second-order solution.

Key idea

There are two arbitrary constants for the coupled pair as a whole. The recovered $y$ must use the same constants as $x$ ; it does not introduce two new independent constants.

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Pc18 - Coupled first order equations

Two quantities sharing two rate equations

Eliminate one variable, then recover it

Initial data must describe one shared state

Continue this lesson.

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