Coupled first order equationsAQA A-Level Further Maths: Revision notes

A coupled system has one independent variable $t$ and two dependent variables $x(t)$ and $y(t)$, each rate of change depending on both: $\frac{dx}{dt}=ax+by,\quad\frac{dy}{dt}=cx+dy.$ The signs of the coefficients tell the story. In a predator-prey model with $x$ prey and $y$ predators, a term $-by$ in $\frac{dx}{dt}$ means predators reduce the prey, a term $+cx$ in $\frac{dy}{dt}$ means more prey help the predators grow, and $-dy$ means predators die off without food. Models of connected tanks, chemical reactions and competing species have the same form. The point $x=y=0$ is an equilibrium, where both rates are zero.

To solve, turn the pair into one second order equation. Take $x'=ax+by$ and $y'=cx+dy$.

1. Rearrange the first equation for $y$: $y=\frac{x'-ax}{b}$ (or differentiate it first).
2. Differentiate the first equation: $x''=ax'+by'$.
3. Substitute $y'=cx+dy$, then replace $y$ using step 1. The result is $x''-(a+d)x'+(ad-bc)x=0$: the coefficient of $x'$ is minus the trace and the constant is the determinant of the coefficient matrix.

Solve the second order equation for $x$ with the auxiliary equation. Then find $y$ from the first equation ($y=\frac{x'-ax}{b}$), not by solving a second equation independently, because that would add two more constants that are not free. Example. $x'=3x+2y$, $y'=x+2y$. Then $x''-5x'+4x=0$, with $m=1,4$, so $x=Ae^{t}+Be^{4t}$. Then $y=\frac12(x'-3x)=-Ae^{t}+\frac12Be^{4t}$. Check: $y'=-Ae^{t}+2Be^{4t}$ and $x+2y=-Ae^{t}+2Be^{4t}$.

Two conditions, such as $x(0)$ and $y(0)$, fix $A$ and $B$. Substitute $t=0$ into both $x$ and $y$ to get two simultaneous equations. Example. With $x'=-2x+y$, $y'=2x-3y$, $x(0)=3$, $y(0)=0$: $x=Ae^{-t}+Be^{-4t}$ and $y=x'+2x=Ae^{-t}-2Be^{-4t}$. Then $A+B=3$ and $A-2B=0$ give $B=1$, $A=2$. If the conditions are given as $x(0)$ and $y(0)$, you can use the equation to convert $y(0)$ into $x'(0)$.

• Real distinct roots: a sum of exponentials, so the populations grow or decay without oscillating. For large $t$ the larger root dominates.
• Complex roots $p\pm iq$: oscillating solutions $e^{pt}(\ldots\cos qt+\ldots\sin qt)$. With $p<0$ the cycles die away to equilibrium; with $p>0$ they grow, which is unrealistic for long times.
• In predator-prey cycles the predator peak lags the prey peak, because predators grow only once prey are plentiful. Always say what the result means in context (populations, masses) and use the correct units. A model with negative populations has passed beyond its limit of validity.