Coupled first-order differential equationsEdexcel A-Level Further Maths: Revision notes

A coupled system has one independent variable $t$ and two dependent variables, with each derivative depending on both: $\frac{dx}{dt}=ax+by+f(t),\qquad\frac{dy}{dt}=cx+dy+g(t).$ You cannot solve either equation alone, because each contains the other variable. The method is to eliminate one variable to get a second-order equation in the other, solve that, then recover the second variable.

Method:

1. Rearrange one equation to give the other variable: from $\dot{x}=2x+y$, $y=\dot{x}-2x$.
2. Differentiate the same equation: $\ddot{x}=2\dot{x}+\dot{y}$.
3. Substitute into the other equation, $\dot{y}=3x+4y$, to get $\ddot{x}-6\dot{x}+5x=0$. The coefficients come from the matrix: $-(a+d)$ on $\dot{x}$ and $(ad-bc)$ on $x$. For this system $a+d=6$ and $ad-bc=8-3=5$. Solve with the auxiliary equation $m^2-6m+5=0$, $m=1,5$: $x=Ae^{t}+Be^{5t}$. Then $y=\dot{x}-2x=-Ae^{t}+3Be^{5t}$.

If $f(t)$ or $g(t)$ is non-zero, the second-order equation is non-homogeneous, so the solution is complementary function + particular integral. Example: $\dot{x}=-2x+y$, $\dot{y}=x-2y+9$. From the first, $y=\dot{x}+2x$. Then $\ddot{x}=-2\dot{x}+\dot{y}=-2\dot{x}+x-2(\dot{x}+2x)+9$, so $\ddot{x}+4\dot{x}+3x=9.$ CF: $m=-1,-3$. PI: $x=3$. So $x=3+Ae^{-t}+Be^{-3t}$ and $y=\dot{x}+2x=6+Ae^{-t}-Be^{-3t}$. The constants $3$ and $6$ are the equilibrium values, where $\dot{x}=\dot{y}=0$. As $t\to\infty$ the exponential terms vanish and $(x,y)\to(3,6)$.

There are two arbitrary constants overall (not four). Write $x$ with $A$ and $B$, obtain $y$ from the first equation, then use $x(0)$ and $y(0)$ to find both constants. For the system above with $x(0)=y(0)=0$: $3+A+B=0$ and $6+A-B=0$, so $A=-\frac92$, $B=\frac32$, giving $x=3-\frac92e^{-t}+\frac32e^{-3t}$ and $y=6-\frac92e^{-t}-\frac32e^{-3t}$. Check by substituting back into one original equation, e.g. $\dot{x}(0)=-2x(0)+y(0)=0$, and confirm the solution gives that.

Let $x$ be the prey and $y$ the predators. In $\dot{x}=3x-2y$, prey grow by themselves ($+3x$) but are eaten ($-2y$). If $\dot{y}=x$, predators increase with prey and have no death term. With $x(0)=7$ and $y(0)=4$ (hundreds): $\ddot{y}-3\dot{y}+2y=0$, so $y=e^{t}+3e^{2t}$ and $x=e^{t}+6e^{2t}$. Then $x-y=3e^{2t}>0$ (prey always outnumber predators), and both grow without limit. Interpreting: identify the roles from the signs of the cross terms (a term that reduces one population and helps the other signals predator-prey). Comment on limitations: unlimited growth (no carrying capacity), constant rates, and only two species.