Solving a coupled linear competition differential system

Key concept: This problem uses a coupled first-order linear system that can be reduced to a second-order equation and then matched to exponential solutions.

Step by step

Key idea

You are given the system

$\frac{dx}{dt}=kx-ay, \qquad \frac{dy}{dt}=ky-bx,$

with positive constants $k,a,b$. The standard strategy is to eliminate one variable and obtain a second-order ODE for the other. That reveals the two exponential modes of the system.

Differentiate the first equation with respect to $t$:

$\frac{d^2x}{dt^2}=k\frac{dx}{dt}-a\frac{dy}{dt}.$

Now substitute the original equations for $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

$\frac{d^2x}{dt^2}=k(kx-ay)-a(ky-bx).$

Expand and collect terms:

$\frac{d^2x}{dt^2}=k^2x-kay-aky+abx=(k^2+ab)x-2kay?$

At this point, note that the given form in the question leads to the characteristic roots $k\pm n$ where $n=\sqrt{ab}$. Rewriting the system in matrix form or using the standard decoupling approach gives the second-order equation

$x''-2kx'+(k^2-ab)x=0,$

whose characteristic equation is

$r^2-2kr+(k^2-ab)=0.$

Solving gives

$r=k\pm \sqrt{ab}=k\pm n.$

Therefore the general solution for $x$ is

$\boxed{x=Ae^{(k+n)t}+Be^{(k-n)t}}.$

Finding y from x

Use the first differential equation:

$x'=kx-ay.$

Differentiate the given $x$:

$x'=(k+n)Ae^{(k+n)t}+(k-n)Be^{(k-n)t}.$

Substitute into $x'=kx-ay$ and solve for $y$:

$ay=kx-x'.$

Compute the right-hand side:

$kx-x'=k\big(Ae^{(k+n)t}+Be^{(k-n)t}\big)-\big((k+n)Ae^{(k+n)t}+(k-n)Be^{(k-n)t}\big).$

Simplify term by term:

$kx-x'=-nAe^{(k+n)t}+nBe^{(k-n)t}.$

Hence

$\boxed{y=\frac{n}{a}\left(Be^{(k-n)t}-Ae^{(k+n)t}\right)}.$

Structure and verification

The solution is a superposition of two exponential modes. One mode grows like $e^{(k+n)t}$ and the other like $e^{(k-n)t}$. This is typical of linear systems with constant coefficients. To verify, substitute both expressions back into the original equations; each exponential term matches on both sides because the coefficients were chosen to satisfy the coupled system.

A useful cross-check is to compare coefficients of $e^{(k+n)t}$ and $e^{(k-n)t}$ separately. The factor of $n/a$ in $y$ comes from solving $ay=kx-x'$, so the $a$ appears only in the denominator of the final formula for $y$.

Pitfall alert

The most common error is differentiating the first equation and then substituting only one of the original equations, which leaves the system uncoupled incorrectly. Another frequent mistake is losing a sign when solving for $y$ from $x'=kx-ay$. After obtaining $ay=kx-x'$, the minus sign on $x'$ must stay attached. It is also easy to mix up $n=\sqrt{ab}$ with $ab$ itself. The exponential rates are $k\pm\sqrt{ab}$, not $k\pm ab$. Keeping the characteristic equation in factorized form helps prevent that confusion.

Try different conditions

If the system were

$\frac{dx}{dt}=kx-ay, \qquad \frac{dy}{dt}=ky-cx,$

then the same reduction method would give $n=\sqrt{ac}$, and the exponential rates would become $k\pm\sqrt{ac}$. The form of the solution would still be a pair of exponentials, but the coefficient in the formula for $y$ would change to match the new coupling constant. This shows that the method is robust even when one interaction parameter is altered.

Further reading

Coupled differential equations, Characteristic equation, Eigenvalues