= Exponential Dulac multiplier for a quadratic planar system
{title2=$\psi=e^{-2(x+y)/(k-2)}$}
For $\dot x=2x-x^2-y$, $\dot y=y-kx$ with $k>2$, take the positive <Dulac function> in the title. More generally, $\psi=e^{a(x+y)}$ gives
$$
\nabla\cdot(\psi(X,Y))=\psi\bigl[3+(-2+(2-k)a)x-ax^2\bigr].
$$
The equal coefficients in the exponential cancel the $y$ term. Setting $a=-2/(k-2)$ also cancels the linear $x$ term, leaving $\psi(3+2x^2/(k-2))>0$. The <Bendixson-Dulac theorem> excludes <periodic orbits> in the open positive quadrant. The axes are not invariant: the vector field points out of the quadrant at every nonzero axis point. A closed orbit staying in the nonnegative quadrant therefore cannot touch an axis, and cannot pass through the equilibrium at the origin by uniqueness. This extends the exclusion to the closed quadrant. Choosing an exponential multiplier turns a differential search for a useful <Dulac function> into cancellation of polynomial coefficients.
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