 Solution

Let $k>2$ and choose the positive Dulac functionMore generally, for $\psi =e^{a(x+y)}$, direct differentiation givesTaking $a=-2/(k-2)$ cancels the linear term and makes the quadratic term nonnegative:The open positive quadrant is simply connected, so the Bendixson-Dulac theorem excludes every nonconstant periodic orbit entirely inside it. Such an orbit cannot touch $x=0,y>0$ while staying in the closed quadrant, since its differentiable coordinate $x$ would have a local minimum there but $\dot{x}=-y\ne 0$. The same argument applies on $y=0,x>0$. It cannot pass through the origin either, by local uniqueness of this smooth autonomous system. Therefore there is no nonconstant closed trajectory in the entire nonnegative quadrant. The calculation illustrates an exponential Dulac multiplier for a quadratic planar system.