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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.

The nullclines meet at the origin and $P=(2-k,k(2-k))$. At the origin, the Jacobian matrix has trace three, determinant $2-k>0$ and discriminant $1+4k>0$, giving an unstable node. At $P$, its determinant is $k-2<0$, giving a saddle equilibrium. The horizontal component is positive below the parabolic nullcline and negative above it; the vertical component is positive above the straight nullcline and negative below it. These signs determine the directional regions of the phase portrait.

The quadrant is not positively invariant: nonzero points on either axis have an outward-pointing component. Restricting a drawing to that quadrant does not keep its trajectories there. For $k>2$, only the saddle at the origin remains in the quadrant, and an exponential Dulac multiplier for a quadratic planar system excludes nonconstant periodic orbits there.