Prove that this system has no periodic orbits Question:
Show that the system
\begin{align}
\frac{dx}{dt} & = 10x-0.1xy-0.02x^2+1 \\
\frac{dy}{dt} & = -10y+0.1xy+1
\end{align}
has no periodic orbits

Attempt:
The first thing I tried was the Bendixson-Dulac Theorem, but I have yet to find the right function for it.
Next, I tried to perform linear analysis. There are three fixed points, one of which is a stable spiral whilst the other two are saddles.
Unfortunately, this is not enough (I think) to conclude that there should be no periodic solutions, since there may be a limit cycle around the stable spiral.
Any hints?
(Note: It doesn't have to be a completely mathematically rigorous proof. A wordy argument providing good intuition would be great. Thanks!)
This is what the phase plane looks like:

The red and green lines are the $x$- and $y$- nullclines respectively.

Making
$$
X = x-t\\
Y = y-t
$$
we have the equivalent system
$$
\dot X = (a+b y + c x)x\\
\dot Y = (d+e x)y
$$
so for $\dot X = 0$ and $\dot Y = 0$ we have that the axes $x=0$ and $y = 0$ are orbits.
now proposing a test function $\phi(\eta,\mu) = \eta^{\alpha}\mu^{\beta}$ and proceeding according to the Bendixson-Dulac tecnique
$$
(\phi(x,y)\dot X)_x + (\phi(x,y)\dot Y)_y = \phi(x,y)\left((2 + \alpha) c + (1 + \beta) e)x+b(1 + \alpha) y+a + a \alpha + d + \beta d\right)
$$
now choosing conveniently $\alpha = -1, \beta = -\frac{c+e}{e}$ we get at
$$
(\phi(x,y)\dot X)_x + (\phi(x,y)\dot Y)_y = -\frac{c d}{e}\phi(x,y) \ne 0
$$
inside the quadrants defined by $x=0$ and $y = 0$. Concluding that there are no limit cycles.
