Solution trajectories of a plane autonomous system I have the plane autonomous system 
$\dfrac{dx}{dt}=x(1-2x-y)$
$\dfrac{dy}{dt}=y(1-x-2y)$
I need to show that the axes of the phase plane and the line $x=y$ are solution trajectories, but I don't know how to do this.
The other part of the question is as follows.
Use the Bendixson-Dulac theorem with $phi=\dfrac{1}{xy}$ to show that there are no closed trajectories in $R=\{(x,y):x>0, y>0\}$ . (I've done this bit) Comment on whether you can prove that there are no periodic functions in the entire phase plane including the origin.
For the last part, I have plotted the phase plane and there are no closed trajectories but how can i prove the last bit properly?
Thanks for any help.
 A: The first part of this question is trivial. The axes of the phase plane correspond to $x=0$ and $y=0$. On substituting $x=0$ in the first equation, we get $\frac{dx}{dt} = 0$. i.e the particle stays on the $y$-axis. Similarly for the $x$-axis. 
For $x=y$ line, we find that $\frac{dx}{dt} - \frac{dy}{dt} = 0$. i.e. $v_x = v_y$ and hence, the particle continues on the line.
I am trying the last part of your problem.
A: To apply the Bendixson-Dulac criteria we should calculate for a suitable $\phi(x,y)$
$$
\psi(x,y)=\frac{\partial}{\partial x}\left(\phi(x,y)\dot x\right)+\frac{\partial}{\partial y}\left(\phi(x,y)\dot y\right)
$$
and then verify if $\psi(x,y)$ changes sign into the quadrant $Q=\{(x,y),x>0,y>0\}$.
Choosing $\phi(x,y) = x^my^n$ we obtain
$$
\psi(x,y) = x^m y^n (y (m+2 n+5)+2 m x+m-n x+n+3 x+2)
$$
and then choosing again $m=-\frac{11}{5},n=-\frac 75$ we get
$$
\psi(x,y) = -\frac 85
$$
which obviously doesn't changes sign into $Q$. Concluding, there are not closed trajectories inside $Q$.
