I have the Lotka-Volterra equation


$\dot{x}=x(1-y),$


$\dot{y}=y(x-1),$


where $x$ and $y$ are non-negative.
It is easy to see that the $x$- and $y$-axis are invariant sets. I can see from plots that a periodic orbit exists. Now I want to prove this by hand calculations according to the Poincare-Bendixon criterion. 


So, is there any invariant set that is a closed and bounded trajectory and contains no equilibrium point, which I can find by hand calculations? 



 Invariant set of Lotka-Volterra equation
To understand the dynamics of this LV equation you do not even need the existence theorem of ODE (Cauchy-Picard-Lindelöf-Lipschitz). Just apply the implicit function theorem to the function $V(x,y)=x-\log x + y - \log y$. It tells you that for any $c > \min V$ the level sets $\{V=c\}$ are closed simple curves that bounds nbds of the minimum point $(1,1)$ of $V$; these curves suitably parametrized are periodic solutions of the LV equations, turning around the equilibrium $(1,1)$. 
(Alternatively, the fact that the level set $\{V=c\}$ are invariant set for the ODE just comes computing $\frac{d}{dt}V\big(x(t),y(t)\big)= 0$ ).


