Existence of periodic orbit of a nonautonomous system Consider the system 
$$\left\{ \begin{array}
& x' & = & y\\
y' & = & -2y-x^3+1/\sqrt{27} + \epsilon \cos t\\
\end{array}\right.$$
Prove that, if $0<\epsilon \ll 1$, there exists a $2\pi$-periodic orbit of the system. 
I don't know how to proceed. 
The professor solved another similar problem, and he used this argument:
Let $\phi (t;0,x_0,y_0,\epsilon)$ be the solution of the system (with initial conditions $x_0,y_0$), then there exists a $2\pi$-periodic orbit with initial condition $(x_\epsilon, y_\epsilon ) = (0,0) $ if the function 
$$G(x_0,y_0,\epsilon)=\phi (2\pi;0,x_0,y_0,\epsilon)-(x_0,y_0)^t$$
has a zero (because the ODE has no fixed points). And then he uses the implicit function theorem.
I don't understand why this works, and I don't know how to use that in this exercise, because I  don't have the initial condition $x_\epsilon, y_\epsilon$.
Any help is welcome

Thanks you. 
So, if $P:\mathbb{R}^3 \to \mathbb{R}^2$, $P(x,y,\epsilon ) =\phi(2\pi,x,y,\epsilon )$  is the Poincare return map and I define $G(x,y,\epsilon )=P(x,y,\epsilon) - (x,y)^t $, then $G(1/\sqrt 3 , 0 , 0 ) = P(1/\sqrt 3,0,0) -(1/\sqrt 3,0)^t=0$ because $(1/\sqrt 3,0)$ is a fixed point for $\epsilon =0$.
With that, (using my professor's method) I check that G verify the Implicit-FT, so there are neighbourhoods of $(1/\sqrt 3 , 0 , \epsilon ) $ such that $G(1/\sqrt 3 , 0 , \epsilon )=0, \forall \epsilon $. Finally, I deduce that there exists a periodic orbit with initial condition $(1/\sqrt 3,0)$.
But actually I don't understand why the existence of zeros of $G$ in a neighbourhood of the fixed point guarantees the existence of a periodic orbit. 
Thanks again for your help
