First of all let us try to solve the ODEs. The you can deduce what you wish from it.
First integration, orbits
$\dot{x}(t)=y(t)$
$\dot{y}(t)= - 4 x(t) + y(t)^2$
The orbits obey the equation
$\frac{d y}{d x}=\frac{y^2-4 x}{y}$
Multiplying by $y$ this can be written as
$\frac{1}{2}\frac{d y^2}{\text{dx}}=y^2-4 x$
and is easily integrated to give
$y^2=c \exp (2 x)+4 x+2$
where $c$ is a constant of integration.
If the initial state is $( x(0) = x0$, $y(0) = y0 )$
then
$c=\left(-4 x0+y0^2-2\right) \exp (- 2 \text{x0})$
and the orbit corresponding to this Initial state is
$y^2=\left(-4 \text{x0}+\text{y0}^2-2\right) \exp (2 (x-\text{x0}))+4 x+2$
Now the sign of the factor in front of the exponential
$s=\left(-4 \text{x0}+\text{y0}^2-2\right)$
defines distinguishes between finite ($s<0$) and unbounded ($s>0$) orbits. We therefore call the orbit $s=0$ the separatrix.
Second integration, time dependence, frequency of oscillation
Because of
$dx/dt = y[x] = Sqrt[2 + 4 x + c Exp[2 x]]$
variables can be separated an we get the time $t$ as a function of $x$:
$t=\int_{\text{x0}}^x \frac{1}{\sqrt{c \exp (2 z)+4 z+2}} \, dz+\text{t0}$