First of all let us try to solve the ODEs. The you can deduce what you wish from it.

$\dot{x}(t)=y(t)$    
$\dot{y}(t)= - 4 x(t) + y(t)^2$

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.