Stability of the system without equilibrium points

Question

What can be said about the stability of the system if it does not have an equilibrium point ? e.g.

$x' = x^2 + y^2 + 1, \tag{1}$ $y' = x^2 - y^2, \tag{2}$

Answer 1

Changing variables

$$ u = x+y\\ v = x-y $$

results the system

$$ u' = \frac 12(u+v)^2+1\\ v' = \frac 12(u-v)^2+1 $$

so $u(t), v(t)$ grow without limit. An unstable system.

Answer 2

This is just a motivation for Cesareo's answer. If you add/subtract both equations

$$\dot{x} + \dot{y} = 2x^2+1$$ $$\dot{x} - \dot{y} = 2y^2+1.$$

It should be obvious that we can introduce $u = x + y \implies \dot{u} = \dot{x}+\dot{y}$ and $v = x-y\implies \dot{v} = \dot{x} - \dot{y}$. The inverse substitution is given by $x = 0.5(u+v)$ and $y=0.5(u-v)$.

We will obtain the differential equations

$$\dot{u} = 0.5(u+v)^2+1$$ $$\dot{v} = 0.5(u-v)^2+1.$$