Why is this system reversible? What does this mean?

Consider the system $$\dot{x}=y,\qquad\dot{y}=-x+y^2.$$ Then, it is said that the system is reversible $(t\to -t, y\to -y)$. What does this mean?

If I put this into the equations, I get $$\dot{x}(-t)=y(-t),\qquad \dot{y}(-t)=-x(-t)+y(-t)^2.$$

So does reversible mean here that when replacing $t$ by $-t$ and $y$ by $-y$, the differential equations still hold?

Notice that $$\dot{x} = \frac{dx}{dt},\dot{y} = \frac{dy}{dt}$$So$$t\rightarrow -t \Rightarrow (\dot{x} \rightarrow -\dot{x})$$
$$t\rightarrow -t \text{ and } y \rightarrow -y \Rightarrow \dot{y} \rightarrow \dot{y}$$ Getting back to the system, making the given substitution yields,
$$-\dot{x} = -y\Rightarrow \dot{x} = y$$ $$\dot{y} = -x+(-y)^2 \Rightarrow \dot{y} =-x+y^2$$
• How do you know that $t \rightarrow -t \Rightarrow (\dot x \rightarrow - \dot x)$? – Oliver G Jun 27 '17 at 19:43