Show that the system $\ddot x+x\dot x+x=0$ is reversible Show that the system $\ddot x+x\dot x+x=0$ is reversible
My attempt was to use a property from the book that if the system 
$\dot x = f(x,y)$
$\dot y = g(x,y)$
is reversible then $f(x,-y)=-f(x,y)$ and $g(x,-y)=g(x,y)$.
I tried to convert the system into a system like the one above by letting
$\dot x=y$
$\ddot x=\dot y=-x-xy$
However, this system does not satisfy the second criteria, namely that $g(x,y)=-x-xy \ne g(x,-y)=-x+xy$
Is there another way to show that this system is reversible?
 A: To combine the (very useful!) comments of @Did and @Evgeny in an answer:
The definition of reversibility I'm using is that of time reversal symmetry, i.e. the motion described by the system backward in time is equivalent to that forward in time (I'm following James D. Meiss, Differential Dynamical Systems, SIAM MM14, 2007, p. 212 here). That means that there exists a diffeomorphism $S$ that conjugates the flow $\phi_t$ of the dynamical system with its inverse, so that $(\phi_{-t} \circ S)(z) = (S \circ \phi_t)(z)$, for a dynamical system $\dot{z} = f(z)$. A bit more concretely in terms of the dynamical system itself: this implies that
\begin{equation}
 -f(S(z)) = D S(z) f(z).
\end{equation}
So, what's this diffeomorphism $S$ in the case at hand? Well, the idea is that we're looking for a coordinate transformation on $(x,y)$ that will give us 'the same' dynamical system as we originally had, before we inverted the time direction.
Inverting time, i.e. mapping $t \mapsto -t$ leads to the system
\begin{align}
-\dot{x} &= y, \\ - \dot{y} &= -x(1+y).
\end{align}
We now see that there's a particularly simple coordinate transformation that will return the original system, namely $(x,y) \mapsto (-x,y)$. Using this coordinate transformation, we obtain
\begin{align}
\dot{x} &= y, \\ - \dot{y} &= x(1+y),
\end{align}
which is indeed equivalent to the original dynamical system. That means we've found our diffeomorphism $S$, which acts as $S(x,y) = (-x,y)$. Note that, as $S$ is linear, $D S = S$. Moreover, $S$ is an involution, i.e. $S \circ S = \text{Id}$.
Admittedly, this is somewhat different from the 'usual' notion of time reversibility. In that case, the system would be invariant under simultaneously changing the direction of time ($t \mapsto -t$) and the direction of the velocity ($y = \dot{x} \mapsto -y$). This is for example the case with the ODE $\ddot{x} + \dot{x}^2 + x = 0$. Also, this holds for every Hamiltonian system.
