Solution
= Solution
With $y=\dot x$, the system is
$$
\dot x=y,
\qquad
\dot y=-3y+4\log\frac{x^2+1}{2}.
$$
At equilibrium $y=0$ and $x^2=1$, so the equilibrium points are $(1,0)$ and $(-1,0)$.
= Solution
With $y=\dot x$, the system is
$$
\dot x=y,
\qquad
\dot y=-3y+4\log\frac{x^2+1}{2}.
$$
At equilibrium $y=0$ and $x^2=1$, so the equilibrium points are $(1,0)$ and $(-1,0)$.