How do I show that there exists a nontrivial periodic solution? How do I show that there exists a nontrivial periodic solution for this:

$z'' + [log(z^2 + 4(z')^2)]z' + z= 0$?

Here's how I tried it:
Set $x=z$ and $y=z'$. If it can be shown that $dy/dx$ is a closed orbit with no equilibrium point on it, then $(x(t),y(t))$ must be periodic, hence $z$ is periodic. However, I have trouble with showing that $(x(t),y(t))$ is a closed curve. How do I prove it?
 A: The term inside the logarithm looks similar to the total energy of a mass-spring system, the sum of potential plus kinetic energy. If we try a solution of the form
$$
z(t) = A \sin(\omega t + \theta),
$$
then
$$
z'(t) = A \omega \cos(\omega t + \theta).
$$
This solution might only be correct if $\log(z^2+4z'^2)=0$, which implies $A=1$ and $\omega=\frac12$, such that
$$
z(t)^2 + 4 z'(t)^2 = \sin^2(\omega t + \theta) + \cos^2(\omega t + \theta) = 1.
$$
However the linearised differential equation would then look like
$$
z'' + z = 0,
$$
which has the general solution
$$
z(t) = C_1 \sin(t) + C_2 \cos(t),
$$
which has a different frequency then the assumed solution. However this does show the range at which the period and amplitude might occur, namely the fundamental frequency will probably be between $1$ and $\frac12$ and an amplitude slightly smaller than $1$, such that the linearised differential equation will have stable eigenvalues roughly one half of its period and unstable eigenvalues the other half.
This is only a partial answer, but does hopefully give you some insights.
