An explicit formula for a flat metric compatible to certain polynomial vector field with center

Question

Let $X$ be the following vector field on the plane:

$$\begin{cases} x'=y\\ y'=-x-x^3\end{cases}\;\;\;\;\;(X)$$

The vector field $ (X)$ has a non isochronous center at the origin.The proof is given in Remark $2$ below. The punctured plane is filled with periodic orbits of $X$. The vector field is geodesible on the punctured plane, in the sense that there is a Riemannian metric such that solutions of $X$ are geodesics of the metric. The reason of geodesibility of $X$ is explained for some similar system in this post. In fact the $1\_$ form $\psi=d\theta $ works for this vector field $X$.That is $d\psi=0$ and $\psi(X)>0$. This is equivalent to geodesibility in dimension $2$.

Question: What is an explicit flat Riemannian metric such that the trajectories of $X$ are unparametrized geodesics?

Remark 1 Note that because of the following interesting comment of Prof. Goodwillie we should not expect that $X$ has a constant length, since the center is not isochronous

Extension of a vector field to an orthonormal frame for a flat metric

Remark 2: The period is not a constant for periodic orbits surrounding the origin. A proof can be given as follows:

If we compute the integral of signed curvature along a closed orbit of period $T$, we obtain $2\pi=T+\int_0^T \frac{y^2(t)x^2(t)}{y^2(t)+(x(t)+x^3(t))^2}dt$, using Gauss Bonnete theorem. Since the integrant of the latter integral is positive and tends zero when closed orbits tend origin , then period $T$ should depend on choosing the periodic orbits, namely the period $T$ can not be a constant. So the center is not isochronous.

Answer 1

Since the metric doesn't have to extend to the origin, take the flat metric $$ g = \frac{\bigl(\mathrm{d}\left(x\sqrt{1+x^2/2}\right)\bigr)^2 + \mathrm{d}y^2}{x^2+x^4/2+y^2}. $$ The level curves $x^2+x^4/2+y^2 = r^2$ are geodesics for the metric, and these are the integral curves of the vector field $X$.