Consider the plane $\mathbb{R}^2$ together with a metric $\mathsf{g}$. Choosing suitable local coordinates $(x_1,x_2)$, $\mathsf{g}$ takes the form : $$ \mathsf{g}=r(x_1,x_2)\left((dx_1)^2+\epsilon (dx_2)^2\right),\quad r>0\quad\textrm{and}\quad\epsilon = \pm1. $$
Question : Assume that $\mathsf{g}$ has constant curvature $K$, is there a closed form expression for $r(x,y)$ ? In particular in [1] (p. 331) it is claimed $\mathsf{g}$ is isometrically equivalent to
$$ \mathsf{g} =\frac{1}{\left(1+\frac{K}{4}\left((y_1)^2+\epsilon(y_2)^2\right)\right)^2}\left((dy_1)^2+\epsilon (dy_2)^2\right) $$
Is this true ? and if so can somebody provides a reference or a proof of that result ?
My work : When trying to solve the question myself, I'm stuck with the following non-linear PDE (Liouville equation) $$ \frac{\partial ^2r}{\partial (x_1)^2}r + \epsilon\frac{\partial ^2r}{\partial (x_2)^2}r +\left(\frac{\partial r}{\partial x_1}\right)^2 - \epsilon\left(\frac{\partial r}{\partial x_2}\right)^2 = - K $$
[1] George R. Wilkens. “Centro-Affine Geometry in the Plane and Feedback Invariants of Two-State Scalar Control Systems”. Proceedings of Symposia in Pure Mathematics. Ed. by G. Ferreyra et al. Vol. 64. Amer- ican Mathematical Society, 1998. doi: 10.1090/pspum/064/1654544.
By Minding's Theorem, surfaces with the same constant curvature are locally isometric, and the metric they give you is the standard model metric for constant curvature $K$.