If $c_{2} \neq 0$, this equation gives $V(t, \psi)$ for any smooth function $F(t, \psi)$, and the metric (\[light-like metric in t-psi coordinates\]) becomes

$$\label{the metric in Gaussian coordinates}
dS^2 = \left(\frac{\dot{F}}{F'}\right)^2 dt^2 - {\bar{{\cal{B}}}}_{i j}(t, \psi)dx^{i} dx^{j} - d\psi^2.$$

If $c_{2} = 0$, then (\[obtaining V from F in Gaussian coord.\]) is an equation for $F$. Integrating it we find $$\label{F in Gaussian coordinates, c2 = 0}
F = \left[l(t) + \psi h(t)\right]^{a^2/(a + b)},$$ where $l(t)$ and $h(t)$ are arbitrary differentiable functions.

#### Synchronous reference system:

The choice $g_{00} = 1$ is usual in cosmology: it corresponds to the so-called synchronous reference system where the coordinate $t$ is the proper time at each point. Thus, setting $\dot{V} = (1/{\bar{{\cal{A}}}}\dot{F})$, the line element (\[light-like metric in t-psi coordinates\]) becomes

$$\label{cosmological homogeneous solution in synchronous coordinates }
dS^2 =  dt^2 - {\bar{{\cal{B}}}}_{i j}(t, \psi)dx^{i} dx^{j} - \left(\frac{F'}{\dot{F}}\right)^2d\psi^2.$$

In these coordinates (\[diagonal 5D metric\]) reduces to $V' = - (F'/{\bar{{\cal{A}}}}{\dot{F}}^2)$ and $(\partial \dot{V}/\partial \psi) = (\partial {V'}/\partial t)$ yields $$\label{obtaining V from F in Synchronous coord.}
a^2\left(c_{1}F + c_{2}V\right) \ddot{F} - 2c c_{1} \dot{F}^2 = 0.$$ Thus, for $c_{2} = 0$ we get $$\label{F in the synchronous coordinates}
F = \left[M(\psi) 