Initial-value problem for non-linear partial differential equation $y_x^2=k/y_t^2-1$ For this problem, $y$ is a function of two variables: one space variable $x$ and one time variable $t$.
$k > 0$ is some constant.
And $x$ takes is value in the interval $[0, 1]$ and $t \ge 0$.
At the initial time, $y$ follows a parabolic profile, like $y(x, 0) = 1 - (x-\frac{1 }{2})^2$.
Finally, $y$ satisfies this PDE:
$$ \left(\frac{\partial y} {\partial x}\right)^2  = \frac{k}{\left(\frac{\partial y} {\partial t}\right)^2} - 1.$$
Does anyone have an idea how to solve this problem (and find the expression of $y(x,t)$) ?
About:
The problem arise in physics, when studying the temporal shift of a front of iron particles in a magnetic field.

Edit:
I solved it numerically on a (badly-designed) 1st-order numerical scheme with a small space & time discretization, with the initial condition I wanted (in Octave/Matlab, in Python and in OCaml + GNUplot). The numerical result was enough to confirm the theory and the experiment (the observation done in the lab), so I did not try any further to solve it analytically.
See here for an animation of the front of iron matter, and here for more details (in French).

Your equation can be written as follows:

$$F(p,q,x,t,y) = (p^2+1)q^2 -k = 0, \quad p = y_x, \quad q = y_t,$$ 

and hence:
\begin{align}
F_p & = 2pq^2, \\
F_q & = 2(1+p^2) q,\\
F_t = F_x = F_y & = 0,
\end{align}
so the Lagrange-Charpit equations read:
$$ \frac{\mathrm{d}x}{2 pq^2 }= \frac{\mathrm{d}t}{2(1+p^2)q} = \frac{\mathrm{d}y}{2p^2q^2 + 2(1+p^2)q^2 } =  -\frac{\mathrm{d}p}{0} = - \frac{\mathrm{d}q}{0},$$
which tells you that $\mathrm{d}p = \mathrm{d}q = 0$ and, thus, $p = A$ (or $q = B$) constant. Since $p = y_x$ we have that $y(x,t) = Ax+f(t)$. Plug this into the original PDE to find:
$$f(t) = \pm \frac{\sqrt{k} t}{\sqrt{1+A^2}} + C, \quad C \in \mathbb{R}, $$ 
so the solution is finally given by:

$$ \color{blue}{y(x,t) = Ax \pm \frac{\sqrt{k} t}{\sqrt{1+A^2}} + C },$$

this is known as a complete integral of your PDE. It remains to set the initial condition $y(x,0)$. Can you take it from here?
Cheers!
