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).
 A: 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!
A: $y_x^2=\dfrac{k}{y_t^2}-1$
$y_t^2=\dfrac{k}{y_x^2+1}$
$y_t=\pm\dfrac{\sqrt k}{\sqrt{y_x^2+1}}$
$y_{tx}=\mp\dfrac{\sqrt ky_xy_{xx}}{(y_x^2+1)^\frac{3}{2}}$
Let $u=y_x$ ,
Then $u_t=\mp\dfrac{\sqrt kuu_x}{(u^2+1)^\frac{3}{2}}$ with $u(x,0)=-2x+1$
$u_t\pm\dfrac{\sqrt kuu_x}{(u^2+1)^\frac{3}{2}}=0$ with $u(x,0)=-2x+1$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$\dfrac{dx}{ds}=\pm\dfrac{\sqrt ku}{(u^2+1)^\frac{3}{2}}=\pm\dfrac{\sqrt ku_0}{(u_0^2+1)^\frac{3}{2}}$ , letting $x(0)=f(u_0)$ , we have $x=\pm\dfrac{\sqrt ku_0s}{(u_0^2+1)^\frac{3}{2}}+f(u_0)=\pm\dfrac{\sqrt kut}{(u^2+1)^\frac{3}{2}}+f(u)$ , i.e. $u=F\left(x\mp\dfrac{\sqrt kut}{(u^2+1)^\frac{3}{2}}\right)$
$u(x,0)=-2x+1$ :
$F(x)=-2x+1$
$\therefore u=-2\left(x\mp\dfrac{\sqrt kut}{(u^2+1)^\frac{3}{2}}\right)+1$
$y_x=\pm\dfrac{2\sqrt kty_x}{(y_x^2+1)^\frac{3}{2}}-2x+1$
$(y_x+2x-1)(y_x^2+1)^\frac{3}{2}=\pm2\sqrt kty_x$
You will get $y_x(x,t)$ which is difficult to express explicitly, so does for $y(x,t)$ .
A: According to Mathematica, if this helps:
$y(x,t) = \pm \frac{x \sqrt{k-C_2^2}}{C_2}+C_2 t+C_1$
Mathematica won't solve it with the boundary condition.
