Actually, this equation came from the other question: What is the transformation matrix between the metric $diag(-x^2,1)$ and the usual Minkowski metric $diag(-1,1)$.
The results seems to be easy, $x'=x cosh(t)$ and $t'=x sinh(t)$. But I find problems when I try to separate the variables.
Is there any methods to solve the equation? Any comments will be much appreciated, thanks!
$$(\frac{\partial f}{\partial t})^2=x^2(1+(\frac{\partial f}{\partial x})^2)$$ Separation of variables, in order to find a family of particular solutions : $$f(x,t)=X(x)T(t)$$ $$(X\,T')^2=x^2(1+(X'T)^2)$$ $$\left(\frac{T'}{T}\right)^2-\left(\frac{xX'}{X} \right)^2-\left(\frac{x}{XT}\right)^2=0$$ The separation of variables is possible only if the mixed term $\frac{x}{XT}$ is function of $t$ only or function of $x$ only.
The case "function of $x$ only" is possible if $T(t)$=constant leading to trivial solution $f(x,t)=i\,x+c$ of no interest.
The case "function of $t$ only" is possible if $\frac{x}{X}$=constant $\quad\implies\quad X(x)=c\,x$ $$\left(\frac{T'}{T}\right)^2-1-\left(\frac{1}{c\,T}\right)^2=0$$ $$(T')^2-T^2=\frac{1}{c^2}$$ Solving the ODE leads to $$T(t)=\frac{1}{c}\sinh(t+C)$$ $$f(x,t)=x\,\sinh(t+C)$$