Use of method of characteristics to solve the given PDE 
How to use the method of characteristics to solve the nonlinear problem:

$$u_x^2 - u_y^2 = x^2 - y, ~~ u(x,0)=x.$$
While studying for the finals I came up with this question. I honestly don't know how to solve this. Any help is much appreciated.
 A: Hint: 
Observe
\begin{align}
u_x^2-u_y^2 = (u_x-u_y)(u_x+u_y) = x^2-y.
\end{align}
then 
\begin{align}
F(u_x, u_y, u, x, y)=u_x^2-u_y^2 - x^2+y \equiv 0.
\end{align}
Since $F(p_1, p_2, z, x, y) = p_1^2-p_2^2-x^2+y$ then the characteristics equations yields the following system of ODE
\begin{align}
\begin{cases}
\dot p_1 =2x, \dot p_2 =-1\\
\dot z = 2(p_1^2-p_2^2)\\
\dot x= 2p_1, \dot y = -2p_2 
\end{cases}.
\end{align}
Using the first two and last two characteristic equations yield
\begin{align}
\ddot x-4x = 0 \ \ \text{ and } \ \ \ddot y= 2
\end{align}
which means
\begin{align}
x(t) = x_0\cosh 2t + p_1^0 \sinh 2t,  \ \ \ y(t) = t^2-2p_2^0 t+y_0
\end{align}
and
\begin{align}
p_1(t) = \frac{\dot x(t)}{2} = x_0\sinh 2t+p_1^0\cosh 2t, \ \ \ p_2(t) = \frac{-\dot y(t)}{2} = p_2^0-t. 
\end{align}
Hence 
\begin{align}
\dot z(t) = 2(x_0\sinh 2t+p_1^0\cosh 2t)^2-2(p_2^0-t)^2 
\end{align}
which means
\begin{align}
z(t) = z_0+\int^t_0 2(x_0\sinh 2s+p_1^0\cosh 2s)^2-2(p_2^0-s)^2\ ds.
\end{align}
To simplify matter, since $u(x, 0) = x$ then it follows $u_x(x_0, 0) = 1 = p_1^0$. Since $u_x(x_0, 0)^2-u_y(x_0, 0)^2-x_0^2 =0$, then it follows $\pm \sqrt{1-x_0^2}= p_2^0$ (we will not choose yet). 
It's kind of late. I will finish the rest when I have time. 
