PDE from first Order $yu_x+xu_y=u^2$ Solving with characteristic lines I need help with the following question:
The given equation is:

$$yu_x+xu_y=u^2$$

Show that the characteristic lines of the equation  are:
$$\begin{cases}
x(t)=C_1e^t+C_2e^{-t}&\\
y(t)=C_1e^t-C_2e^{-t}&\\
u(t)=\frac{1}{C_3-t} &&or&& u(t)\equiv0
\end{cases}$$
My idea is to solve the Following ODE's system:
$$\begin{cases}
x_t=y&\\
y_t=x&\\
u_t=u^2\end{cases}$$
i still not figure out how i need to to that.that's not a classic ODE.
tried to integrae
$x_t=y$ and $y_t=x$ separately but then i don't get the right answer.
tried to solve ODE first-order system of X and Y but I found that the eigenvalue is equal to 0.
Any ideas?
thanks:)
 A: Write ( I use s instead of t) :
$$\frac {dx}{ds}=y$$
Differentiate wrt s: 
$$\frac {d^2x}{ds^2}=\frac {dy}{ds}$$
Note that we have
$$\frac {dy}{ds}=x$$
So that we have, 
$$\frac {d^2x}{ds^2}-x=0$$
It's linear of second order
$$r^2-1=0 \implies r=\pm 1 \implies x(s)=c_1e^s+c_2e^{-s}$$
for $y(s)$
$$\frac {dx}{ds}=y \implies y(s)=c_1e^s-c_2e^{-s}$$
For the last one
For $u=0$ as @holo pointed out in the comment we have $u=c$ 
And for $u \ne 0$ we have that 
$$\frac {du}{ds}=u^2 \implies \int \frac {du}{u^2}=\int ds$$
$$\implies u(s)=\frac 1 {c_3-s}$$
A: You can see that
$$
x_t+y_t=x+y
$$
and 
$$
y_t-x_t=-(y-x)
$$
which are both scalar equations that lead to the solution of the first two equation.
Or you could consider that
$$
x_{tt}=y_t=x
$$
which again gives the given solution for $x$ and $y=x_t$.

Choosing the way of the Lagrange equations 
$$
\frac{dx}y=\frac{dy}x=\frac{du}{u^2}
$$
results in the constants along characteristics $y^2-x^2=c_1$ and $\ln|x+y|+\frac1u=c_2$ which are dependent, $c_2=\Phi(c_1)$, which implies the general solution form
$$
u=\frac1{\Phi(y^2-x^2)-\ln|x+y|}.
$$
