Can't find the solution to a quasilinear PDE $y{u_{x}}$ + $x{u_{y}}$ = 0 I get that $x'' - x = 0$ and similar for $y$, and that I need to solve these ODEs to start. 
This means that I get $x = Ae^t + Be^t$ and $y = Ae^t - Be^t$, however, 
the solution I have been given gives these to be $x = A \cosh(t) + B \sinh(t)$ and $y = A \sinh(t) - B \cosh(t)$. 
I don't get how these have come out of the hyperbolic expressions since $\sinh(x) = \dfrac{e^{x} - e^{-x}}{2}$ and  $\cosh(x) = \dfrac{e^{x} = e^{-x}}{2}$
Can someone explain how these terms have come from the quasilinear PDE? I know how to solve the rest, I just don't get how these terms are coming out.
 A: Both expressions are equivalents..
$$x = A \cosh(t) + B \sinh(t) $$
$$x=\frac A2(e^t+e^{-t})+\frac B 2(e^t-e^{-t})$$
$$x=e^t(\frac A2+\frac B2)+e^{-t}(\frac A 2-\frac B2)$$
With $C=\frac 12(A+B)$ and $D=\frac 12 (A-B)$
$$x(t)=Ce^t+De^{-t}$$
This is exactly what you have. Do the same for: $$y(t) = A \sinh(t) - B \cosh(t)$$
Edit:
As @LutzLehmann  pointed out in the comments:
$$x'' - x = 0 \implies r^2-1=0 \implies r=1,r=-1$$
$$x(t) =Ce^t+De^{-t}$$
A: The Method of Characteristics is useful here.
Since $\frac{\mathrm{d}u}{\mathrm{d}t}=u_x\frac{\mathrm{d}x}{\mathrm{d}t}+u_y\frac{\mathrm{d}y}{\mathrm{d}t}$, if we consider the path where
$$
\frac{\mathrm{d}x}{\mathrm{d}t}=y\quad\text{and}\quad\frac{\mathrm{d}y}{\mathrm{d}t}=x\tag1
$$
then the given equation says
$$
\frac{\mathrm{d}u}{\mathrm{d}t}=0\tag2
$$
That is, $u$ is constant along such a path. $(1)$ is solved by $x^2-y^2=c$. That is,
$$
u(x,y)=g\!\left(x^2-y^2\right)\tag3
$$
Applying some initial data to $(3)$ will give a solution propagated along the hyperbolas $x^2-y^2=c$.
