General integral of an PDE Consider the PDE
$$
\frac{\partial u}{\partial t}+y\frac{\partial u}{\partial x}-a^2x\frac{\partial u}{\partial y}=0
$$
To find the general integral by the method of characteristics, I construct the system
$$
\frac{dt}{1}=\frac{dx}{y}=\frac{dy}{-a^2x}
$$
It is expected to find two constant relations among the variables $C_1=\varphi(t,x,y)$ , $C_2=\psi(t,x,y)$ , so the general solution is $u=f(C_1,C_2)$ .
From the last two equations, I get $-a^2x~dx=y~dy$ , which leads to the relation
$C_1=\varphi(t,x,y)=a^2x^2+y^2$ . But I'm stuck at finding the other one. Any idea? (I know that I can solve explicitly the system in terms of a parameter $s$ and get $t(s)=t+t_0$ , $x(s)=\dfrac{y_0}{a}\sin as+x_0\cos as$ , $y(s)=y_0\cos as-ax_0\sin as$ . I have tried to isolate $s$ from the linear system given by the $(x,y)$ variables and the initial conditions $(x_0,y_0)$ , but with no luck).

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\begin{cases}\dfrac{dx}{ds}=y\\\dfrac{dy}{ds}=-a^2x\end{cases}$
$\therefore\dfrac{d^2x}{ds^2}=\dfrac{dy}{ds}=-a^2x$
$x=C_1\sin as+C_2\cos as$
$\therefore y=C_1a\cos as-C_2a\sin as$
$x(0)=x_0$ , $y(0)=y_0$ :
$\begin{cases}C_2=x_0\\C_1=\dfrac{y_0}{a}\end{cases}$
$\therefore\begin{cases}x=\dfrac{y_0}{a}\sin as+x_0\cos as\\y=y_0\cos as-ax_0\sin as\end{cases}$
$\begin{cases}x_0=\dfrac{ax\cos as-y\sin as}{a}=\dfrac{ax\cos at-y\sin at}{a}\\y_0=ax\sin as+y\cos as=ax\sin at+y\cos at\end{cases}$
$\dfrac{du}{ds}=0$ , letting $u(0)=f(x_0,y_0)$ , we have $u(x,y,t)=f(x_0,y_0)=f\left(\dfrac{ax\cos at-y\sin at}{a},ax\sin at+y\cos at\right)$
