Solve the Cauchy problem for the linear PDE I have the linear PDE
$$yu_x - xu_y = 0 \qquad x^2 + y^2 < a^2 \\ u(0,y) = (a^2-y^2)^{\frac{1}{2}} \qquad y \in(-a,a)$$
where $a > 0$ is a constant.
So what I have done is to say that 
$$a_1 = y \quad a_2=-x \quad b=0 \\ \gamma(x_0(s), y_0(s)) = \gamma(0,s) = (a^2-s^2)^{\frac{1}{2}}$$
with $\gamma(0,s)$ being the Cauchy curve. 
From this I say that
$$\widetilde{x}_\tau = \widetilde{y} \qquad \widetilde{x}(0,s) = 0  \\ \widetilde{y}_\tau = -\widetilde{x} \qquad \widetilde{y}(0,s) = s \\ 
\widetilde{z}_\tau = 0 \qquad \widetilde{z}(0,s) =  (a^2-s^2)^{\frac{1}{2}}$$
Now I see that 
$$ \widetilde{x}_{\tau\tau} = \widetilde{y}_\tau = -\widetilde{x}$$
this gives
$$\widetilde{x} = A\,cos(\tau) + B\,sin(\tau)$$
and from the initial boundary condition it can be shown that $A=0$.
Similarly,
$$ \widetilde{y}_{\tau\tau} = \widetilde{x}_\tau = -\widetilde{y}$$
giving 
$$\widetilde{y} = C\,e^{\tau} + B\,e^{\tau}$$
which using the boundary condition gives
$$s = C + D$$
now I know I have made a mistake somewhere because I get stuck here. I don't know how to find the values of $B,\, C,\,D$. But I don't know what I ahve done wrong.
I know that there is a second method where you multiply $x_\tau$ and $y_\tau$ together but I don't want to do it that way.
 A: $$yu_x-xu_y=0$$
$$\frac{\partial u}{x\partial x}-\frac{\partial u}{y\partial y}=0$$
With $X=x^2$ and $Y=y^2$ :
$$\frac{\partial u}{\partial X}-\frac{\partial u}{\partial Y}=0$$
The general solution is well known :
$$u=F(X+Y)$$
$$u(x,y)=F(x^2+y^2) \tag 1$$
$F$ is an arbitrary differentiable function, to be determined according to the boundary condition.
Condition : $\quad u(0,y)=\sqrt{a^2-y^2}$
$$F(0^2+y^2)=\sqrt{a^2-y^2}$$
Let $X=y^2$
$$F(X)=\sqrt{a^2-X}$$
Now, the function $F(X)$ is determined. We put it into the above general solution Eq.$(1)$, where $X=x^2+y^2$. Thus $F(x^2+y^2)=\sqrt{a^2-(x^2+y^2)}$
$$u(x,y)=\sqrt{a^2-(x^2+y^2)}$$
A: Even with your method it should work
$$
\begin{cases}
yu_x - xu_y = 0 \qquad x^2 + y^2 < a^2 \\ 
u(0,y) = (a^2-y^2)^{\frac{1}{2}} \qquad y \in(-a,a)
\end{cases}
$$
$$dz=0 \implies z=c_2$$
For the second integral curve
$$\frac {dx}{y}=\frac {dy}{-x} \implies -xdx-ydy=0$$
$$ \implies -\frac {x^2}2-\frac {y^2}2=c_1 \implies   {x^2}+ {y^2}=c_1 $$
$$f(c_1)=c_2 \implies f(x^2+y^2)=z$$
Therefore
$$u(x,y) =f(x^2+y^2)$$
with IC we have that
$$u(0,y) =f(y^2)=(a^2-y^2)^{1/2}$$
So that
$$u(x,y) =(a^2-y^2-x^2)^{1/2}$$
