Solve the Goursat problem $xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0$ 
Solve the following Goursat problem 
$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0,$
$u(x,y)=f(x) \; \; on \; \; y^2+x^2=16 \; \; for \; \; 0 \leq x \leq 4
> $-- eq 1
$u(x,y)=g(y) \; \; on \; \; x=0 \; \; for \; \; 0 \leq y \leq4$ -- eq 
  2
$f(0) = g(4)$

My attempt :
Using $\alpha = y^2 - x^2 \; \; and \; \; \beta = y^2 +x^2 $ I've reduced the given problem to 
$u_{\alpha \beta} = 0 \Rightarrow u= \phi(\alpha) + \gamma(\beta)$
$ \Rightarrow u(x,y) = \phi(y^2-x^2) + \gamma(y^2+x^2)$ - eq 3
Using 1 $\Rightarrow f(x) = \phi(16 - 2x^2) + \gamma(16) \Rightarrow \phi(x)= f(\sqrt{\frac{16-x}{2}}) - \gamma(16)$-- eq 4
Using 2 $\Rightarrow g(y) = \phi(y^2) + \gamma(y^2) \Rightarrow \gamma(y) = g(\sqrt{y}) - \phi(y) = g(\sqrt(y)) - f(\sqrt{\frac{16-x}{2}}) + \gamma(16)$ -- eq 5
$f(0)= \phi(16) + \gamma(16) = g(4)$ -- eq 6
using 4 and 5
$u(x,y)= \phi(y^2-x^2) + \gamma(y^2-x^2) = f(\sqrt{\frac{16-y^2+x^2}{2}}) - \gamma(16) + g(\sqrt{y^2+x^2}) - f(\sqrt{\frac{16-x^2-y^2}{2}}) + \gamma(16)$
$\Rightarrow u(x,y)= f(\sqrt{\frac{16-y^2+x^2}{2}})  - f(\sqrt{\frac{16-x^2-y^2}{2}})  +  g(\sqrt{y^2+x^2})$
Is this correct ?
I didn't get to use the given condition f(0)=g(4) anywhere.

$$xy^3u_{xx} - x^3yu_{yy} -y^3u_x + x^3u_y=0$$
HINT :
We observe that the change of $x$ into $-x$ doesn't change the equation, as well as the change of $y$ into $-y$. This draw us to change of variables :
$\begin{cases}X=x^2\\Y=y^2\end{cases} \quad;\quad u_x=2xu_X\quad;\quad u_y=2yu_Y$
$u_{xx}=2u_X+4x^2u_{XX}\quad;\quad u_{yy}=2u_Y+4y^2u_{YY}$
$$xy^3(2u_X+4x^2u_{XX}) - x^3y(2u_Y+4y^2u_{YY}) -y^3(2xu_X) + x^3(2yu_Y)=0$$
$$Y(2u_X+4Xu_{XX}) - X(2u_Y+4Yu_{YY}) -2Yu_X + 2Xu_Y=0$$
$$u_{XX}-u_{YY}=0$$
The well-known general solution is :
$$u(X,Y)=F(X+Y)+G(X-Y)$$
with arbitrary functions $F$ and $G$.
$$u(x,y)=F(x^2+y^2)+G(x^2-y^2)$$
Then, determine the functions $F$ and $G$ according to the boundary conditions.
