Soving PDE $u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0$ I have some problems with solving PDEs.

  
*
  
*\begin{cases}
\ u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \\[2ex]
u(x,x)=1,\quad u(1,y)=y
\end{cases}

What I've done: $$u(x,y)=\frac{1}{x^2}v(x,y)$$
$$u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \Rightarrow v_{xx}-v_{yy}=0$$
$$v(x,y)=C_1(x+y)+C_2(x-y)$$
$$u=\frac{1}{x^2}(C_1(x+y)+C_2(x-y))$$
From $u(x,x)=1, u(1,y)=y$ we get
$$\begin{cases}
\ C_1(2x)+C_2(0)=1 \\[2ex]
C_1(1+y)+C_2(1-y)=y
\end{cases}$$
And this is where I stopped. No idea how to solve such a system. Any ideas?


  
*\begin{cases}
\ u_{xy}-\frac{1}{x-y}(u_x-u_y)=1 \\[2ex]
u(x,x)=0,\quad u(2,y)=2+2y+\frac{1}{2}y^2
\end{cases}

$$u(x,y)=\frac{1}{x-y}v(x,y)$$
$$u_{xy}-\frac{1}{x-y}(u_x-u_y)=1 \Rightarrow v_{xy}=x-y$$
$$v=\frac{1}{2}x^2y-\frac{1}{2}y^2x+C_1(x)+C_2(y)$$
$$u=\frac{1}{x-y} \left(\frac{1}{2}x^2y-\frac{1}{2}y^2x+C_1(x)+C_2(y)\right)$$
$$\begin{cases}
\ C_1(x)+C_2(-x)=x^3 \\[2ex]
C_1(2)+C_2(y)=4-\frac{y^3}{4}
\end{cases}$$
And again the same problem. What's wrong? Thank you.

The equation $C_1(2x)+C_2(0)=1$ shows that $C_1(2x)$ doesn't vary with $x$, i.e. $C_1$ is a constant function, say $C_1(x) = c.$
Inserting this in $C_1(1+y)+C_2(1-y)=y$ we get $C_2(1-y) = y-c$ i.e. $C_2(y) = (1-y)-c = (1-c)-y.$
Thus,
$$
\begin{cases}
C_1(x) = c \\
C_2(x) = (1-c)-y
\end{cases}
$$
