Method of characteristics for $x u_y + u_x=u$ 
I am trying to solve the PDE
  $$x\frac{\partial u}{\partial y}+\frac{\partial u}{\partial x}=u,$$ subject to the conditions $u(x,0)=0$ and $u(0,y)=y$.

The characteristic equations are:
$$\frac{dx}{dt}=1, \ \frac{dy}{dt}=x, \ \frac{du}{dt}=u.$$
Solving these ODEs gives
\begin{align}
x(t,s)&=t+A(s) \\
y(t,s)&=\frac{t^2}{2}+A(s)t+B(s) \\
 u(t,s)&=C(s)e^t.
\end{align}
Case $1$, $u(x,0)=0$: 
\begin{align}
x(0,s)&=s\implies x(t,s)=t+s \\
y(0,s)&=0\implies y(t,s)=\frac{t^2}{2}+st \\
u(0,s)&=0\implies u(t,s)=0 \\
\therefore u(x,y)&=0
\end{align}
Case $2$, $u(0,y)=y$: 
\begin{align}
x(0,s)&=s\implies x(t,s)=t \\
y(0,s)&=0\implies y(t,s)=\frac{t^2}{2}+s \\
u(0,s)&=0\implies u(t,s)=se^t \\
\therefore u(x,y)&=\left(y-\frac{x^2}{2}\right)e^x
\end{align}
The solutions say that case $1$ is valid for $y\leq x^2/2$ and case $2$ is valid for $y>x^2/2$. Have absolutely no idea why this.
I solved this problem using a Laplace transform and I arrived at the same conditions. But how can this possibly be interpreted when using the method shown.
 A: $$x\frac{\partial u}{\partial y}+\frac{\partial u}{\partial x}=u$$
I agree with your calculus. A different presentation of the Charpit-Lagrange equations :
$$\frac{dy}{x}=\frac{dx}{1}=\frac{du}{u}=dt$$
A first characteristic equation comes from $\frac{dy}{x}=\frac{dx}{1}$ :
$$y-\frac{x^2}{2}=c_1$$
A second characteristic equation comes from $\frac{dx}{1}=\frac{du}{u}$ :
$$e^{-x}u=c_2$$
General solution of the PDE on implicit form $c_2=F(c_1)$ :
$$e^{-x}u=F\left(y-\frac{x^2}{2}\right)$$
$$u(x,y)=e^xF\left(y-\frac{x^2}{2}\right)$$
$F(X)$ is an arbitrary fonction to be determined according to the boundary conditions.
First condition alone :
$u(x,0)=0=e^xF\left(0-\frac{x^2}{2}\right)\quad\implies\quad F(X)=0\quad\text{on}\quad X<0$
Putting $F=0$ into the above general solution leads to 
$$u(x,y)=0$$
Of course this trivial solution was obvious by inspection of the PDE.
Second condition alone :
$u(0,y)=y=e^0F\left(y-\frac{0^2}{2}\right)=F(y)=y$
The function $F$ is determined : $F(X)=X$. We put it into the above general solution where $X=y-\frac{x^2}{2}$ .
$$u(x,y)=e^x (y-\frac{x^2}{2})$$
Both conditions together :
In order to satisfy both conditions, necessarily the function $F(X)$ is a piecewise function now defined on two domains :
$$F(X)=\begin{cases}
0 && X<0\\
X && X>0
\end{cases}$$
Since $X=y-\frac{x^2}{2}$ the boundary between the two domains is $y=\frac{x^2}{2}$.
$$u(x,y)=\begin{cases}
0 && y<\frac{x^2}{2}\\
e^x (y-\frac{x^2}{2}) && y>\frac{x^2}{2}
\end{cases}$$
This is in agreement with your own results.
