PDE first order solution Consider the following PDE  $x_2 u_{x_1} + u_{x_2}= u$ in $\{x_1>1\} \times\mathbb{R}$.Can someone give me a hint how to solve this equation ?
 A: Let us rewrite your equation as
$$
yu_x+u_y=u,
$$
which is a bit more nicely-looking. Also, what it seems necessary to assume is either $y > 0,$ or $y < 0.$ Let's consider the case when $y > 0,$ that is, work to obtain the general solution of the PDE on $\mathbf R \times (0,+\infty).$   
We apply the method of characteristics for first-order
semilinear equations, equations of the form
$$
a(x,y)u_x + b(x,y) u_y =c(x,y,u).
$$
The main steps are as follows (I'd recommend my book on PDEs to see how to treat first-order semilinear PDEs.) 
1) Solve the characteristic ODE
$$
y'= \frac{b(x,y)}{a(x,y)}
$$
in your case it is the separable ODE
$$
y'=\frac 1y \iff T(x,y)=y^2-2x=C
$$
where $C$ is an arbitrary constant.
2) Next, according to the general statement
of the method the change of variables
$$
\begin{cases}
s=x,\\
t=T(x,y)
\end{cases}
$$
in your case
$$
\begin{cases}
s=x,\\
t=y^2-2x
\end{cases}
$$
transforms your PDE to
$$
a(x,y) w_s =w
$$
or to
$$
y w_s =w,
$$
which in new terms is
\begin{equation*} \tag{1}
\sqrt{2s+t} w_s =w.
\end{equation*}
This PDE is reducible to the ODE
$$
\sqrt{2s+t} z'(s)=z(s) \iff z(s)= C e^{\sqrt{2s+t}}
$$
which means that the general solution of the PDE (1) is
$$
w(s,t)= f(t) e^{\sqrt{2s+t}}
$$
where $f$ is a continuously differentiable function.
3) Accordingly, the general solution of the original PDE is
$$
u(x,y) =//\, f(t) e^{\sqrt{2s+t}}\,// = f(y^2-2x) e^{\sqrt{y^2}} =
f(y^2-2x) e^y
$$
(rememebering our assumption on $y$).
A: You can use the so-called method of characteristics. For example you can check:
https://en.wikipedia.org/wiki/Method_of_characteristics
A: The solution to this problem is given by $u(x,t)=\frac{x_2}{ln(x_1)+1}$
