Help with PDE $xu_x+yu_y=2u$ PDE looks like this:
$$xu_x+yu_y=2u$$ $$u(x,1)=x^2$$
$u$ is a function of two variables $u(x,y)$
I'm new at this part of mathematics so I need help. 
My first idea is to use transport PDE $u_t+cu_x=0$(little change o variable $t$ with $y$) but the coefficients with $u_x$ isn't constant so I have another problem. 
 A: Hint:


*

*Oversve that this is an equation of the form $a(x,y,u)u_x+b(x,y,u)u_y=f(x,y,u)$

*The solution to this equation can be found by comparing below ratios:
$$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{2u}$$


*Find two linearly independent solutions $h(x,y,u)=c_1$ and $g(x,y,u)=c_2$
The general solution to given equation is given by $c_2=F(c_1)$

*After that use initial condition to get final solution.
A: The left-hand side $x u_x + y u_y$ is a directional derivative in the direction $(x,y)$, i.e., the radial direction, which suggests that problem will become easier if expressed in polar coordinates.
A: If you stare at this long enough, you'll see that $u = x^2$ is a solution.
(Or you can use the method of characteristics...)
A: By chance is the initial condition $u(x,0) = x^2$? This is due to:
\begin{align}
\frac{du}{2 \, u} &= \frac{d(x + y)}{x+y} \\
\ln u &= 2 \, \ln(x + y) + \ln(c_{0}) \\
u(x,y) &= c_{0} \, (x+y)^{2}.
\end{align}
Check:
\begin{align}
x \, u_{x} + y \, u_{y} &= x \, 2 c_{0} (x+y) + y \, c_{0} (x+y) \\
&= 2 \, c_{0} (x+y)^{2} \\
&= 2 u.
\end{align}
If $u(x,0) = x^2$ then the solution is as is. If $u(x,1) = x^2$ then there is a potential issue. 
