Partial Differential Equations $xu_x+yu_y=1$ I am trying to solve the PDE $xu_x+yu_y=0$. I reach a point where, by setting $\frac{\partial{y}}{\partial{x}}=\frac{y}{x}$ which ends up implying that y=Ax, for some constant A. From there, I am trying to show that , since the derivative of any general solution $u(x,y)$ is $0$, then $u(x,Dx)=u(0,0)=u(x,y)$. From here I don't where to go to have a general solution. Any suggestions? Thanks in advance!
 A: The equation
$$
\begin{align}
0
&=x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\\
&=(x,y)\cdot\nabla u
\end{align}
$$
means that the gradient of $u$ is perpendicular to $(x,y)$. This means that in any radial direction, $u$ is constant. In other words, $u$ is a function of $\arg(x,y)$. This is a bit more general than $u$ being a function of $\frac yx$ since we can have $u(x,y)\ne u(-x,-y)$. For example, $u(x,y)=\sin(\arg(x,y))$.
Since $u(x,y)=\frac12\log\left(x^2+y^2\right)$ is a particular solution of $xu_x+yu_y=1$, the general solution is
$$
u(x,y)=\frac12\log\left(x^2+y^2\right)+f(\arg(x,y))
$$
where $f$ should be $2\pi$-periodic if we want $u$ to be smooth away from $(0,0)$.
Note that in polar coordinates, this becomes
$$
u(r,\theta)=\log(r)+f(\theta)
$$

Example
$$
u(x,y)=\log\left(\sqrt{x^2+y^2}\right)+\frac y{\sqrt{x^2+y^2}}
$$
Is an example of a solution which is smooth away from $(0,0)$ and cannot be written as $\log|x|+f\left(\frac yx\right)$ since
$$
u(3,4)=\log(5)+\frac35\ne u(-3,-4)=\log(5)-\frac35
$$
but no matter what $f$ is
$$
\log|3|+f\left(\frac43\right)=\log|{-}3|+f\left(\frac{-4}{-3}\right)
$$
A: $$xu_x+yu_y=1$$
First, we look for one (any one) particular solution. The simplest way is to look for a solution on the form $u=f(x)$ for example :
$$xf'(x)=1$$
$$f(x)=\ln|x|$$
Then let $u(x,y)=\ln|x[+U(x,y)$
Bringing it back into $xu_x+yu_y=1$ leads to :
$$xU_x+xU_y=0$$
which general solution is easy to find thanks to the change of variables $X=\ln[x[$ and $Y=\ln|y|$. The resulting PDE is $U_X+U_Y=0$ which general solution is $U(X,Y)=F(X-Y)$, any derivable function $F$. 
$U(x,y)=F(\ln|x|-\ln|y|)$
$$u(x,y)=\ln|x|+F(\ln|x|-\ln|y|)$$
where $F$ is any derivable function.
This is the same as $u(x,y)=\ln|y|+G(\ln|x|-\ln|y|)$ where $G$ is any derivable function.
This general solution can be written on many different equivalent forms, for example : $u(x,y)=\ln|x|+\Phi\left(\frac{y}{x}\right)=\ln|y|+\phi\left(\frac{x}{y}\right)=\ln|y|+\varphi\left(\frac{y}{x}\right)=...$ where $\Phi$, $\phi$, $\varphi$ are any derivable functions.
Another way consists to transform the PDE given in Cartesian coordinates to a PDE in polar coordinates. This can be donne very easily as robjohn smartly did.
Below, a more hard work is shown, just for the fun :

It is interresting to compare the solutions found in polar and in Cartesian coordinates and try to answer to the question : in which conditions are they equivalent or not ?

