How to recover the general solution of a PDE This PDE, $xu_x+yu_y=4u$, or some variant of it has been solved several times in this site. Sometimes it's given as solution
$$u(x,y)=f\left(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}}\right)(x^2+y^2)^2$$
There are for me three questions about this equation and its solutions.
1.- I get this two families of solutions in cartesian using the method of characteristics:
$$u(x,y)=f(y/x)x^4$$
$$u(x,y)=f(y/x)y^4$$
It's said it doesn't account for all possible solutions. It's true: $u(x,y)=g\left(\frac{x}{\sqrt{x^2+y^2}}\right)x^4$ is a solution and it's not reducible to one of the form I found. 
What's the flaw in the characteristics method?
2.- I wrote the equation in polars $ru_r=4u$ and solved with the same method. I got $u(x,y)=h(y/x)(x^2+y^2)^2$. It's, say, halfway between the previous ones.
The appearing of the quadratic term as sum has something to do with linearity?
3.- What method has to be used to get the more general solution, the first one in this post?
 A: $$u(x,y)=x^4f(y/x)$$
Of course, the solution can be expressed of a lot of equivalent forms, for example :
$$u(x,y)=y^4g(y/x)\quad  
\text{where} \quad g(X)=\frac{f(X)}{X^4}$$
because $\quad y^4g(y/x)=y^4\frac{f(y/x)}{(y/x)^4}=x^4f(y/x)$
Do not use the same symbole $f$ for all those forms of functions. They are related, but not equal. Of course, each one is an arbitrary function. But when we compare one to the other, they must not be confused. So it's better to use different symbols.
Other example : 
$$f(X)=G\left(\frac{1}{\sqrt{1+X^2}}\right)\quad\to\quad f(y/x)=G\left(\frac{1}{\sqrt{1+(y/x)^2}}\right)=G\left(\frac{x}{\sqrt{x^2+y^2}}\right)$$
$$x^4f(y/x)=x^4G\left(\frac{x}{\sqrt{x^2+y^2}}\right)$$
$x^4f(y/x)\quad\text{and}\quad x^4G\left(\frac{x}{\sqrt{x^2+y^2}}\right)\quad$ are two equivalent forms to express the same general solution, with arbitrary functions $f$ and $G$, which are related functions but different functions.
Other example :
$$f(X)=(1+X^2)^2h(X) \quad\to\quad f(y/x)=  \left(1+(y/x)^2\right)^2h(y/x)$$
$$x^4f(y/x)=x^4\left(x^2+y^2\right)^2x^{-4}h(y/x)=\left(x^2+y^2\right)^2h(y/x)$$
$x^4f(y/x)\quad\text{and}\quad \left(x^2+y^2\right)^2h(y/x)\quad$ are two equivalent forms to express the same general solution, with arbitrary functions $f$ and $h$, which are related functions but different functions.
This shows that all the examples that you raised are equivalent. There is no method to give a more general solution because each of the forms of solution obtained express the general solution and each method used leads to the same general solution, but expressed on various equivalent forms.
