Existence of a non-constant polynomial solution for $ y\partial_x u + (-3y-2x)\partial_yu = 0$ in $\mathbb{R}^2$ I have the following problem:
a) Show that the only $C^1$ colution defined in all $\mathbb{R}^2$ for the equation $y\partial_x u + (-3y-2x)\partial_yu = 0$ is the constant solution.
b) Show an open set and a non-constant solution defined in this set.
c) Show that there is a non-constant polynomial solution defined in some open set.
My attempt to answer:
a) It is easy to find the integral curves of the vector field $A(x, y) = (y, -3y-2x)$ of the coefficients of the operator. With this we find that all such curves go to $(0,0)$ when the prameter goes to $\infty$. Since any solution must be constant along these curves, there an only be constant $C^1$ solutions. Indeed, if $u$ is a non-constant solution, near the origin we will have that the derivatives of this function blows up to $\infty$.
b) Since the only problem is in at the origin, we only need an open set that does NOT have $(0,0)$ as one of it's accumulation points and a function $u(x, t) = f(\gamma)$, being $\gamma$ the integral curve.
I have, however, no idea on how to solve the item c). Any hints and/or solutions will be the most appreciated. 
Thank you very much.
 A: I see that the answer is already given. That is well, but I suppose that someone would be interested to know what is the general solution of the PDE. Also, looking at the relationship between the answer already given and the explicit general solution should be of interest.
$$ y\partial_x u + (-3y-2x)\partial_yu = 0 \tag 1$$
The system of characteristic ODEs is :
$$\frac{dx}{y}=\frac{dy}{-3y-2x}=\frac{du}{0} \tag 2$$
A first family of characteristic curves comes from $\quad du=0\quad\to\quad u=c_1\tag 3$.
A second family of characteristic curves comes from $\quad \frac{dx}{y}=\frac{dy}{-3y-2x} \tag 4$
$\frac{dx}{y}=\frac{dy}{-3y-2x}=\frac{adx+dy}{ay-3y-2x}=\frac{adx+dy}{-2x+(a-3)y}\qquad \tag 5$ any constant $a$.
We chose $a$ so that it becomes integrable: $\quad \frac{a}{-2}=\frac{1}{a-3}\quad\to\quad a^2-3a+2=0 \tag 6$
Two values of $a$ are possible : $a_1=1$ and $a_2=2$
$$\frac{a_1dx+dy}{-2x+(a_1-3)y}=\frac{a_2dx+dy}{-2x+(a_2-3)y}=\frac{dx+dy}{-2x-2y} = \frac{2dx+dy}{-2x-y}=\frac{d(x+y)}{-2(x+y)}=\frac{d(2x+y)}{-(2x+y)}\tag 7$$
Integrating leads to :
$\quad -\frac{1}{2}\ln|x+y|=-\ln|2x+y|+\text{constant}\quad\to\quad  2\ln|2x+y|-\ln|x+y|=\text{constant}\tag 8$. 
Second family of characteristic curves : $\quad\frac{(2x+y)^2}{x+y}=c_2 \tag {9}$.
The general solution of the PDE is:
$$u(x,y)=F\left(\frac{(2x+y)^2}{x+y}\right)\tag {10}$$
$F$ is any differentiable function.
Since $F$ is arbitrary (until a boundary condition allows to determine it), this includes the trivial case of constant function $\to u(x,y)=c$. 
Check :
$\frac{\partial u}{\partial x}=\frac{(2x+y)(2x+3y)}{(x+y)^2}F'\qquad ; \qquad
\frac{\partial u}{\partial y}=\frac{(2x+y)(y)}{(x+y)^2}F'$
$y\frac{\partial u}{\partial x} + (-3y-2x)\frac{\partial u}{\partial y}=
y\frac{(2x+y)(2x+3y)}{(x+y)^2}F'-(2x+3y)\frac{(2x+y)(y)}{(x+y)^2}F' = 0$
A: a) Looks good, although I would say that "since all characteristic curves of the equation approach $(0,0)$, the value of $u$ on each of those curves is $u(0,0)$ by the continuity of $u$", without talking about derivatives blowing up.

c) Show that there is a non-constant polynomial solution defined in some open set. 

This can't be true because if a polynomial solves that PDE in an open set, it solves it everywhere. Indeed, if $u$ is a polynomial, then so is $y\partial_x u + (-3y-2x)\partial_yu $. And if a polynomial vanishes on an open set, it vanishes identically. 
Perhaps part c) should be "Show that there is NO non-constant polynomial solution defined in an open set" 
