Let $\Omega=\{(x,y) \in \mathbb{R}^2: x^2+y^2<1\}$, and let $u\in C^1(\bar{\Omega})$ satisfy
$$\alpha(x,y)u_x+\beta(x,y)u_y =-u \hspace{0.2in} \forall (x,y)\in\bar{\Omega}, $$
where $\alpha$ and $\beta$ are continuous functions on $\bar{\Omega}$ such that
$$\alpha(x,y)x+\beta(x,y)y >0 \hspace{0.2in} \forall (x,y)\in\bar{\Omega}, $$
Prove that $u\equiv 0$.
Let $X_0$ be a maximum of $u$.
If $X_0$ is interior, then $ u(X_0)=-\alpha(X_0) u_x(X_0) -\beta(X_0) u_y(X_0) =0 $ since $\nabla u(X_0)=0$, and therefore $u\le 0$.
If $X_0$ is in the boundary, then since $v=(\alpha(X_0),\beta(X_0))$ is an exterior vector to the ball at $X_0$ (i.e. $v\cdot X_0>0$), we have $\nabla u(X_0)\cdot v \ge 0$. That is $u(X_0)\le 0$ and thus again $u\le0$.
Similarly you can see that $u\ge 0$.
Suppose not. Then either $\max_{\overline{\Omega}} u>0$ or $\min_{\overline{\Omega}} u<0$. Assume it's the first one. The PDE implies that $u$ cannot have an interior critical point with nonzero value. Therefore, the maximum is at some boundary point $(x,y)$. But then the derivative along the (inward-pointing) vector $-(\alpha,\beta)$ is positive at that point, which is impossible.
