Uniqueness of a solution to a 1st order PDE I'm asked to discuss the uniqueness of the solution for $$u_y+u_x=u-x-y$$ $$u(x,-2)=x$$ I've found the solution to be $$u(x,y)=x+y+2$$ But I don't know how to prove the uniqueness of the solution.

Take
$$
v(x,y)=u(x,y)-x-y-2.
$$
We shall show that $v\equiv 0$. The function $v$ satisfies
$$
v_y+v_x=v, \quad v(x,-2)=0.
$$
Fix $s\in\mathbb R$ and set $f(t)=v(s+t,t)$. Then
$$
f'(t)=v_x+v_y=v=f(t),
$$
and hence $$
f(t)=\mathrm{e}^{t+2} f(-2)=\mathrm{e}^{t+2} v(s-2,-2)=0.
$$
Thus
$v(s+t,s)=0$, for all $s,t\in\mathbb R$, and hence $v\equiv 0$. 
