This question is related to another one (asked by myself) on the Physics forum.
The "more mathematical" part of the question requires the solution of the following differential equation:
$$(z+a) \rho'(z) = 0,$$
where $a$ is a real number and $\rho$ is a real function of $z$.
To me, it seems quite obvious that $\rho(z) = \rho$ represents all the possible solution to this kind of problem. Indeed, $(z+a) \rho'(z)$ must be $0$ for all values of $z$, and then $\rho'(z) = 0$.
For each $z\in (-\infty,-a)\cup (-a,+\infty)$ we have $\rho'(z)=0$. Hence $$\rho(z)=\begin{cases}\rho_1 & z<-a\\ \rho_2 & z>-a\end{cases}\qquad \rho_1,\rho_2\in \mathbb{R} $$ On the other hand $\rho(z)$ is assumed to be differentiable and hence continuous. Therefore $\rho_1=\rho(-a)=\rho_2$ and $\rho(z)$ is constant on $\mathbb{R}$.