Suppose let's say I know $\tau > 0$ and $\rho > 0$. I am looking for a non-trivial solution $\eta, \delta, \omega$ all constants (independent of $t$) and greater than zero, such that:
$$ \tau e^{-\rho t} = \eta +\delta e^{-{\omega t}}. $$
Please note that $(\eta, \delta, \omega) = (0,\tau,\rho)$ will constitute a trivial case and I am not looking for that. And to reiterate, I need a solution independent of $t$. Is a solution possible at all? My apologies if the question sounds silly.
I guess you want some $(\eta,\delta,\omega)$ such that $\tau e^{-\rho t}-\delta e^{-\omega t}=\eta$ for all $t$. In particular, the identity must pass to derivatives, which means that for all $t$ $$-\tau\rho e^{-\rho t}+\omega\delta e^{-\omega t}=0\\ e^{(\omega-\rho)t}=\frac{\omega\delta}{\tau\rho}$$ Since $e^{at}$ is a constant function if and only if $a=0$, we know that $\omega=\rho$ and thus $\delta=\tau$. Putting this back in the first equation, $\eta=0$ as well.
Consider $f(t)=e^{-\tau t}, g(t)=e^{-\omega t}.$ We have that the Wronskian of $f$ and $g$ is
$$W(f,g)(t)=\begin{vmatrix}e^{-\rho t}& -\rho e^{-\rho t}\\ e^{-\omega t} & -\omega e^{-\omega t}\end{vmatrix} =(\rho -\omega)e^{-(\rho+\omega)t}.$$
It is different from zero unless $\rho=\omega.$ That is, the functions $f,g$ are linearly independent (and thus there is no solution) unless $\rho=\omega.$
Assume $\omega=\rho.$ Then we have
$$\tau e^{-\rho t} = \eta +\delta e^{-{\rho t}}\iff (\tau-\delta)e^{-\rho t}=\eta. $$ Since $e^{-\rho t}$ is not constant it must be $\tau =\delta$ and, in such a case, $\eta=0.$ Thus $(\eta, \delta, \omega) = (0,\tau,\rho)$ is the only solution.
