Vtyjkyuk

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 $t$ $$-taurho e^-rho t+omegadelta e^-omega t=0\ e^(omega-rho)t=fracomegadeltataurho$$ 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

$$W(f,g)(t)=beginvmatrixe^-rho t& -rho e^-rho t\ e^-omega t & -omega e^-omega tendvmatrix =(rho -omega)e^-(rho+omega)t.$$

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

$$tau e^-rho t = eta +delta e^-rho tiff (tau-delta)e^-rho t=eta. $$ 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.