where $c>0$ and $x,y \geq 0$.
Is there an explicit way (closed form) to solve the equation in the title?
I believe the answer is that it must be that $x=y$, at least according to my logic below, but I am wondering if a closed form solution can be obtained.
My approach is as follows: Suppose there is a solution with $x>y$, then $e^{-\frac{(x/y)-1}{(x/y)+1}} <1$, which means that the $ \frac{y-c}{y}e^{-\frac{(x/y)-1}{(x/y)+1}}< \frac{y-c}{y}$
But since $\frac{y-c}{y}== \frac{x-c}{x}$ IFF $x=y$, then for $$\frac{y-c}{y}e^{-\frac{(x/y)-1}{(x/y)+1}} == \frac{x-c}{x}$$ requires $y>x$, a contradiction
A similar argument shows that there cannot be a solution with $y>x$
Hint: Using the substitution $$x=ty$$ we get $$ty-c=(ty-ct)e^{-\frac{t-1}{t+1}}$$ this can be solved for $y$: $$y={\frac {c}{t} \left( t{{\rm e}^{-{\frac {t-1}{t+1}}}}-1 \right) \left( {{\rm e}^{-{\frac {t-1}{t+1}}}}-1 \right) ^{-1}} $$ and you have also an equation for $$x$$ with the parameter $$t$$