Consider that the functions $u(t)$ and $v(t)$ are solutions to a first order system of differential equations, and define $y(t)$ such as:
$$y(t)=\frac{v(t)}{v(t)+u(t)}$$
$u(t)$ and $v(t)$ are solutions to the following system of ODE:
$$u'(t)=b(u+v)-\mu_1u-\beta\frac{uv}{u+v}$$ and $$v'(t)=-\mu_2v+\beta\frac{uv}{u+v}$$
I tried to differentiate $y(t)$ which after using the chain rule gives that $y'(t)={y^2(t)}\cdot (v/u)'$. I have absolutely no idea how I can go on from here. I also tried with partial integration, but this did not give anything either, anybody knows if this is some kind of special problem, and therefore has a very special solution?
Let us start with $$y' = \frac{v'(v+u) - v(v'+u')}{(v+u)^2} = y^2\, \frac{v'u - vu'}{v^2}\, .$$ We substitute the expressions of $u'$, $v'$ in terms of $u$, $v$, so that $$ \frac{v'u - vu'}{v^2} = -b - (b - \mu_1 + \mu_2)\frac{u}{v} + \beta\frac{u}{u+v} + \beta\frac{u^2}{v^2}\frac{v}{u+v} \, . $$ Now, let us make the substitutions $\frac{v}{u+v} = y = 1-\frac{u}{u+v}$ and $\frac{u}{v} = \frac{1}{y}-1$ to obtain the Bernoulli ODE -- a particular type of Riccati ODE -- satisfied by $y$: $$ y' = (\beta-b-\mu_2+\mu_1)y + (-\beta + \mu_2-\mu_1)y^2 . $$ The solutions are $$ y(t) = \frac{\beta-b-\mu_2+\mu_1}{\beta-\mu_2+\mu_1 - c_2 e^{-t(\beta-b-\mu_2+\mu_1)}} \, . $$ In the case $\mu_1 = \mu_2$, one can proceed differently. Indeed, rewriting the initial system in terms of $u$ and $w = u+v$ gives straightforwardly $w(t) = c_1 e^{t(b-\mu)}$. Then, one solves a Riccati equation to obtain $u$. Finally, $y = 1-u/w$ gives $y(t) = (\beta-b)/(\beta - c_2 e^{-t(\beta-b)})$, which is consistent with the previous result.