Limiting Ratio of Solutions to Ordinary Differential Equations I'm trying to find the limit of the ratio of two functions
$ \lim_{t \rightarrow \infty} \frac{f(t)}{g(t)} $ but only have the initial conditions and the differential equations they solve, but the equations don't have easy closed form solutions:
$f'(t) = -\beta f(t)^2 + (\beta - \gamma) f(t) + \gamma e^{-\delta t} \\
g'(t) = -\beta g(t)^2 + (\beta - \gamma) g(t) $
$f(0)=1, g(0)=1$ and $f(t) \rightarrow 0$ and $g(t) \rightarrow 0$.
L'hopital's is no use, and trying to numerically solve this for long time by solving the top and bottom first is no good, it breaks down after a point. I've consider the method of dominant balance, but I'm not sure it is applicable. I don't know what else to try, I'm a statistician out of his depth.
 A: Your differential equations do have closed-form solutions. $f(t)$ is a rather complicated expression involving Bessel functions, while
$$   g \left( t \right) ={\frac {{{\rm e}^{- \left( \gamma-\beta \right) t
}} \left( \gamma-\beta \right) }{{\gamma-{\rm e}^{- \left( \gamma-\beta
 \right) t}}\beta}}
$$
Since you want $g(t) \to 0$ (presumably as $t \to + \infty$), I suppose you want
$\gamma > \beta$.  I'll also assume $\beta, \gamma, \delta, c > 0$.  As $t \to \infty$ we  have $$g(t) \sim 
e^{-(\gamma-\beta)t} (1 - \beta/\gamma)$$
Now suppose $f(t) = u(t) g(t)$. I get
$$ u'(t) =  -{\frac { \left( u \left( t \right)  \right) ^{2} \left( \gamma-\beta
 \right) \beta}{\gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}-
\beta}}+{\frac {u \left( t \right)  \left( \gamma-\beta \right) \beta
}{\gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}-\beta}}-{\frac {c
 \left( 2\,{{\rm e}^{ \left( \gamma-\beta-\delta \right) t }}\gamma\,
\beta-{{\rm e}^{\left( 2\,\gamma-2\,\beta-\delta \right) t}}{\gamma}^
{2}-{{\rm e}^{-\delta\,t}}{\beta}^{2} \right) }{ \left( \gamma-\beta
 \right)  \left( \gamma\,{{\rm e}^{ \left( \gamma-\beta \right) t}}-
\beta \right) }}
$$
For large $t$, the coefficients of the first two terms on the right go to $0$ exponentially, while the largest contribution to the third is 
$$ \dfrac{c \gamma}{\gamma-\beta} e^{(\gamma-\beta-\delta)t}$$
So if $\gamma - \beta - \delta > 0$, $u$ will grow exponentially, while
if $\gamma - \beta - \delta < 0$, $u'$ will decay exponentially and $u$ will approach a constant.   
One nice special case is $\beta = \gamma/2$, $\delta = \gamma$, where 
I get
$$ \lim_{t \to \infty} \dfrac{f(t)}{g(t)} = \dfrac{4\,\sqrt {c\gamma}\cosh \left( {\sqrt{2c/\gamma}}
 \right) +4\,\sqrt {2}c\,\sinh \left( {\sqrt{2c/\gamma}} \right) 
}{\sqrt{2}\,\gamma\sinh \left( {\sqrt{2c/\gamma}} \right) +2\,\sqrt {c\gamma}\cosh \left( {\sqrt{2c/\gamma}} \right) 
}
$$
A: This expands on the answer given by Robert Israel. Using the closed form solution in terms of Bessel functions and Mathematica, I found the following expression for $\lim_{t\to\infty} f(t)e^{(\gamma-\beta)t}$: $$\frac{\sqrt{\gamma } (\gamma -\beta ) \delta ^{\frac{2 (\beta -\gamma
   )}{\delta }} (\beta  \gamma )^{-\frac{2 \beta +\delta -2 \gamma }{2 \delta }}
   \Gamma \left(\frac{\beta +\delta -\gamma }{\delta }\right) \left(\sqrt{\beta }
   I_{\frac{\beta -\gamma }{\delta }}(p)+\sqrt{e^{-\delta }} \sqrt{\gamma }
   I_{\frac{\beta +\delta -\gamma }{\delta }}(p)\right)}{\sqrt{\beta }
   \Gamma \left(\frac{-\beta +\delta +\gamma }{\delta }\right)
   \left(\sqrt{e^{-\delta }} \sqrt{\gamma } I_{-\frac{\beta +\delta -\gamma
   }{\delta }}(p)+\sqrt{\beta } I_{\frac{\gamma -\beta }{\delta
   }}(p)\right)},$$
where $$p=\frac{2\sqrt{\beta\gamma e^{-\delta}}}{\delta}.$$
