I found a good approximation to a differential equation, but I can't see why it works.
Given this differential equation:
$$\frac{dk}{dt}=\kappa(1-r_0)-\kappa s - (\gamma+\kappa)k$$
where $\kappa$, $\gamma$, and $r_0$ are all constants and ideally, $s=s_0e^{\frac{-\beta k}{\kappa}}$, with $\beta$ a constant.
Realistically, $\beta$ varies in such away that using this exponential form of $s$ fits the true $s$ vs. $k$ curve with r^2$\approx$0.9.
Further, both $k$ and $s$ are positive reals less than 1. Typical values have, $\kappa\approx 0.164393$, $\gamma \approx 5.385094$, $\beta \approx 5.564689$, and $0<r_0<0.001$.
For the right choice of parameters, $$k=k_\infty e^{\frac{p}{q}e^{qt}}$$
where q is negative and $q \approx -0.0459$, $p \approx 10.06$, and $k_\infty \approx 0.000388$.
By direct substitution, the equation isn't solved by this function, however, it is often possible to pick parameters in $k$ that fit the curve described by the differential equation with $r^2 >0.99$.
How is it possible to have a good approximation here but without actually being a solution? I'm thinking that $\ln{|1-k|} \approx k $ applies somewhere, but that hasn't gotten me anywhere.
For example, the equation for k suggests $\ln{\frac{d(lnk)}{dt}}=p+qt$.
By manipulating the above differential equation, we have
$$\ln{\frac{d(\ln \ k)}{dt}}=\ln{|\kappa(1-r_0)-\kappa s - (\gamma + \kappa)k|}-\ln{k}$$
From here I'm not sure how to show the expression is approximately linear for values of $k$ under 0.02.
Any thoughts?