Answer
$c(t) = c_s(1-e^{-\alpha t^{1-b}}~)~~$ is a solution.
The differential equation shows that the rate is a maximum at the start when $c_s-c$ is a maximum, but the rate decreases toward zero as $c$ gets closer to $c_s$
Work Step by Step
$c(t) = c_s(1-e^{-\alpha t^{1-b}})$
We can verify that this function is a solution of the differential equation:
$\frac{dc}{dt} = (-c_se^{-\alpha t^{1-b}})[-\alpha\cdot (1-b)\cdot t^{-b}]$
$\frac{dc}{dt} = (c_se^{-\alpha t^{1-b}})[\frac{k}{1-b}\cdot (1-b)\cdot t^{-b}]$
$\frac{dc}{dt} = \frac{k}{t^b}(c_se^{-\alpha t^{1-b}})$
$\frac{dc}{dt} = \frac{k}{t^b}(c_s-c_s+c_se^{-\alpha t^{1-b}})$
$\frac{dc}{dt} = \frac{k}{t^b}[c_s-(c_s-c_se^{-\alpha t^{1-b}})]$
$\frac{dc}{dt} = \frac{k}{t^b}[c_s-c_s(1-e^{-\alpha t^{1-b}})]$
$\frac{dc}{dt} = \frac{k}{t^b}(c_s-c)$
Therefore $~~c(t) = c_s(1-e^{-\alpha t^{1-b}}~)~~$ is a solution.
The differential equation shows that the rate is a maximum at the start when $c$ is far from $c_s$, but the rate decreases toward zero as $c$ gets closer to $c_s$
