For constants $v,K$ and a function $C(t)$, can you prove that if :
$$ \frac{dc}{dt} = - \frac{vc(t)}{K + c(t)},~\text{with } c(0) = c_0 $$
Then the solution:
$$ \left[ K \ln c(t) + c(t) \right]_{c_0}^{c(t)} = [-vt]_0^t $$
Implies the closed form:
$$c(t) = K \cdot W\left(\frac{c_0}{K}exp(\frac{-vt + c_0}{K})\right) $$
Where $W$ is the Lambert W function.
I can prove $$ce^c = c_0 exp(\frac{-vt + c_0}{K}) $$ so i'm close, but clearly missing some $K $ terms.
Try using the sub $c = Ky$ then you will get $$ Ky' = -v\frac{Ky}{K+Ky} = -v\frac{y}{1+y} $$ Thus $$ \ln y + y = -\frac{v}{K}t + \lambda $$
or $$ y\mathrm{e}^y = A\mathrm{e}^{-\frac{v}{K}t} $$
Now $y= W(x)$ corresponds to $$ y\mathrm{e}^y = x $$ Thus letting $$ x = A\mathrm{e}^{-\frac{v}{K}t} $$
We find that $$ y = W\left(A\mathrm{e}^{-\frac{v}{K}t}\right) $$ Then finally $$ c = K W\left(A\mathrm{e}^{-\frac{v}{K}t}\right) $$ Use conditions now. So the only bit that would have helped you was the definition of the lambert function.