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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.

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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.

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