# Lambert W function identity from differential equation

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.