Differential Equation involving Lambert W function I was wondering whether there is an explicit solution to the following differential equation
$$f'(x) = g'(t)\left(f(t)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t) \lambda}\left( 1+ W_{-1}(-e^{-1-\lambda f(t)}  \right)    \right) $$
All the functions are well-behaved, $W_{-1}$ is the -1 branch of the Lambert W function. I am not sure how to solve this one.
Thanks a lot for any help!
 A: $$f'(t) = g'(t)\left(f(t)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t) \lambda}\left( 1+ W_{-1}(-e^{-1-\lambda f(t)}  \right)    \right) $$
Let :  $W_{-1}( -e^{-1-\lambda f(t) })=-y(t) \quad\to\quad -ye^{-y}=-e^{-1-\lambda f(t)}\quad\to\quad f(t)=\frac{1}{\lambda}\left( y-\ln(y)-1\right)$ 
$f'(t)=\frac{1}{\lambda}\left( 1-\frac{1}{y}\right) y'(t)$
$$y'(t)=\lambda \frac{y}{y-1} g'(t)\left(\frac{1}{\lambda}\left( y-\ln(y)-1\right)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t) \lambda}\left( 1-y  \right)    \right)$$
$$y'(t)=\frac{y}{y-1} g'(t)\left(\left( y-\ln(y)-1\right)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t)}\left( 1-y  \right) \right)$$
$$y'(t)=y\: g'(t)\left(\left( 1-\frac{\ln(y)}{y-1}\right)\left(\frac{a}{g(t)} -1 \right)+\frac{a}{g(t)} \right)$$ 
So, we got rid of the Lambert function.
Supposing that we could solve this non-linear ODE, the result $y(t)$ gives :
$$f(t)=\frac{1}{\lambda}\left( y-\ln(y)-1\right)$$
Without knowing what exactly is the function $g(t)$ it seems not possible to go further.
