I am searching for a function $u: \mathbb{R} \to \mathbb{R}$ satisfying the differential equation: $$0=g(u') u' + g'(u') u'' u + c_1 u g(u') + c_2 u $$
Edit: we also need a starting condition: $ u(0) = c > 0 $
Where $g:\mathbb{R} \to \mathbb{R}$ is smooth and strictly monotone and $c_1,c_2 \in \mathbb{R}$ are some constants. I have a hunch that this might be a really tough question, even assuming $g$ to be linear (read: you're welcome to assume $g$ linear). Any solution/approximation technique would surprise me very much in a positive way :)
As to how this equation popped up: There was a conversation about drinking coffee continuously, but at a speed changing with the temperature of the coffee. $u$ is the volume of coffee and $g$ is the inverse of a function describing the speed at which you a drink coffee at a certain temperature ($u' = g^{-1}(T)$). Using some physics: $E = c_3 u T$, $E' = c_4 A (T - T_s)$, where $A = 2 \pi r^2 + r^{-1} u$ is area of the surface of the coffee (a cylinder with height proportional to $u$) and $E$ is the thermic energy of remaining coffee, the formula above was obtained.
To further detail: The height of the cylinder is proportional to the volume as $u = h \pi r^2 \iff h = u/(\pi r^2)$ so $ A = 2 \pi r^2 + h 2 \pi r = 2 \pi r^2 + r^{-1} u$ Since $E = c_3 u T$ $$ E' = c_3 (u' T + u T') = c_4 (c_5 + c_6 u) $$ Choosing the constants so that it fits. Combining with $T = g(u') \implies T' = g'(u') u''$ we can put it all together to get the formula in question.
Hint:
Let $v=\dfrac{du}{dt}$ ,
Then $\dfrac{d^2u}{dt^2}=\dfrac{dv}{dt}=\dfrac{dv}{du}\dfrac{du}{dt}=v\dfrac{dv}{du}$
$\therefore vg(v)+uvg'(v)\dfrac{dv}{du}+c_1ug(v)+c_2u=0$
$-uvg'(v)\dfrac{dv}{du}=(c_1g(v)+c_2)u+vg(v)$
$((c_1g(v)+c_2)u+vg(v))\dfrac{du}{dv}=-vg'(v)u$
This belongs to an Abel equation of the second kind.