Proving that a solution to a differential equation is monotonic The answer that ws given on a previous question of mine, stated that the solution to this DE:
$$x(t)\cdot r+x'(t)\cdot l+a\cdot\ln\left(1+\frac{x(t)}{b}\right)=0\space\Longleftrightarrow\space x(t)=\dots\tag1$$
Must be monotonic.
Is there a way to proof that that is the case, that the function $x(t)$ is monotonic?!

Background:
I've to find (the average of a function over a particular interval, where $t_1>0$, $t_2>0$ and $t_2>t_1$):
$$\frac{1}{t_2-t_1}\int_{t_1}^{t_2}x(t)dt\tag2$$
Where $x(t)$ in equation $(2)$ is the solution to the DE in equation $(1)$.
 A: Considering the DE
$$
l x'+r x+a\ln\left(1+\frac xa\right)=0
$$
taking it's derivative (assuming $x$ is twice diferentiable)
$$
\frac{a x'}{b \left(\frac{x}{b}+1\right)}+l x''+r x' = 0
$$
after substituting $x'$ from the first equation we get
$$
l x''= \frac{(a+b r+r x) \left(a \ln \left(\frac{b+x}{b}\right)+r x\right)}{l (b+x)}
$$
and now solving for $x$ the condition $lx'' = 0$ we obtain
$$
x^* = \{\frac ar W\left(\frac{b e^{\frac{b r}{a}} r}{a}\right)-b,-\left(b+\frac ar\right)\}
$$
also regarding the first equation we have
$$
l x'=-\left(r x+a\ln\left(1+\frac xa\right)\right) = 0\Rightarrow x^* = \frac ar W\left(\frac{b e^{\frac{b r}{a}} r}{a}\right)-b
$$
so the only stationary point has null $x''$ so it is an inflexion point indicating that $x$ is monotonic..
A: A differentiable function is not monotonic if there are two different points whose derivatives have different signs. Since $x' = -\frac{rx + a\ln ( 1 + x/b)}{l}$ is a composition of differentiable functions, it's differentiable and thus continuous; if not monotonic, $x'$ it must have a root; and at least one root is not an inflection point. Simbolically, $$\begin{cases} x'(t) = 0 \\ x''(t) \neq 0\end{cases}$$ for some $t$. But $x'' = -x'\left(r + \frac{a}{b + x}\right)$ so the second derivative is zero on all the roots of the first derivative. So every root is an inflection point and thus, in fact, $x$ is monotonic.
