In my work, I encountered the following equation:
$$
(a'(x)+1)^2+k^2(x) a^2(x)=1,\;\;k(x)=2 {\mbox{sech}}(x).
$$
I would like to know as much as possible about the solution. More particularly, I would like to say that if $a(x)$ is bounded as $x\rightarrow \infty$, then $a(x)\rightarrow 0$. The same as $x\rightarrow -\infty$. If more information can be obtained, I would happily welcome it. 



Behavior of a Solution of a Nonlinear ODE
There are really two separate cases: 
$$\eqalign{a' + 1 &= +\sqrt{1 - k^2 a^2}\cr
 a' + 1 &= -\sqrt{1 - k^2 a^2}\cr} $$
 You could switch continuously from one to the other only on the curve $1 - k^2 a^2 = 0$ (but at almost all points on that curve the trajectories will be either entering or leaving the forbidden region $1 - k^2 a^2 < 0$).


Let's consider the $+$ case, with an initial condition $a(x\_0) = a\_0$ where $x\_0 > 0$ and $0 < a\_0 k(x\_0) < 1$. For $x > x\_0$ we will have $0 < a(x) < a\_0 < 1/k(x\_0)$, and 
$$ 1 - \dfrac{k^2 a^2}{2} > \sqrt{1-k^2 a^2} > 1 - k^2 a^2 $$
so $$-\dfrac{k^2}{2} > \dfrac{a'}{a^2} = - \left(\frac{1}{a}\right)' > -k^2$$
Thus for $x > x\_0$
$$ - 2 \tanh(x) + 2 \tanh(x\_0) > \dfrac{1}{a(x\_0)} - \dfrac{1}{a(x)} > -4 \tanh(x) + 4 \tanh(x\_0) $$ 
Of course, as $x \to +\infty$ we have $\tanh(x) \to 1$, and we find that the limit of $a(x)$ must be nonzero.


