Does the following ODE:
$$d^2u/dx^2 + u/A = 0 \quad (\text{or } \ C),$$
have a solution with the conditions:
$$ \left.\frac{d^2u}{dx^2}\right|_{x=0} = 0, $$ $$u(x=0) = B$$ and $$ \left.\frac{du}{dx}\right|_{x->\infty} = 0, $$
$A$, $B$, and $C$ are constants.
Despite I don't know the answer but the taylor series of following equation and its second derivative is close to the result and it matches above condition: $$u(x) = \frac{(e^{x/a}-e^{-x/a})}{(1+e^{2x/a})} $$
For $A > 0$, the second boundary condition won't be met, because $u(x)$ will oscillate around $AC$. So let's assume $A$ is negative.
We'll try to solve when $C$ is included (and shortly see why it must equal zero...) i.e. we have the equation: $$\frac{d^2u}{dx^2} +\frac{u}{A} -C = 0 \Rightarrow \frac{d^2u}{dx^2}+\frac{u-AC}{A}=0.$$ Make the substitution $v=u-AC$: $$\frac{d^2v}{dx^2} +\frac{v}{A}= 0.$$ This has solutions of the form $v(x)= ae^{\frac{x}{-A}}+be^\frac{-x}{-A}$, for some $a$ and $b$, so $u(x)=ae^{\frac{x}{-A}}+be^\frac{-x}{-A}+AC$. Now we just need to find $a$ and $b$. For the second condition to be met (i.e. $\lim_{x \to \infty} u(x) = 0$), we require $a =0$, and $C=0$. Finally, for $u(0)=B$, we require $b=B$.
Therefore, whenever $A < 0$ and $C =0$, we have the solution: $$u(x) = Be^{-\left(\frac{x}{-A}\right)}.$$