Counter example for uniqueness of second order differential equation I have a second order differential equation, 
\begin{eqnarray}
\dfrac{d^2 y}{d x^2} = H\left(x\right) \hspace{0.05ex}y \label{*}\tag{*}
\end{eqnarray}
where, $\,H\left(x\right) = \dfrac{\mathop{\rm sech}\nolimits\left(x\right) \mathop{\rm sech}\nolimits\left(x\right)}{x + \ln\big(2\cosh\left(x\right)\big)}$.
Plot of function $\,H\left(x\right) $ is shown below:

I need to find solution of the  equation  $\eqref{*}$ for the boundary conditions
$$\begin{aligned}
y\left(x\right)\bigg\rvert_{-\infty} &= 0, &
\left.\dfrac{d\hspace{0.1ex}y\left(x\right)}{d\hspace{0.1ex}x}\right\rvert_{ -\infty}  &= 0 
\end{aligned} \label{**}\tag{**}$$
Obvious solution of the problem is $\,y=0$.
But $\,y = x + \ln\big(2 \cosh\left(x\right)\big)\,$  also satisfies  differential equation $\eqref{*}$, and  satisfies boundary conditions $\eqref{**}$. Plot of $y\left(x\right)$ is shown below:

As far as I know there cannot be two solution of the differential equation satisfying given boundary conditions. What am I missing here? Is uniqueness theorem not valid if the boundary conditions are applied at $\,\pm\infty$.
EDIT
Thanks to the comment by Santiago, appearance contradiction is better seen:
Differential eq.
$%\begin{align}
y'\left(x\right) = y\left(x\right)
%\end{align}
$
with boundary condition $\displaystyle\lim_{x \to \infty}y\left(x\right) = 0$. There are infinitely many solution to this problem all of the form $y\left(x\right) = k\exp\left(x\right)$, where $k$ is some constant. 
Post Edit
Is it possible to generalize observation above that, boundary conditions at $\pm \infty$ may not yield unique solution?
 A: Consider the constant  coefficient first order linear dynamical system of dimension $n$ \begin{equation}{\bf{x}}'=A{\bf{x}},\end{equation} where $A$ is an $n\times n$ constant matrix and ${\bf{x}}$ is an $n$-dimensional vector. If $A$ has $k$ eigenvalues with positive real parts, then system above has a $k$ dimensional space of solutions satisfying $\displaystyle{\lim_{t\rightarrow-\infty}{\bf{x}}=0}$. 
More generally, we now consider a class of equations which the system (*) in the question is part of. More precisely, we consider systems of the form 
\begin{equation}{\bf{x}}'=A(t){\bf{x}},\end{equation}
where $A$ is $t$-dependent and such that the limit $A^{-\infty}\equiv \displaystyle{\lim_{t\rightarrow -\infty} A(t)}$ exists. Assume also that $A(t)$ approaches $A^{-\infty}$ exponentially fast as $t\rightarrow -\infty$. Then the solutions of the system above behave like the solutions of the constant coefficient system 
$$
{\bf{x}}'=A^{-\infty}{\bf{x}},
$$
as $t\rightarrow -\infty$, which in particular implies that If $A^{-\infty}$ has $k$ eigenvalues with positive real parts, then the non-constant system above has a $k$ dimensional space of solutions satisfying $\displaystyle{\lim_{t\rightarrow -\infty}{\bf{x}}=0}$. 
