Behaviour of solutions of nonautonomous ODE Consider the differential equation
$\ddot x = a(t) x + \cos t, \tag 1$
where $a$ is a smooth monotonically increasing function, $a(0) = -1$, $a(t) \to 1 \;\; t \to +\infty$ (if it's important, let's think that $a(\frac\pi 4) = 0$).
Whether the equation has unbounded solutions and whether it has nonzero bounded solutions?
It seems that the answer is yes to both questions: the equation "tends" to the equation $\ddot x = x + \cos t$, for which the answer is yes. But that's too naive.
If I'd be able to show that there is such solution that $x(t_0) > 0$, $a(t_0) x(t_0) > 1$ for some $t_0 > 0$ than it would give the answer for the first question as the derivative of $x$ would stay positive and would not tend to zero.
It seems, it should be somewhat benefit from that "limit ode" is simple: $\ddot x = x + \cos t$, but I don't see how.
UPDATE:
If it's important, I'm particularly interested in the case $a(t) = 2 \left( \frac{e^{2t} - 1}{e^{2t} + 1} \right)^2 - 1$.
 A: In your particular example, the homogeneous equation $\ddot{x} = a(t) x$ has closed-form solutions $$\eqalign{v_1(t) &= \frac{e^t}{e^{2t}+1}\cr
v_2(t) &= \frac{e^{3t} + 4 t e^t - e^{-t}}{e^{2t}+1}\cr}$$
where of course $v_1(t) \to 0$ as $t \to +\infty$ while $v_2(t) \to +\infty$.
Since you can always add a constant multiple of $v_2(t)$ to any solution of the inhomogeneous equation, there certainly are unbounded solutions. 
Whether there are bounded solutions is a more delicate question, I think.
Using variation of parameters, you can express the general solution of the inhomogeneous equation as
$$ x(t) = u_1(t) v_1(t) + u_2(t) v_2(t) $$
where
$$ \eqalign{\dot{u_1}(t) &= -\frac{1}{2} v_2(t) \cos(t)\cr
           \dot{u_2}(t) &= \frac{1}{2} v_1(t) \cos(t)\cr} $$
Now
$v_1(t) = e^{-t} + O(e^{-3t})$ while $v_2(t) = e^t + O(t e^{-t})$.
Thus if I'm not mistaken we should have $$
\eqalign{u_1(t) &= -\frac{e^t}{4} (\cos(t)+\sin(t)) + const + O(t e^{-t})\cr
u_2(t) &=\frac{e^{-t}}{4} (\sin(t)-\cos(t))  + const + O(e^{-3t})}$$
so that, with constants $= 0$, we should have a solution that is asymptotically $-\cos(t)/2$, and thus is indeed bounded.
EDIT: As requested, here are some more details on the asymptotics of $u_1$.
$$v_2(t) = e^t - e^{-t} + 4 t \left(e^{-t} - e^{-3t} + e^{-5t} - \ldots\right) $$
Thus we get
$$ u_1(t) = u_1(\pi) -\frac{1}{2} \int_\pi^t  v_2(s) \cos(s)\; ds $$
I'm starting at $\pi$ rather than $0$ because I want to stay well inside the convergence region of the series.  The $e^t$ term gives us
$$ const - \frac{e^t}{4}(\cos(t)+\sin(t))$$ 
the $e^{-t}$ contributes
$$ const + \frac{e^{-t}}{4}(-\cos(t)+\sin(t))$$
and for the other terms,
$$  \int_{\pi}^t t e^{-ks} \cos(s) \; ds = A(k) e^{-\pi k} + ( (B_0(k) + B_1(k) t) \cos(t) + (C_0(k)+C_1(k)t) \sin(t)) e^{-kt} $$
where $A(k), B_i(k), C_i(k)$ are rational functions of $k$.  Using the exponential factors,  you get nice enough bounds to conclude that indeed
$$ u_1(t) = -\frac{e^t}{4} (\cos(t)+\sin(t)) + const + O(t e^{-t})$$
