Stability theory: Every solution of the scalar equation: $\ddot{x}+\left [a+b(t) \right ]x=0$ is bounded in $\left [t_0, +\infty \right )$. EDITED
Prove that: If $a>0$ and $$\int_{t_0}^{\infty} |b(t_1)|\mathrm{d}t_1<+\infty$$ then every solution of the scalar equation: $$\ddot{x}+\left [a+b(t)  \right ]x=0$$ is bounded in $\left [t_0, +\infty  \right )$.
I have thought about my problem, but I still no solution :( . Can anyone have an idea or a solution? Any help will be appreciated! Thanks.
 A: Noting that solutions of $\ddot{y} + a y = 0$ have $\dot{y}^2 + a y^2$ constant,
I suspect you might start by looking at $E(t) = \dot{x}^2 + a x^2$ for solutions of your equation.
EDIT:  With $E(t)$ defined as above, you get $$\dot{E}(t) = 2 \dot{x} (\ddot{x} + a x) = - 2 b(t) x \dot{x} \le \frac{|b(t)|}{\sqrt{a}} (\dot{x}^2 + a x^2) = 
\frac{|b(t)|}{\sqrt{a}} E(t)$$
Thus $$\dfrac{d}{dt} \log E(t) = \dfrac{\dot{E}(t)}{E(t)} \le \dfrac{|b(t)|}{\sqrt{a}}$$
Can you take it from there?
A: Write as a 2-dimensional differential equation in matrix form by defining $y = \dot x$. Then Use the Gronwall Lemma (or the version known as the Bellman-Gronwall Lemma) to treat the time-dependent "small" part that depends on $b(t)$ as a perturbation of the time-invariant stable system. (A textbook in adaptive control such as Sastry & Bodson, Ioannou & Sun, or Narendra & Annaswamy well set you in the right direction on the use of the Bellman-Gronwall for stability analysis of your problem).
Then let me know whether your system is indeed stable for any $b(t)$. I suspected not, because the time-invariant unperturbed system is merely stable, not asymptotically so, and the time-varying perturbation could be chosen to resonate with its modes, which are $i\sqrt{a}$ and $-i\sqrt{a}$, but Artem's argument shows that it is indeed stable. Thanks!
A: I will give you a general result.
Your equation can be rewritten as a system of two first order equations, in the general form
$$
\dot u=(A+B(t))u,\quad u=(u_1,u_2)^{\top}.
$$
Assume that the system 
$$
\dot u=Au
$$
has the fundamental matrix solution 
$$
\Phi(t)=\exp(At).
$$
Since the eigenvalues of $A$ have zero real parts, it can be shown that
$$
\|\Phi(t)\|\leq C
$$
for some constant $C$.
Now look for a solution of the original system in the form $u=\Phi(t)z$. After simplifications you should be able to find
$$
\dot z=\Phi^{-1}(t)B(t)\Phi(t)z.
$$
Hence the solution to the system can be written as
$$
u(t)=\Phi(t)u_0+\int_0^t\Phi(t-\tau)B(\tau)u(\tau)d\tau.
$$
From the last, we have
$$
\|u\|\leq C\|u_0\|+\int_{0}^tC\|B(\tau)\|\|u(\tau)\|d\tau.
$$
By Grownwall's inequality this is the same as
$$
\|u\|\leq C\|u_0\|\exp( C\int\|B(\tau)\|d\tau),
$$
which proves that the solutions are bounded.
A: EDITED


*

*By leting: $u_1=\dot{x}$, $u_2=\dot{u_1}$ then we have:
$$\begin{cases}
 & \text{  } \dot{u_1}=u_2 \\ 
 & \text{  } \dot{u_2}=-[a+b(t)]u_1 
\end{cases}$$
Hence,
$$\begin{bmatrix}
\dot{x}\\  \ddot{x}\\
\end{bmatrix}=\begin{bmatrix}
 &0&1\\ 
 &-(a+b(t)& 0
\end{bmatrix}\begin{bmatrix}
x\\  \dot{x}\\
\end{bmatrix} \implies \frac{dv}{dt}=A(t)v, (2)$$.


Now, We need to show that all the solutions of the scalar equation:$$\ddot{x}+\left [a+b(t)  \right ]x=0, (1)$$ are bounded. 


*

*We have $\dfrac{dv}{dt}=A(t)v$ can be rewrite:
$$\begin{bmatrix}
 &0&1\\ 
 &-(a+b(t)& 0
\end{bmatrix} =\begin{bmatrix}
 &0&1\\ 
 &-a& 0
\end{bmatrix} +\begin{bmatrix}
 &0&0\\ 
 &-b(t)& 0
\end{bmatrix}=A+B(t)$$.
The solution $v(t)$ of (2) such that:
$$v(t)=e^{At}\cdot v(t_0)+\int_{0}^{t} e^{(t-s)A}\cdot B(s) \cdot v(s)\mathrm{d}s$$.

*Now, since Artem's hints, implies $v(t)$ is bounded! 
I did exactly as what you say! Is it correct?

Here's my solution:
Aplying the Theorem:

Assume that: 
  
  
*
  
*The system: $$\dfrac{dx}{dt}=Ax$$ is stable.
  
*And $$\int_{0}^{\infty}\|B(s)\|\mathrm{d}s<\infty$$ Whence, system: $$\dfrac{dy}{dt}=[A+B(t)]y$$ is stable.

Now, $A=\begin{bmatrix}
 &0&1\\ 
 &-a& 0
\end{bmatrix}$; and $B=\begin{bmatrix}
 &0&0\\ 
 &-b(t)& 0
\end{bmatrix}$. QED.
