Proving Nonhomogeneous ODE is Bounded I am trying to prove the following:
Let $x(t)$ be a solution of the IVP 
$$
\dot x=A(t)x+h(t),
$$ 
where $A(t), h(t)$ continuous on $1\le t<\infty$. Further assume that
$$
\int_1^\infty \| A(t)\|\,dt < \infty\quad \text{and}\quad \int_1^\infty \|h(t)\|\,dt < \infty.
$$
Prove that $x(t)$ is bounded for $t\ge1$.
I have some work, but I am doubtful on its legitmacy.  I wrote the solution $x(t)$ in integral form so:
$$
x(t)=\frac1\mu\left(C+\int_\tau^t \mu(s)h(s)\,ds\,\right).
$$
Then using the intial condition that $x(\tau)=\xi$ we have $C=\xi\mu(\tau)$, so the previous formula is now:
$$
x(t,\tau,\xi)=\exp\Big(\int_\tau^t A(s)\,ds\Big)\xi
+\int_\tau^t \exp\Big(\int_s^t A(w)\,dw\Big)\,h(s)\,ds.
$$
Then clearly $\|x(t)\|$ is bounded only when it's integral equation is bounded which requires both $\int_1^\infty \|A(t)\|\,dt < \infty$ and $\int_1^\infty \|h(t)\|\,dt < \infty$.  
I am just not sure if this is the correct way to go about this proof.  
Any guidance would be greatly appreciated.
 A: Let me try to provide a shorter answer:
First observation. It suffices to show that the solution $\varPhi(t;\tau)$ of the initial value problem
$$
\left\{\begin{array}{lc}X'=A(t)X, \\ X(\tau)=I,\end{array}\right. \qquad\qquad (\star)
$$ 
where $I$ is the identity matrix in $\mathbb R^{n\times n}$, and $X\in\mathbb R^{n\times n}$ is uniformly bounded for $1\le \tau <\infty$. 
Indeed, if $\|\varPhi(t;\tau)\|\le M$, and as the solution of 
$$
x'=A(t)x, \quad x(1)=\xi,
$$
is expressed as 
$$
x(t)=\varPhi(t;1)\xi+\int_1^t \varPhi(t;s) \,h(s)\,ds,
$$
then
$$
\|x(t)\|\le M\|\xi\|+M\int_1^\infty \|h(s)\|\,ds.
$$
Proof of the uniform boundedness of $\varPhi(t;\tau)$. Let $a(t)=\|A(t)\|$, the $L^2-$norm (operator norm) and for fixed $\tau\ge 1$, $y(t)=\|\varPhi(t;\tau)\|$. Then $\varPhi$ satisfies the the equivalent to the $(\star)$ integral equation 
$$
\varPhi(t;\tau)=I+\int_\tau^t A(s)\,\varPhi(s;\tau)\,ds,
$$
implies that
$$
y(t)\le 1+\int_\tau^t a(s)\,y(s)\,ds, \qquad\qquad (\star\star)
$$
and as in dealing with Gronwald's type inequalities, multiplying both sides of the above by $a(t)\exp\big(-\int_\tau^t a(s)\,ds\big)$, we obtain
$$
a(t)y(t)\exp\big(-\int_\tau^t a(s)\,ds\big)-a(t)\exp\big(-\int_\tau^t a(s)\,ds\big)\int_\tau^t a(s)\,y(s)\,ds\\ \le a(t)\exp\big(-\int_\tau^t a(s)\,ds\big)
$$
or
$$
\left(\exp\big(-\int_\tau^t a(s)\,ds\big)\int_\tau^t a(s)\,y(s)\,ds+
\exp\big(-\int_\tau^t a(s)\,ds\big)\right)'
\le 0,
$$
and integrating in $[\tau,t]$ we get
$$
\exp\big(-\int_\tau^t a(s)\,ds\big)\int_\tau^t a(s)\,y(s)\,ds+
\exp\big(-\int_\tau^t a(s)\,ds\big)\le 1,
$$
or
$$
\int_\tau^t a(s)\,y(s)\,ds+
1\le \exp\big(\int_\tau^t a(s)\,ds\big) \qquad\qquad (\star\star\star).
$$
Combining $(\star\star)$ with $(\star\star\star)$ we obtain that
$$
\|\varPhi(t;\tau)\|=y(t) \le \exp\big(\int_\tau^t a(s)\,ds\big)\le 
\exp\big(\int_\tau^\infty \|A(s)\|\,ds\big),
$$
and hence $\varPhi(t;\tau)$ is uniformly bounded.
