Roots operator “algebraic” equation and bounded solutions of differential equation of second order in banach space

In the work the question of invertebelity of the differential operator investigated using square roots operator “algebraic” equations with specic properties. Let $End(E)$ - Banach algebra of linear bounded operators acting in a Banach space $E$, $C_b=C_b(\mathbb{R}, E)$ - the Banach space of continuous bounded functions $x:\mathbb{R}\to E$ with the norm $||x||_c=sup_{t\in \mathbb{R}}||x(t)||$.
The problem of bounded solutions of differential equations
\begin{equation}
\ddot{x}+B\dot{x}+Cx=g,
\tag{1}\label{eq:1}
\end{equation}
where $B, C \in End(E)$, $g \in C_b$, and the corresponding operator $L:D(L) \subset C_b \to C_b$, acting on the rule $Lx=\ddot{x}+B\dot{x}+Cx$. Its domain of definition $D(L)$ - linear subspace of $C_b^2=C_b^2(\mathbb{R}, E)$ of function $x \in C_b$.Then (\ref{eq:1}) is equivalent to operator $Lx=g$. The study of equation (\ref{eq:1}) reduces to the study the first order equation over the Cartesian product $E^2=E\times E$. In the space $C_b(\mathbb{R}, E^2)$ is entered, the operator $\mathscr L$ with domain of definition $D(\mathscr L)=C_b^1(\mathbb{R}, E^2) \subset C_b(\mathbb{R}, E^2)$, acting on the rule $\mathscr Ly=\dot{y}-\mathscr Ay$.
Putting in equation (\ref{eq:1}) $x=y_1$, $\dot{x}=y_2$, we get the following equation the first order
\begin{equation}
\dot{y}=Ay+f,
\tag{2}\label{eq:2}
\end{equation}
where $f=(0,g)$ and the operator $A \in End(E^2)$ is determined by the operator the matrix $\mathscr A=$$\begin{pmatrix}
0 & I\\
-C & -B
\end{pmatrix}$$
$.
Used research method of equation (\ref{eq:1}) based on studying in a Banach algebra $End(E)$ "algebraic" operator equation
\begin{equation}
X^2+BX+C=0.
\tag{3}\label{eq:3}
\end{equation}
Two roots $\Lambda_1$ and $\Lambda_2$ of the equation (\ref{eq:3}) are called $separated$ if the operator $\Lambda_1-\Lambda_2$ is reversible in $End(E)$.
$\mathbf{Theorem 1.}$ If the equation (\ref{eq:3}) has two roots $\Lambda_1,\Lambda_2 \in End(E)$, which are separated, the operator $A \in End(E^2)$ such a diagonal operator $\Lambda \in End(E^2)$, determined by matrix $\Lambda=$$\begin{pmatrix}
\Lambda_1 & 0\\
0 & \Lambda_2
\end{pmatrix}$$.$This is the relation
\begin{equation}
A=U^{-1}\Lambda U,\; where
\tag{4}\label{eq:4}
\end{equation}

\begin{equation}
U=\begin{pmatrix}
I & I\\
\Lambda_1 & \Lambda_2
\end{pmatrix} \;
U^{-1}=\begin{pmatrix}
-(\Lambda_1-\Lambda_2)^{-1}\Lambda_2 & (\Lambda_1-\Lambda_2)^{-1}\\
(\Lambda_1-\Lambda_2)^{-1}\Lambda_1 & -(\Lambda_1-\Lambda_2)^{-1}
\end{pmatrix}.
\tag{5}\label{eq:5}
\end{equation}

$\mathbf{Theorem 2.}$ Let $\Lambda_1,\Lambda_2$ - separated by a pair of roots of the equation (\ref{eq:3}). Then for the invertibility of the operators $L$ and $\mathscr L$ is necessary and sufficient that

\begin{equation}
[\sigma(\Lambda_1)\cap\sigma(\Lambda_2)]\cup i\mathbb{R}=\varnothing
\tag{6}\label{eq:6}
\end{equation}
Obtained the representation of the inverse operator in terms of the roots square operator “algebraic”, equations with specific properties. An example in which the roots are expressed using the coefficients of the differential equation.
Key words: differential equations, bounded solutions, Banach space.