Bounded and unique solution of a first order ODE Let $x'(t)=Ax(t)+p(t)$ with $A \in \mathbb{R}^{2 \times 2},$ diagonalizable matrix and real eigenvalues $\lambda_1, \lambda_2$ with  $sgn(\lambda_1) \neq sgn(\lambda_2).$ Let $p(t)=(p_1(t),p_2(t)), $
$ p_i \colon \mathbb{R} \to \mathbb{R}$ such that
$$\left\|p_{i}\right\|_{\infty}:=\sup _{(-\infty, \infty)}\left|p_{i}(t)\right|<+\infty, \quad i \in\{1,2\}$$
show that the system has a unique and bounded solution.
I know the general solution has the following form $x(t)=x_p(t)+x_c(t)$ where $x_p(t)=\Phi(t) \int \Phi(t)^{-1} \boldsymbol{p}(t) d t$ with $\Phi(t)$ the fundamental matrix.
How can I ensure the uniqueness of the solution? What does it mean that the solution is bounded?
 A: Let $\lambda_i$ be the eigenvalues of $A$ with eigenvectors $v_i$. Write $S = \begin{pmatrix} v_1 & v_2 \end{pmatrix}$, then
$$ \Phi(t) = S \begin{pmatrix} e^{\lambda_1 t} & 0 \\ 0 & e^{\lambda_2 \ t}\end{pmatrix} S^{-1}$$
and so
$$\Phi(t) \int^t_0 \Phi^{-1}(s) p(s) ds = S \begin{pmatrix} e^{\lambda_1 t} & 0 \\ 0 & e^{\lambda_2 \ t}\end{pmatrix} \int^t_0 \begin{pmatrix} e^{-\lambda_1 s} & 0 \\ 0 & e^{-\lambda_2 \ s}\end{pmatrix} q(s) ds, $$
where $q(s) = S^{-1}p(s)$
So
$$ x_p(t) = \left( \int_0^t e^{-\lambda_1 s} q_1(s) ds\right) e^{\lambda_1 t} v_1 +  \left( \int_0^t e^{-\lambda_2 s} q_2(s) ds\right) e^{\lambda_2 t} v_2 $$
By assumption, we let $\lambda_1 >0$ and $\lambda_2 <0$. Then
$$ C_1 = \int_0^\infty e^{-\lambda_2 s} q_2(s) ds, \ \ C_2 = -\int^0_{-\infty} e^{-\lambda_2 s} q_2(s) ds.$$
Then we claim
$$\tag{1} x(t) = x_p(t) - C_1 e^{\lambda_1 t} v_1 - C_2 e^{\lambda_2 t} v_2$$
is the unique bounded solution. Indeed,
$$ x(t) = -\left( \int_t^\infty e^{-\lambda_1 s} q_1(s) ds\right) e^{\lambda_1 t} v_1 +  \left( \int_{-\infty}^t e^{-\lambda_2 s} q_2(s) ds\right) e^{\lambda_2 t} v_2 $$
Note that
$$ \left| \int_t^{\infty} e^{-\lambda_1 s} q_1(s) ds\right| \le C \int_t^{\infty} e^{-\lambda_1 s}ds= \frac{C}{\lambda_1} e^{-\lambda _1 t}$$
and similarly
$$ \left| \int_{-\infty}^t e^{-\lambda_2 s} q_2(s) ds\right| \le C \int_{-\infty}^t e^{-\lambda_2 s} ds= \frac{C}{-\lambda_2} e^{-\lambda _2 t},$$
which shows that $x(t)$ in (1) is bounded. Lastly, this is the only solution which is bounded: all solutions are of the form
$$\tag{2} x(t) + c_1 e^{\lambda_1 t} v_1+ c_2 e^{\lambda_2 t} v_2.$$
if $c_1 \neq 0$, then (2) tends to infinity as $t\to +\infty$, and when $c_2\neq 0$, (2) tends to infinity when $t\to -\infty$.
