On the existence of a particular solution for an ODE

On the existence of a particular solution for an ODE

The problem asks to find a bounded $u(\cdot) \in
\mathcal{C}^2(\mathbb{R})$ such that $$u''+u'-2u=f$$ where $f$ is a
bounded continuous function on the real line.
[Observations] We can reduce the 2-order ODE to first-order one simply by
letting $v=(u\ u')^t\in\mathbb{R}^2$, then we find $${dv \over
dt}=Av+\bar{f}$$ where $A=\begin{pmatrix} 0 & 1\\2 & -1 \end{pmatrix}$,
and $\bar{f}=\begin{pmatrix} 0 \\ f \end{pmatrix}$. Since the eigenvalue
of $A$ is $1,-2$, we can associate some invertible $P$ such that
$A=P^{-1}\begin{pmatrix}1 & 0\\0 & -2\end{pmatrix}P$. It's an easy
calculation that $$P=\begin{pmatrix} 1 & 1\\1 & -2\end{pmatrix},\quad
P^{-1}={1 \over 3}\begin{pmatrix} 2 & 1\\1 & -1\end{pmatrix}$$ then we
find \begin{equation*} \begin{split} v(t)&=P^{-1}\begin{pmatrix}e^t &
0\\0& e^{-2t}\end{pmatrix}Pv(0)+\int_0^t P^{-1}\begin{pmatrix}e^{t-s} &
0\\0 &e^{-2(t-s)}\end{pmatrix}P\bar{f}(s)\ ds\\ &= P^{-1}\begin{pmatrix}
e^t v_1+\int_0^t e^{t-s}f(s)\ ds\\e^{-2t} v_2-2\int_0^t e^{-2(t-s)}f(s)\
ds\end{pmatrix} \end{split} \end{equation*} where $Pv(0)=(v_1\ v_2)^t$.
Hence $$3u(t)=2\underline{\left(e^t v_1+\int_0^t e^{t-s}f(s)\
ds\right)}_{(1)}+\underline{\left(e^{-2t} v_2-2\int_0^t e^{-2(t-s)}f(s)\
ds\right)}_{(2)}$$ Now I want to choose some $v_1,v_2$ making $(1)$ stays
bounded as $t \to +\infty$ and $(2)$ stays bounded as $t \to -\infty$.
Since we are considering linear ODEs, we may assume $f\geq 0$. My question
is how to choose such $v_1,v_2$ making it possible to lead boundedness of
$u(\cdot)$ over the whole real line? Or is there any other approach that
can be used to show the existence of some bounded $u$?