Find the general solution of the system of ODEs
\begin{equation*}
\left\{\begin{aligned}
&x'_t = -\frac{5}{4} x + \, \frac{3}{4} y + \frac{2}{1+e^t} , \\
&y'_t =\,\hphantom{-}\frac{3}{4} x - \frac{5}{4} y\ .
\end{aligned}\right.
\end{equation*}
Solution
Characteristic equation is
\begin{equation*}
\left| \begin{matrix} -\frac{5}{4}-r & \frac{3}{4}\\ \frac{3}{4} & \frac{5}{4}-r\end{matrix}\right| =( r+\frac{5}{4})^2-\frac{9}{16} =0
\end{equation*} with characteristic roots $r_{1,2}=-\frac{5}{4}\pm \frac{3}{4} $, $r_1=-\frac{1}{2}$, $r_2=-2$.
Finding corresponding eigenvectors: (a) $r_1=1$,
\begin{equation*}
\begin{pmatrix} -\frac{3}{4} & \frac{3}{4}\\ \frac{3}{4} & -\frac{3}{4}\end{pmatrix}\begin{pmatrix}\alpha \\ \beta\end{pmatrix}=0
\end{equation*}
and then $\alpha=\beta=1$ and eigenvector is $\mathbf{e}_1=\begin{pmatrix} 1 \\ 1\end{pmatrix}$.
$r_2=2$ and $\mathbf{e}_2=\begin{pmatrix} 1 \\ -1\end{pmatrix}$ (since matrix is symmetric eigenvectors are orthogonal).
Therefore the general solution of the homogeneous system is
\begin{equation}
\begin{pmatrix} x^* \\ y^*\end{pmatrix}=C_1 \begin{pmatrix} 1 \\ 1\end{pmatrix} e^{-\frac{1}{2}t} + C_2\begin{pmatrix} 1 \\ -1\end{pmatrix}e^{-2t}.
\label{eq-3-1}
\end{equation}
To solve inhomogeneous system we use method of variation of parameters leading to
\begin{equation*}
\begin{pmatrix} e^{-\frac{t}{2}} & e^{-2t}\\ e^{-\frac{t}{2}} & -e^{-2t}\end{pmatrix}\begin{pmatrix} C'_1 \\ C'_2 \end{pmatrix}=
\begin{pmatrix} \frac{2}{1+e^{t} }\\ 0\end{pmatrix}\implies\\
\begin{aligned}
&C'_1=\frac{ e^{\frac{t}{2}}} { 1+e^t }\implies C_1= \int \frac{ e^{\frac{t}{2}}} { 1+e^t } \,dt = 2\arctan (e^{\frac{t}{2}})+ c_1, \\
&C'_2= \frac{e^{2t}}{1+e^{2t}} \implies C_2= \int \frac{ e^{2t}} { 1+e^t } \,dt = \int \Bigl(e^t - \frac{ e^{t}} { 1+e^t } \bigr)\,dt=
e^t - \ln (1+e^t)+c_2\end{aligned}
\end{equation*}
where the first integral is taken by substitution $u=e^{\frac{t}{2}}$ and the second by substitution $u=1+e^{t}$.
Thus
\begin{equation*}
\begin{pmatrix} x \\ y\end{pmatrix}=\bigl(2\arctan (e^{\frac{t}{2}})+ c_1\bigr) \begin{pmatrix} 1 \\ 1\end{pmatrix} e^{-\frac{1}{2}t} +
\bigl(e^t - \ln (1+e^t)+c_2\bigr)\begin{pmatrix} 1 \\ -1\end{pmatrix}e^{-2t}.
\end{equation*}
