Solve the initial value problem
$$Y'=\begin{bmatrix}2&1\\-1&2\end{bmatrix}Y
+\begin{bmatrix}e^x\\0\end{bmatrix},
\quad Y(0)=\begin{bmatrix}1\\1\end{bmatrix}$$
not sure but we find eigenvalues next in the A part
$Y'=AY+G$
Solve the initial value problem
$$Y'=\begin{bmatrix}2&1\\-1&2\end{bmatrix}Y
+\begin{bmatrix}e^x\\0\end{bmatrix},
\quad Y(0)=\begin{bmatrix}1\\1\end{bmatrix}$$
not sure but we find eigenvalues next in the A part
$Y'=AY+G$
Let $\begin{bmatrix}u(x) \\ v(x)\end{bmatrix}= Y$. Then $u'= 2u+ v+ e^x$ and $v'= -u+ v$ with initial condition u(0)= 1, v(0)= 1. Differentiating the first equation again, $u''= 2u'+ v'+ e^x= 2u'+ (-u+ v)+ e^x$. From the first equation, $v= u'- 2u- e^x$ so $u''= 2u'- u+ u'- 2u- e^x+ e^x= 3u'- 3u$. So we want to solve the equation $u''- 3u'+ 3u= 0$. The characteristic equation is $r^2- 3r+ 3= 0$ which has roots $r= \frac{3\pm i\sqrt{3}}{2}$. The general solution to the homogeneous equation is $u(x)= e^{3x/2}\left(cos\left(\frac{\sqrt{3}}{2}x\right)+ sin\left(\frac{\sqrt{3}}{2}x\right)\right)$.
And $v(x)= u'- 2u- e^x$.