(a)
To find eigenvalues, $(1-\lambda x)(-\lambda)-2=0$,we can get $\lambda = 2$ or $\lambda = -1$
To find eigenvectors, when $\lambda=2$,
$\begin{pmatrix}
-1 & 2 \\
1 & -2
\end{pmatrix}
~
\begin{pmatrix}
1 & -2 \\
0 & 0
\end{pmatrix}$
let $x_2=t, x_1=2t$, so the eigenvector is \begin{bmatrix}1 \\ 2\end{bmatrix}
when $\lambda = -1$
$\begin{pmatrix}
2 & 2 \\
1 & 1
\end{pmatrix}$
~
$\begin{pmatrix}
1 & 1 \\
0 & 0
\end{pmatrix}$
let $x_2=-t, x_1=t$, so the eigenvector is \begin{bmatrix}1 \\ -1\end{bmatrix}
Therefore, the general solution is y=$c_1 \begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t}$ + $c_2 \begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$
(b)
from (a), we can know $\phi(t)=
\begin{pmatrix}
e^{2t} & e^{-t} \\
2e^{2t} & -e^{-t}
\end{pmatrix}$
then
$\begin{pmatrix}
e^{2t} & e^{-t} \\
2e^{2t} & -e^{-t}
\end{pmatrix}
\begin{bmatrix}u_1'\\ u_2'\end{bmatrix}
=
\begin{bmatrix} 0 \\ \frac{6e^{3t}}{e^{2t}+1}\end{bmatrix}$
By ref form, we can know $u_1'= \frac{6e^t}{e^2t+1}, u_2' = 0$, then by integrating, we can get $u_1=6arctan(e^t)+c_1, u_2=c_2$
then the general solution is $x(t)=(6arctan(e^t)+c_1)\begin{bmatrix}1 \\ 2\end{bmatrix} e^{2t} + c_2\begin{bmatrix}1 \\ -1\end{bmatrix} e^{-t}$
