It looks like the system is supposed to be
[tex]\mathbf x'=\begin{bmatrix}0&-1\\2&3\end{bmatrix}\mathbf x+\begin{bmatrix}e^t\\e^{-t}\end{bmatrix}[/tex]
with the initial condition, [tex]\mathbf x(0)=\begin{bmatrix}5&4\end{bmatrix}^\top[/tex].
Compute the eigensystem for the coefficient matrix:
[tex]\begin{vmatrix}-\lambda&-1\\2&3-\lambda\end{vmatrix}=\lambda^2-3\lambda+2=(\lambda-1)(\lambda-2)=0[/tex]
[tex]\lambda_1=1\implies\begin{bmatrix}-1&-1\\2&2\end{vmatrix}\mathbf v_1=\mathbf 0\implies\mathbf v_1=\begin{bmatrix}1\\-1\end{bmatrix}[/tex]
[tex]\lambda=2\implies\begin{bmatrix}-2&-2\\2&1\end{bmatrix}\mathbf v_2=\mathbf 0\implies\mathbf v_2=\begin{bmatrix}1\\-2\end{bmatrix}[/tex]
So we have the characteristic solution,
[tex]\mathbf x_c=C_1e^{\lambda_1t}\mathbf v_1+C_2e^{\lambda_2t}\mathbf v_2[/tex]
[tex]\mathbf x_c=C_1e^t\begin{bmatrix}1\\-1\end{bmatrix}+C_2e^{2t}\begin{bmatrix}1\\-2\end{bmatrix}[/tex]
For the non-homogeneous part, we can guess a particular solution of the form
[tex]\mathbf x_p=te^t\begin{bmatrix}a\\b\end{bmatrix}[/tex]
(We might have started with [tex]e^t[/tex] instead, but that is already accounted for in the first characteristic solution.)
Its derivative is
[tex]{\mathbf x_p}'=e^t\begin{bmatrix}a\\b\end{bmatrix}+te^t\begin{bmatrix}a\\b\end{bmatrix}=e^t\begin{bmatrix}at+a\\bt+b\end{bmatrix}[/tex]
Substitute [tex]\mathbf x_p[/tex] into the system and solve for [tex]a,b[/tex]:
[tex]e^t\begin{bmatrix}at+a\\bt+b\end{bmatrix}=te^t\begin{bmatrix}0&-1\\2&3\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}+e^t\begin{bmatrix}1\\-1\end{bmatrix}[/tex]
[tex]\begin{bmatrix}at+a\\bt+b\end{bmatrix}=\begin{bmatrix}-bt+1\\(2a+3b)t-1\end{bmatrix}[/tex]
[tex]\implies a=1,b=-1[/tex]
and so the particular solution is
[tex]\mathbf x_p=te^t\begin{bmatrix}1\\-1\end{bmatrix}[/tex]
The general solution to the system is then
[tex]\mathbf x=\mathbf x_c+\mathbf x_p[/tex]
[tex]\mathbf x=C_1e^t\begin{bmatrix}1\\-1\end{bmatrix}+C_2e^{2t}\begin{bmatrix}1\\-2\end{bmatrix}+te^t\begin{bmatrix}1\\-1\end{bmatrix}[/tex]
Use the given initial conditions to solve for the remaining coefficients.
[tex]\mathbf x(0)=\begin{bmatrix}5\\4\end{bmatrix}[/tex]
[tex]\implies\begin{bmatrix}5\\4\end{bmatrix}=C_1\begin{bmatrix}1\\-1\end{bmatrix}+C_2\begin{bmatrix}1\\-2\end{bmatrix}[/tex]
[tex]\implies C_1=14,C_2=-9[/tex]
Then the solution to the initial value problem is
[tex]\mathbf x=14e^t\begin{bmatrix}1\\-1\end{bmatrix}-9e^{2t}\begin{bmatrix}1\\-2\end{bmatrix}+te^t\begin{bmatrix}1\\-1\end{bmatrix}[/tex]
[tex]\mathbf x=\begin{bmatrix}14e^t-9e^{2t}+te^t\\-14e^t-18e^{2t}-te^t\end{bmatrix}[/tex]
