Methods for solving nonhomogeneous system of ODEs I have the system
$$ \mathbf{x}'=\begin{pmatrix}2&1\\1&2\end{pmatrix}\mathbf{x}+\begin{pmatrix}e^t\\t\end{pmatrix}. $$
The eigenvalues of the matrix are 1 and 3 with eigenvectors $(1,-1)^T$ and $(1,1)^T$ respectively. Thus, we have the fundamental matrix
$$
\Psi(t)=\begin{pmatrix}e^t & e^{3t}\\-e^t & e^{3t}\end{pmatrix}.
$$
From here I used the method of variation of parameters to assume there's a solution $\mathbf{x}=\Psi(t)\mathbf{u}(t)$, where $\mathbf{u}(t)$ satisfies $\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t)$. I solved this matrix equation and multiplied to find
$$
\mathbf{x}=c_1\begin{pmatrix}1\\-1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\1\end{pmatrix}e^{3t}+\frac{1}{2}\begin{pmatrix}1\\-1\end{pmatrix}te^t-\frac{5}{12}\begin{pmatrix}1\\1\end{pmatrix}e^t-\frac{1}{2}\begin{pmatrix}1\\-1\end{pmatrix}t+\frac{1}{9}\begin{pmatrix}4\\-5\end{pmatrix}.
$$
I wanted to do this same problem using the Laplace transform given $\mathbf{x}(0)=\mathbf{0}$.
When I take the Laplace transform of the system and solve for $\mathbf{X}(s)$ I get
$$
\mathbf{X}(s)=
\begin{pmatrix}
s-2&-1\\
-1&s-2
\end{pmatrix}^{-1}
\begin{pmatrix}
(s-1)^{-1}\\
s^{-2}.
\end{pmatrix}.
$$
Performing this computation and taking the inverse Laplace transform, I get
$$
\mathbf{x}=\begin{pmatrix}
\frac{1}{8}e^t+\frac{1}{8}e^{5t}-\frac{1}{4}e^t+t\\
e^t+\frac{2}{5}t+\frac{1}{2}e^{-t}+\frac{1}{50}e^{5t}-\frac{13}{25}.
\end{pmatrix}
$$
At this point I realized that there's no way this is the same answer. Does someone see where I'm going wrong or what I misunderstood about one of the methods?
 A: For the first part, I will use this approach. We are given
$$\mathbf{x}'=\begin{pmatrix}2&1\\1&2\end{pmatrix}\mathbf{x}+\begin{pmatrix}e^t\\t\end{pmatrix}$$
The eigenvalues/eigenvectors are
$$\lambda_1 = 1, v_1 = \begin{pmatrix}-1 \\ 1 \end{pmatrix} \\ \lambda_2 = 3, v_2 = \begin{pmatrix}1 \\ 1 \end{pmatrix} $$
The Fundamental matrix is
$$X = \begin{pmatrix}-e^t & e^{3t} \\ e^t & e^{3t} \end{pmatrix}$$
We have $$x_p = X \int X^{-1} g~dt = \begin{pmatrix}
 \dfrac{e^t t}{2}+\dfrac{t}{3}-\dfrac{e^t}{4}+\dfrac{4}{9} \\
 -\dfrac{e^t t}{2}-\dfrac{2 t}{3}-\dfrac{e^t}{4}-\dfrac{5}{9} \\
\end{pmatrix}$$
We can now write the solution as
$$\mathbf{x}(t) = c_1e^{\lambda_1 t} + c_2 e^{\lambda_2 t} + x_p = c_1 e^t\begin{pmatrix}-1 \\ 1 \end{pmatrix} + c_2 e^{3t}\begin{pmatrix}1 \\ 1 \end{pmatrix} + \begin{pmatrix}
 \dfrac{e^t t}{2}+\dfrac{t}{3}-\dfrac{e^t}{4}+\dfrac{4}{9} \\
 -\dfrac{e^t t}{2}-\dfrac{2 t}{3}-\dfrac{e^t}{4}-\dfrac{5}{9} \\
\end{pmatrix}$$
Using the IC, $\mathbf{x}(0)=\mathbf{0}$, this reduces to
$$x(t) = \frac{e^t t}{2}+\frac{t}{3}+\frac{11 e^{3 t}}{36}-\frac{3 e^t}{4}+\frac{4}{9}\\
y(t) = \frac{e^t t}{2}-\frac{2 t}{3}+\frac{e^t}{4}+\frac{11 e^{3 t}}{36}-\frac{5}{9}$$
For the Laplace method, we have
$$\mathbf{X}(s)=
\begin{pmatrix}
s-2&-1\\
-1&s-2
\end{pmatrix}^{-1}
\begin{pmatrix}
\dfrac{1}{s-1}\\
\dfrac{1}{s^2}
\end{pmatrix} = \begin{pmatrix}
 \dfrac{s-2}{(s-1) \left(s^2-4 s+3\right)}+\dfrac{1}{s^2 \left(s^2-4 s+3\right)} \\
 \dfrac{s-2}{s^2 \left(s^2-4 s+3\right)}+\dfrac{1}{(s-1) \left(s^2-4 s+3\right)} \\
\end{pmatrix}$$
The final result from the inverse Laplace Transform is
$$x(t) = \frac{e^t t}{2}+\frac{t}{3}+\frac{11 e^{3 t}}{36}-\frac{3 e^t}{4}+\frac{4}{9}\\
y(t) = \frac{e^t t}{2}-\frac{2 t}{3}+\frac{e^t}{4}+\frac{11 e^{3 t}}{36}-\frac{5}{9}$$
Those two match $\Large\color{\green}{\checkmark}$.
