Solving for the particular solution of a system of differential equations Consider the IVP 
$\vec{y}'= \begin{bmatrix}0 & -1\\-1 & 0\end{bmatrix}\vec{y} + \begin{bmatrix}t \\e^{2t}\end{bmatrix}$
$\vec{y}(0) = \begin{bmatrix}1 \\1\end{bmatrix}$
The complementary solution of the linear system $\vec{y}′=Ay$ is
$\vec{y}_C(t)= \begin{bmatrix}e^{-t} & -e^t\\e^{-t} & e^t\end{bmatrix} \begin{bmatrix}c_1 \\c_2\end{bmatrix}$
The particular solution is supposed to appear thusly
$\vec{y}_P(t) = \vec{a}e^{2t}+\vec{b}t+\vec{c}$
$\vec{y}_P(t)$ is supposed to appear as a $2 \times 1$ vector.
I tried to look at the system as two separate equations, but because of the initial $A$ matrix something like
$y'_1+y_2 = t$
$y'_2 + y_1 = e^{2t}$
I tried plugging in the supposed particular solution and the derivative as well, but this is confusing because there are separate $a_1$ and $a_2$ values and such that I can't combine. What am I doing wrong here? 

We have a linear system 
$$\dot y = A y + u$$
with initial condition $y (0) = y_0$. What you call the complementary solution is what is usually known as the homogeneous solution, or, natural response. We now assume zero initial conditions and look for the forced response. Laplace-transforming both sides, we get
$$s Y(s) = A Y (s) + U (s)$$
Hence,
$$Y (s) = (s I_2 - A)^{-1} U(s) = \dfrac{1}{s^2 - 1} \begin{bmatrix} s & -1\\ -1 & s\end{bmatrix} \begin{bmatrix} \frac{1}{s^2}\\ \frac{1}{s-2}\end{bmatrix}$$
Now compute $Y (s)$ and then find the inverse Laplace transform of its two entries. Use a table of Laplace transforms, not the definition.
