Update: I was able to solve it on my own after all.
This problem is actually an example in my differential equations textbook which I cannot seem to duplicate. My textbook states to use the "elimination method" to crack this. The final answer uses $t$ as the independent variable which both $x$ and $y$ are dependent on. This is a linear nonhomogeneous equation however I only need assistance with deriving the characteristic equation of the homogeneous part of the general solution.
The textbook states that the general solution for both $x$ and $y$ are:
$$\begin{align*} &x(t) = c_1e^{-t}+c_2e^{2t}-3\cos t-\sin t &\\ &y(t) = c_1e^{-t}+4c_2e^{2t}-7\cos t+\sin t & \end{align*}$$
The system consists of the following two linear ordinary differential equations:
$$\begin{align*} &(1) \: x' = -2x + y&\\ &(2) \: y' = -4x + 3y + 10\cos t& \end{align*}$$
The textbook states that the general solution in terms of $x$ should look like this (this is where I am having difficulty):
$$ x'' - x' -2x = 10\cos t $$
That was the part of the problem that I cannot seem to duplicate. Here is what I get when I attempted to work this example problem:
From $(1)$ above I solve for $y$ to get:
$$y = x' + 2x$$
Furthermore I differentiate for $y$ producing:
$$y' = x'' + 2x'$$
Now I am ready to insert $y$ and $y'$ (substitutions shown within parenthesis) into $(2)$ from above resulting in the following steps to get an equation in terms of $x$ and $t$:
$$ (x'' + 2x') = -4x + 3(x' + 2x) + 10\cos t \\ x'' + 2x' = -4x + 3x' + 6x + 10\cos t $$ This yields the following:
$$ x'' - x' - 2x = 10\cos t $$
This time my solution worked. This happens often with me. I don't get the correct solution while writing down on paper but it comes to me when posting it on Mathematics StackExchange. No further help required. I apologize if I wasted your time.
The solution is shown in the body of the post. The answer to this problem is: $$\begin{align*} &x(t) = c_1e^{-t}+c_2e^{2t}-3\cos t-\sin t &\\ &y(t) = c_1e^{-t}+4c_2e^{2t}-7\cos t+\sin t & \end{align*}$$