Every resource I have seen on undetermined coefficients for inhomogeneous systems uses only examples in which the components of the forcing term have a common replication under differentiation. Thus the solution techniques suggest guesses of forms such as $\vec{x_p} = \vec{a}e^{2t}$ or $\vec{x_p} = \vec{a}t^2 + \vec{b}t + \vec{c}$, where each component of one of the undetermined vectors plays the same role (i.e., $b_1$ and $b_2$ are the coefficients of the $t$ terms).
Can undetermined coefficients also be used to solve systems with forcing terms in which each component has a different replication under differentiation, by making the usual guesses component-wise? For example, if the forcing term is $\begin{bmatrix} .5t \\ \sin(3t) \end{bmatrix}$, the guess would be $\vec{x_p} = \begin{bmatrix} at + b \\ c\cos(3t) + d\sin(3t) \end{bmatrix}$. For each component, the guess is made which would be made if that component were the forcing term in a second-or-higher-order single ODE. Is this a legitimate way to generalize undetermined coefficients for systems, or does the technique only work if all components share a common replication under differentiation?
The first step is to come as close as possible to decoupling the variables. For example, consider $$\begin{bmatrix} \dot{x_1}\\ \dot{x_2}\end{bmatrix}=\begin{bmatrix}5&4\\-2&1 \end{bmatrix} \begin{bmatrix}x_1\\x_2 \end{bmatrix} +\begin{bmatrix} f(t)\\g(t)\end{bmatrix} $$, The matrix $\begin{bmatrix} 5&4\\-2&1\end{bmatrix} $ has eigenvalues $3\pm 2i$. There exists a real non-singular matrix $P$ such that $$P^{-1}\begin{bmatrix} 5&4\\-2&1\end{bmatrix}P=\begin{bmatrix}3&-2\\2&3 \end{bmatrix}. $$ Make the change of variable $$\begin{bmatrix} x_1\\x_2\end{bmatrix} =P\begin{bmatrix} u_1\\u_2\end{bmatrix} $$ Then $$\begin{bmatrix}\dot{u_1}\\ \dot{u_2}\end{bmatrix}=\begin{bmatrix}3&-2\\2&3\end{bmatrix} \begin{bmatrix}u_1\\u_2\end{bmatrix} +P^{-1}\begin{bmatrix}f(t)\\g(t)\end{bmatrix} $$.Let us suppose, for the sake of illustration, that $$P^{-1}\begin{bmatrix}f(t)\\g(t)\end{bmatrix}=\begin{bmatrix}2t+1\\5\sin(2t)\end{bmatrix} $$ which gives us $$\begin{bmatrix}\dot{u_1}\\ \dot{u_2}\end{bmatrix}=\begin{bmatrix}3&-2\\2&3\end{bmatrix} \begin{bmatrix}u_1\\u_2\end{bmatrix} +\begin{bmatrix}2t+1\\5\sin(2t)\end{bmatrix} $$. Our second step is to use the method of undetermined coefficients to find particular solutions $u_{1p}$ and $u_{2p}$ of these equations. Your question is what expressions to use. Both for $u_1$ and $u_2$ we take any expression that appears in either one of the equations. So we set $$u_{1p}=at+b+r\cos(2t)+s\sin(2t), $$ $$ u_{2p}=a’t+b’+r’\cos(2t)+s’\sin(2t). $$. Substitute these expressions for $u_{1p}$ and $u_{2p}$ into $\dot{u_1}-(3u_1-2u_2)$and $\dot{u_2}-(2u_1+3u_2)$ respectively and obtain expressions $(…)t+(…)+(…)\cos(2t)+(…)\sin(2t)$. Set each (…) equal to 0, giving 8 linear equations in 8 unknowns. Solve these. The complementary solutions, i.e. the solutions of the homogeneous equations for $u_1$ and $u_2$ are $$u_{1c}=Ce^{3t}\cos(2t)+De^{3t}\sin(2t), u_{2c}=-De^{3t}\cos(2t)+Ce^{3t}\sin(2t) $$ The general solution for $u_1$ and $u_2$ is $$u_1=u_{1c}+u_{1p}, u_2=u_{2c}+u_{2p}$$. From initial conditions on $x_1$ and $x_2$ , we multiply by $P^{-1}$ to obtain initial conditions on $u_1$ and $u_2$, which we use to find $C$ and $D$. Having completely found $u_1$ and $u_2$,we multiply by $P$ to find $x_1$ and $x_2$.