Method of Undetermined Coefficients using X's on left hand side

I have a couple questions on this question.

The question is asking me to find the general solution to

$$x'' + 6x' + 9x = \cos(2t) + \sin(2t).$$

Solving for the general solution, I got $$Y_c = C_1e^{-3x}+ C_2xe^{-3x}.$$

I was wondering if the fact that the left hand side uses $x$'s instead of $y$'s matters? For almost every question, it uses $y$'s instead of $x$'s.

Also, for $Y_p$, I got $Y_p = A\cos(2t) + B\sin(2t)$.

I ultimately got the answer to be $$C_1e^{-3x}+ C_2xe^{-3x} - (7\cos(2t)/169) +(17\cos(2t)/169),$$ which I do not feel to be correct.

In this problem, $x$ is the dependant variable, and $t$ is the independant variable. It looks like for your homogenous solution, you solved the homogenous ODE

$$y_c''(x)+6y_c'(x)+9y_c(x)=0$$ With the result

$$y_c(x)=C_1e^{-3x}+C_2xe^{-3x}$$

All the work you did in solving this is totally valid, but you renamed both variables for the ODE. Using the original variables from the problem, we get, $$x_c''(t)+6x_c'(t)+9x_c(t)=0$$ $$\Rightarrow x_c(t)=C_1e^{-3t}+C_2te^{-3t}$$

Your $y_p(t)$ is calculated correctly, but again should actually be named $x_p(t)$.

You can then get the final answer by adding $x_c(t)$ and $x_p(t)$.