Steady State Differential Equation I am trying to find the steady-state solution of the following ODE:
$$(D^2+D+4.25I)y=22.1\cos(4.5t)$$
The answer from the back of my textbook is:
$$y_p = -1.28\cos(4.5t)+0.36\sin(4.5t)$$
I found
$$\begin{align}
y_p&=k_1\sin(4.5t)+k_2\cos(4.5)t\\
y_p'&=4.5k_1\cos(4.5t)-4.5k_2\sin(4.5t)\\
y_p''&=-20.25k_1\sin(4.5t)-20.25k_2\cos(4.5t)
\end{align}$$
But I am not sure how to plug back into initial equation to find the answer $y_p$. I don't understand how $D$ and $I$ work.

You are almost there, we choose the particular solution:
$$y_p(x) = a \cos(4.5 t) + b \sin(4.5 t)$$
We take:
$$(D^2 + D + 4.25 I)y_p = y_p'' + y_p' + 4.25 y_p = 22.1 \cos(4.5t)$$
Using your derivatives and adding all these terms and simplifying, we get:
$$(-4.5 a-16 b) \sin (4.5 t)+(-16 a+4.5 b) \cos (4.5 t) = 22.1 \cos(4.5t)$$
Equating like terms, we have to solve the $2x2$ system:
$$-4.5 a-16 b  = 0 \\-16 a +  4.5 b = 22.1$$
Solving this using your favorite approach (Cramer's, Substitution, Gaussian, ...), we get:
$$a=-1.28,   b=0.36$$
Now you have:
$$y(x) = y_h(x) + y_p(x)$$
Now all that is left is finding the steady-state.
