Let the following circuit:
I was looking at some old youtube videos, explaining how obscure this simple looking circuit, here is my question, suppose that the AC voltage source \$ v_i(t) = A \sin(\omega t) \$, I want to understand how this circuit works when \$ \omega \neq \frac{1}{RC}\$, so what I did is find the expression of \$ v_o\$, using basic KVL and laplace's transform, I done this: \$v_o (t) + RC \frac{dv_o}{dt} = A\sin(\omega t) \\ \$
** EDIT **
Using sir Periblepsis approach of annihilators :
\$[D+\frac{1}{\tau}] v_o = A \sin(\omega t) \iff [D^2 + \omega^2][D+\frac{1}{\tau}] v_o = 0 \$
First lets do a quick proof :\$\begin{align*}[D^2 + a^2][D+b] f(t) = 0 \implies f(t) = A_0 \exp{(-bt)} + A_1 \cos{at} + A_2 \sin{at}\end{align*} \$
I will use a formula i think its true : \$ \prod_{i=1}^n [D+a_i] \cdot \sum_{k=1}^n e^{-a_k t} = 0\$
So \$ \begin{align*}[D^2 + a^2][D+b] f(t) = 0 \iff [D+ia][D-ia][D+b] f(t) = 0 \end{align*} \$\$\begin{align*}\implies f(t) = Ae^{-bt} + Be^{iat} + Ce^{-iat} \end{align*}\$\$ \begin{align*} = Ae^{-bt} + \frac{B+C}{2} [e^{iat}+e^{-iat}] + \frac{B-C}{2}[e^{iat}-e^{-iat}] \end{align*} \$\$ \begin{align*} = A_0 e^{-bt} + A_1 \cos(at) + A_2 \sin(at) \end{align*} \$
Thus : \$ \begin{align*} v_o(t=0) = 0 = A_0 + A_1 \land \frac{\partial v_o}{\partial t} (0) = 0 = A_2 \omega - \frac{A_0}{\tau} \land \end{align*} \$\$ \begin{align*} |v_o(t \rightarrow \infty)|^2 = 0 = A_1 ^2 + A_2^2 \end{align*} \$