RC Circuit Example

Q: Suppose you have a circuit consisting of a voltage source $v_{in}(t)=v_{in}sin(\omega t)$ in a series RC circuit.
What is the complete solution for the current $i(t)$ through the circuit?

A: By KVL:
$v_{in}(t)=v_R+v_C$
$v_{in}(t)=Ri(t)+v_C$
Differentiating both sides:
$\frac{dv_{in}(t)}{dt}=R\frac{di(t)}{dt}+\frac{dv_C}{dt}$
From the definition of a capacitor:
$\frac{dv_{in}(t)}{dt}=R\frac{di(t)}{dt}+\frac{1}{C}i(t)$
$C\frac{dv_{in}(t)}{dt}=RC\frac{di(t)}{dt}+i(t)$
Defining a time constant $\tau \equiv RC$:
$C\frac{dv_{in}(t)}{dt}=\tau \frac{di(t)}{dt}+i(t)$
$\tau \frac{di(t)}{dt}+i(t)=C\frac{dv_{in}(t)}{dt}$

$i(t)=?$
$i(t)=i_h(t)+i_f(t)$

$i_h(t)=?$
$\tau \frac{di_h(t)}{dt}+i_h(t)=0$
Suppose $i_h(t)=Ae^{st}$:
$\tau sAe^{st}+Ae^{st}=0$
Since $e^{st} \neq 0$:
$\tau sA+A=0$
Since we are looking for a non-trivial solution - i.e. $A \neq 0$:
$\tau s + 1 =0$
$s = -\frac{1}{\tau}$
$i_h(t)=Ae^{-\frac{t}{\tau}}$

$i_f(t)=?$

In this post, a basic method will be used to solve the zero-state solution for current. But a method using complex numbers is examined in another post.
$\tau \frac{di_f(t)}{dt}+i(t)=C\frac{dv_{in}(t)}{dt}$
$\tau \frac{di_f(t)}{dt}+i(t)=v_{in}\omega Ccos(\omega t)$
For cleaner notation, let $K \equiv v_{in}\omega C$
$\tau \frac{di_f(t)}{dt}+i(t)=Kcos(\omega t)$

Suppose $i_f(t)=Dcos(\omega t)+Esin(\omega t)$
 $-D \omega \tau sin(\omega t)+E \omega \tau cos(\omega t)+Dcos(\omega t)+Esin(\omega t)=Kcos(\omega t) $
 $(E-D \omega \tau) sin(\omega t)+(D+E \omega \tau) cos(\omega t)=Kcos(\omega t) $

$\begin{cases} E-D \omega \tau = 0 \\ D+E \omega \tau=K \end{cases}$

$E=D \omega \tau$
$D+D \omega^2 \tau^2=K$
$D=\frac{K}{\omega^2 \tau^2}$

$i_f(t)=\frac{K}{\omega^2 \tau^2}(cos(\omega t)+\omega\tau sin(\omega t))$
$i_f(t)=\frac{v_{in}\omega C}{\omega^2 \tau^2}(cos(\omega t)+\omega\tau sin(\omega t))$