Find the resonance frequencies of a circuit governed by a given differential equation

The current in an electric circuit obeys the differential equation

$I''(t) + RI'(t) + I(t) = \sin (\omega t)$

for positive real numbers $R$ and $\omega$ . The solutions to this equation can be expressed as

$I(t) = F(t) + A \sin (\omega t + a),$

where $A>0$ and $a$ are constants which depend on $R$ and $\omega$ and

$\lim_{t \to \infty} F(t) = 0.$

If there exists a value of $\omega$ which maximizes $A$ then the value $\frac{\omega}{2\pi}$ is called a resonant frequency of the circuit.

Find all resonant frequencies when $R = 1$ .
Find all values of $R$ such that the frequency has a resonant frequency.

The homogeneous equation $I'' + I' + I = 0$ is of the form $y'' + ay' + by = 0$ where $a = 1$ and $b = 1$ . This gives us $d = a^2 - 4b = -3$ . Therefore, $k = \frac{1}{2}\sqrt{-d} = \frac{\sqrt{3}}{2}$ . Hence, the general solution of the homogeneous equation is
$I = e^{-\frac{t}{2}} \left( A \cos \frac{\sqrt{3}}{2} + B \sin \frac{\sqrt{3}}{2} \right).$

Now, to find a particular solution of $I'' + I' + I = \sin (\omega t)$ we let

$\begin{align*} && I(t) &= c_1 \cos (\omega t) + c_2 \sin (\omega t) \\ \implies && I'(t) &= \omega c_2 \cos (\omega t) - \omega c_1 \sin (\omega t) \\ \implies && I''(t) &= -\omega^2 c_2 \sin (\omega t) - \omega^2 c_1 \cos (\omega t). \end{align*}$

Plugging these into the differential equation we have

$-\omega^2 c_2 \sin (\omega t) - \omega^2 c_1 \cos (\omega t) + \omega c_2 \cos (\omega t) - \omega c_1 \sin (\omega t) + c_1 \cos (\omega t) + c_2 \sin (\omega t) = \sin (\omega t).$

Evaluating at $t = \frac{\pi}{2}$ and $t = 0$ we obtain the two equation

$\begin{align*} -c_2 \omega^2 - c_1 \omega + c_2 &= 1, \\ -c_1 \omega^2 + c_2 \omega + c_1 &= 0. \end{align*}$

Solving for $c_1$ and $c_2$ we obtain

$c_1 = \frac{-\omega}{\omega^2 + (\omega^2-1)^2}, \qquad c_2 = \frac{1-\omega^2}{\omega^2 + (\omega^2 -1)^2}.$

Thus, we have the particular solution

$I_1 = \frac{1-\omega^2}{(\omega^2-1)^2 + \omega^2} \sin (\omega t) - \frac{\omega}{\omega^2 + (\omega^2-1)^2} \cos (\omega t).$

Therefore, the general solution is

$\begin{align*} I(t) &= e^{-\frac{t}{2}} \left( A \cos \frac{\sqrt{3}}{2} t + B \sin \frac{\sqrt{3}}{2} t \right) - \frac{\omega}{\omega^2 + (\omega^2 -1)^2} \cos (\omega t) + \frac{1-\omega^2}{(\omega^2 -1)^2 + \omega^2} \sin (\omega t) \\ &= F(t) + C \left( -\omega \cos (\omega t) + (1-\omega^2) \sin (\omega t) \right). \end{align*}$

Where,

$F(t) = e^{-\frac{t}{2}} \left( A \cos \frac{\sqrt{3}}{2} t + B \sin \frac{\sqrt{3}}{2} t \right).$

This implies

$F(t) \to 0 \quad \text{as} \quad t \to + \infty.$

Furthermore, from a previous exercise (Section 2.8, Exercise #9) we know for $C_0 = \sqrt{\omega^2 + (1-\omega)^2}$ we have

$\begin{align*} I(t) &= F(t) - C_0 \cdot C (\sin (\omega t + \alpha)) \\ &= F(t) - A \sin (\omega t + \alpha) \end{align*}$

where

$A = \frac{\sqrt{\omega^2+ (1-\omega^2)^2}}{\omega^2 + (1-\omega^2)^2} = \frac{1}{\sqrt{\omega^2 + (1-\omega^2)^2}}.$

So, to maximize $A$ we minimize $\sqrt{\omega^2 + (1-\omega^2)^2}$ . Setting the derivative equal to 0 we have

$\begin{align*} && \frac{d}{d \omega} \left( \omega^2 + (1-\omega^2) \right) &= 0 \\ \implies && 2 \omega - 4 \omega (1-\omega^2) &= 0 \\ \implies && \omega = \frac{1}{\sqrt{2}}. \end{align*}$

Hence, the resonance frequency is

$\frac{\omega}{2 \pi} = \left( \frac{1}{\sqrt{2}} \right) \left( \frac{1}{2 \pi} \right).$
In order to have a resonance frequency, we must have a value of $\omega$ such that $A$ is maximized. Hence,
$D \left( (R \omega)^2 + (1 - \omega^2)^2 \right) = 0 \quad \implies \quad R < \sqrt{2}.$