How to solve the differential equation for driven series RLC circuit?

[omoplata]

Homework Statement

It is the driven series RLC circuit. It is given in the following images.

It is from the section 12.3 in this note.

Homework Equations

The differential equation, as given by 12.3.3, is $L \frac{d^2 Q}{d t^2} + R \frac{d Q}{d t} + \frac{Q}{C} = V_0 \sin{(\omega t)}$.

It says that one possible solution is $Q(t) = Q_0 \cos{(\omega t - \phi)}$, where $Q_0 = \frac{V_0}{\omega \sqrt{R^2 + (\omega L - 1 / \omega C)^2}}$ and $\tan \phi = \frac{1}{R} \left( \omega L - \frac{1}{\omega C} \right)$.

The Attempt at a Solution

So, assuming $Q(t) = Q_0 \cos{(\omega t - \phi)}$, I get$$\frac{d Q}{d t} = -Q_0 \omega \sin{(\omega t - \phi)}\\\frac{d^2 Q}{d t^2} = -Q_0 \omega^2 \cos{(\omega t - \phi)}$$
Substituting,$$-L Q_0 \omega^2 \cos{(\omega t - \phi)} - R Q_0 \omega \sin{(\omega t - \phi)} + \frac{Q_0}{C} \cos{(\omega t - \phi)} = V_0 \sin{(\omega t - \phi)}$$

I cannot figure out show to eliminate $t$ and $\phi$ and solve this for $Q_0$.

I try applying$$\cos{(\omega t - \phi)} = \cos{(\omega t)} \cos \phi + \sin{(\omega t)} \sin \phi\\\sin{(\omega t - \phi)} = \sin{(\omega t)} \cos \phi - \cos{(\omega t)} \sin \phi$$

But still there's no way cancel $t$ and $\phi$ and solve for $Q_0$, or cancel $Q_0$ and $t$ and solve for $\phi$.

 

[gneill]

The equations should hold for any value of t. Choose wisely