gimak said:
Yes, I know that they plugged in Q(t) = Aeiwt into the differential equation & found A. Then they took Re(Aeiwt) & plugged that into energy equation for part a. I still don't understand this: Why can't you plug in Aeiwt into the energy equation as opposed to its real part?
There's a little more to it than that. Recognize that your instructor's solution
first converted/transformed the differential equation from the time domain to the phasor domain by replacing V' \cos(\omega t) with V' e^{i \omega t}. The steady-state solution of the differential equation was then found in the phasor domain. The solution was then converted
back to the time domain (from the phasor domain) by taking the projection of the solution onto the real axis, Q_0 A e^{i \omega t + \phi} \rightarrow \Re \left[ Q_0 A e^{i \omega t + \phi} \right] = \Re [Q_0 A] \cos(\omega t + \phi).
In summary, the process is:
1) Convert (i.e.,
transform) to the phasor domain. (Convert from real sinusoidal to complex spiral)
2) Solve the special case, steady-state solution to the differential equation (ignoring all transients).
3) Convert back to the time domain by finding the real projection of the complex solution.
Why do this? Why convert to the phasor domain in the first place? Answer: it just makes the math a little easier. It's easier to work with e^{i \omega t} than it is with \cos(\omega t). That's all.
You
could have solved the differential equation in terms of cos(\omega t + \phi) from the beginning, but it would have been a lot harder. It's easier to first transform V' \cos(\omega t) \rightarrow V' e^{i \omega t}, solve for the steady-state solution, then transform back via Q_0 A e^{i \omega t + \phi} \rightarrow \Re \left[ Q_0 A e^{i \omega t + \phi} \right] = \Re [Q_0 A] \cos(\omega t + \phi).
By the way, this process of solving the steady-state solution by way of the phasor domain transformation is done by electrical engineers all the time. They even have special notation where Ae^{i \omega t + \theta} = A \angle \theta. Your instructor's solution does not use this notation, but mathematically it's the same thing.
[Edit: The use of the "phasor domain" as described here can be seen as a stepping stone toward solving steady-state solutions of differential equations by means of the "frequency domain" and Fourier analysis. But I'm getting ahead of myself here. Full conversion to the frequency domain transformation was not done in your instructor's solution; all that was done was to transform from real sinusoidal to complex spiral and back.]
