ACDC (Part 2)

Differential equation of the circuit is: $v_{S}(t)-Ri-L\frac{di}{dt} - V_{R} = 0$ This is a first-order linear non-homogeneous DE. We have two "inputs" to the equation, one is a constant - $V_{R}$ - and the other is sinusoidal - $v_{S}$ . Since the equation is linear, we can solve separately for each input and then superpose solutions to get the solution to our two-input equation. These inputs are very common, so I will just write out the final solution without derivation: $i(t) = \underbrace{Ce^{-\frac{R}{L}t}}_{\text{hom. part}} ~+~\underbrace{-\frac{V_{R}}{R}}_{\text{const. input}} ~+~\underbrace{\frac{V_{S}}{\left | Z \right |}\sin{(\omega t - \phi)}}_{\text{sinusoidal input}}$ where $\left | Z \right | = \sqrt{R^{2}+\omega^{2}L^{2}}$ and $\phi = \arctan{\frac{\omega L}{R}}$ . Constant C is determined from initial condition $i(t = 0) = 0$ which is implied since any amount of current will produce a magnetic field inside the inductor, contrary to what was said in the question. Once we determine C and plug in given values, we get: $\begin{aligned} i(t) &= \left(1+\frac{1}{\sqrt{2}} \right )e^{-120\pi t} + \sin{\left(120\pi t - \frac{\pi}{4}\right)} - 1 \\ \frac{di}{dt} &= 120\pi\left[-\left(1+\frac{1}{\sqrt{2}} \right )e^{-120\pi t}+\cos{\left(120\pi t - \frac{\pi}{4}\right)} \right ]\end{aligned}$ By numerical means we find that first time that time-derivative of the current is 0 is at $t\approx 0.001486$ and we find current $i \approx -0.248379$ .

I assumed that the starting polarity of AC source is the opposite of the DC source, but this should've been explicitly marked in the question. Also, in writing out DE I assumed the direction of the current to be clockwise.