Either it works for the device, or it doesn't.

In this case, what part about V(t) = L di/dt reads as subject to waveform?

Suppose you had a physically impossible, perfect squarewave switching between -5 and 5V at 100kHz. The period is 10us which means 5us on, and 5us off. And we have a physically impossibly-perfect 10uH inductor.

At time 0+, we apply 5V across the inductor. We can divide both sides by L to find the slew rate of the current:

5V/10u = di/dt

500,000A/s = di/dt

Now, multiply both sides by the differential dt

5*10^5 dt = di

Integrate both sides

5*10^5*t + C= i(t)

Let C=0 since we're just energizing it. After a perfect 5us, we have 5*10^5 * 5 * 10^-6 = 25 * 10^-1 or 0.25A stored in the inductor after 5us.

Now, repeat the same thing but with -5V as applied voltage. Since you have an initial current of 0.25A, use that as your constant after integration. You will find that the current has returned to 0 after 10us.

This is only for the perfect system as described, and is "pretty close" for a real world system.