Boundary conditions for a purely inductive load in an AC circuit

[rude man]

Uhm. Right, but the standard derivation of a purely inductive circuit never mentions any resistor, and
<br /> I(0) = \frac{V_0}{\omega L}<br />
It seems hard to make contact with your answer.

The resistor is there just to set up the initial condition. When you apply your voltage $V_0 sin(\omega t)$ it no longer forms any part of the inductor current.

You have set up an initial current and have then applied your voltage directly across the inductor.