PD=potential difference.
WHERE did I apply the boundary conditions?

... write out the general solution (from the DE) for the kind of quantity you have to solve when you do linear circuit analysis, then compare with the result you get when you use that analysis, and you'll be able to see what the boundary conditions were.
You did explicitly assume I(0)=0 and Q(0)=a (Q(0) where?).
When you draw the voltage source in the diagram, you specify a phase direction. When you draw the PD arrows on the diagram  you use that phase direction to decide if the potential is gained or dropped. Those contribute to the boundary conditions. Since the current will be sinusoidal, it actually doesn't matter what the initial phase is  so we have picked one that makes the math easy.
You'll see it clearly when you do the calculus.