Solving for in the first part of your solution yields
We get This is the initial condition for the next region's DE.
So now you're solving subject to Ideas?
Hmm. I suggest you take a look at Chris's DE tutorial. You can find the technique you need there.
Reply to chisigma:
I'm not following you. What do you mean by "queue" and "head"? I would agree that the LT is a very convenient method of solution for this problem (non-homogeneous IVP's, especially with respect to piece-wise defined functions). However, as Sambit appears not to be familiar with DE's in general, I think the problem is more basic than that.
I would argue that the way I've set the problem up, you will theoretically get a continuous solution on the desired interval. I mean, the initial condition for the region x >= 3 is as I've given, correct?
Why not? Are there no values of C which can make that happen? Incidentally, in post # 6, I was confused by your assertion that C = 4. How did you arrive at that? Is there a part of the problem statement that you haven't posted?...but it does not hold here...
Right, but you've already found the solution for the region 0 < x < pi/2. You're now finding a solution for pi/2 < x < infty. You'll have a piece-wise defined solution, just like you have a piece-wise defined RHS of the DE. Therefore, the condition y(0) = 2 does not apply to the piece you're trying to find now.