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?
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.