$$y''+4y=sin\;t(1-u(t-2\pi))$$ $$L\lbrace y''\rbrace+L\lbrace 4y\rbrace= L\lbrace sin\;t \rbrace - L\lbrace sin\;t\;u(t-2\pi)\rbrace$$ $$s^2Y(s)-s\;y(0)-y'(0)+4Y(s)={1\over s^2+1}-e^{-2\pi s}L\lbrace{sin(t+2\pi)\over sin\;t}\rbrace$$ $$={1\over s^2+1}-{e^{-2\pi s}\over s^2+1}$$ The LHS doesn't give me a problem, the Laplace of just $sin\;t$ doesn't give me a problem, but the steps taken for the $L\lbrace sin\;t\; u(t-2\pi)\rbrace$ I really don't understand. I have a feeling that this Laplace transform : $L\lbrace f(t-a)u(t-a)\rbrace(s)=e^{-as}F(s)$ has something to do with it, but I don't know how to get there. Where did the ${sin(t+2\pi)\over sin\;t}$ come from?
I think that maybe wherever you got this, it is the author's attempt to say that $\sin(t + 2 \pi) = \sin(t)$... i.e. "We are taking the Laplace transform of $\sin(t + 2\pi)$... but that is just the Laplace transform of $\sin(t)$."
It is not only the only thing that seems to make sense in terms of where that came from and the formula that you gave with the Heaviside step function, but also with how when the author takes the Laplace transform of $\sin(t + 2\pi)/ \sin(t) = 1$ and gets $\frac{1}{s^2 + 1}$, the Laplace transform of $\sin(t)$.
• Yes, it was a friend of mine who did this problem (I was using his test as a study guide) and his professor gave him full credit for that problem. I asked him later and he said that instead of ${sin(t-\pi)\over sin(t)}$ he meant to have it as $sin(t)$. So I suppose that the professor just missed that error. – matryoshka Jun 14 '17 at 18:36