Extra negative in in Laplace transform solution

[abalmos]

I have the correct solution of this problem (cos(t)+u_{3\pi}[1-cos(t-3\pi)]). As you can tell from my solution below I am off by a extra negative sign on the cos(t) coefficient. I have had this same issue on several other problems, so I believe I must be missing something on the procedure. Any guidance would be much appreciated!

Solve IVP y'' + y = u_{3\pi}(t) \left\{y(0)=1, y'(0)=0

My solution:
s^{2}Y(s) + sY(0) + Y'(0) + Y(s) = \frac{e^{-3\pi s}}{s}
Y(s)(s^{2} +1) + s = \frac{e^{-3\pi s}}{s}
Y(s) = \frac{e^{-3\pi s}}{s(s^{2}+1)} - \frac{s}{(s^{2}+1)}

H(t) = \frac{1}{s(s^{2}+1)}
h(t) = Y(H(t))

y = u_{3\pi}h(t-3\pi}) - cos(t)

h(t) = \frac{1}{s} - \frac{s}{s^{2}+1}
h(t) = 1 - cos(t)

so ...

-cos(t) + u_{3\pi}[1-cos(t-3\pi)]


I have seen in other examples (http://tutorial.math.lamar.edu/Classes/DE/IVPWithStepFunction.aspx , example 2) which have similar setups, and I noticed that when the little h is introduce (the inverse transformation to the u_{3\pi} term) the sign of the cos(t) term changes but I do not understand why.

Thank you so much for help and guidance you can provide!

- Andrew Balmos