# Extra negative in in Laplace transform solution

1. Nov 4, 2009

### 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