P Peleus Sep 2007 81 6 May 11, 2010 #1 Hi all, Sorry I hate posting urgent questions up here but either way I need some help. I'm looking for the laplace of \(\displaystyle \cosh^2{t}\) but I've got no idea how to go about it, can anyone help out? Cheers.

Hi all, Sorry I hate posting urgent questions up here but either way I need some help. I'm looking for the laplace of \(\displaystyle \cosh^2{t}\) but I've got no idea how to go about it, can anyone help out? Cheers.

D dwsmith MHF Hall of Honor Mar 2010 3,093 582 Florida May 11, 2010 #2 Peleus said: Hi all, Sorry I hate posting urgent questions up here but either way I need some help. I'm looking for the laplace of \(\displaystyle \cosh^2{t}\) but I've got no idea how to go about it, can anyone help out? Cheers. Click to expand... \(\displaystyle cosh=\frac{e^x+e^{-x}}{2}\)

Peleus said: Hi all, Sorry I hate posting urgent questions up here but either way I need some help. I'm looking for the laplace of \(\displaystyle \cosh^2{t}\) but I've got no idea how to go about it, can anyone help out? Cheers. Click to expand... \(\displaystyle cosh=\frac{e^x+e^{-x}}{2}\)

chisigma MHF Hall of Honor Mar 2009 2,162 994 near Piacenza (Italy) May 11, 2010 #3 Is... \(\displaystyle \cosh^{2} t = \frac{1}{2} + \frac{e^{2t} + e^{-2t}}{4}\) (1) ... so that... \(\displaystyle \mathcal{L} \{\cosh^{2} t\} = \frac{1}{2s} + \frac{1}{4(s-2)} + \frac{1}{4(s+2)}\) (2) Kind regards \(\displaystyle \chi\) \(\displaystyle \sigma\)

Is... \(\displaystyle \cosh^{2} t = \frac{1}{2} + \frac{e^{2t} + e^{-2t}}{4}\) (1) ... so that... \(\displaystyle \mathcal{L} \{\cosh^{2} t\} = \frac{1}{2s} + \frac{1}{4(s-2)} + \frac{1}{4(s+2)}\) (2) Kind regards \(\displaystyle \chi\) \(\displaystyle \sigma\)