Question

give the laplace transform of f(t)=(sinh3t)^2 --------> The answer is (1/2)[s/(s^2-36)-s^-1], but can someone please explain to me how you get there??

Answer 1

\[f(t)=\sinh^2{3t}=\left(\frac{e^{3t}-e^{-3t}}{2}\right)^2=\frac{e^{6t}+e^{-6t}-2}{4}\]\[L(e^{at})=\frac{1}{s+a}\]\[L(f(t))=\frac{1}{4}\left(L(e^{6t})+L(e^{-6t})-L(2)\right)=\frac{1}{4}\left(\frac{1}{s+6}+\frac{1}{s-6}-\frac{2}{s}\right)=\]\[=\frac{1}{4}\left(\frac{2s}{s^2-36}-\frac{2}{s}\right)=\frac{1}{2}\left(\frac{s}{s^2-36}-\frac{1}{s}\right)\]

Answer 2

thank you kind sir/madam!