
Benjamin T. answered 01/20/23
Physics Professor, and Former Math Department Head
Probably the best thing you can do to start this problem is rewrite, (sinh t + cosh t)2. Using the definition of
sinh(t) = (et-e-t)/2 and cosh(t) = (et+e-t)/2,
(sinh t + cosh t)2=e2x.
Given L[tneat] = n!/(s-a)n+1, then
L[t(sinh t + cosh t)2] = 1/(s-2)2.