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Laplace transforms of functions

Short answer

a) t e^{-2t} cos t: ((s+2)^2-1)/(( (s+2)^2+1)^2 = (s^2+4s+3)/(s^2+4s+5)^2. b) sinh at cos t: Use exponential definition, result a(s^2-a^2-1)/(((s-a)^2+1)((s+a)^2+1).

Find the Laplace transform of the following: a) t e^(−2t) cos t b) sinh at cos t.

a) L[t e^(−2t) cos t]

L[cos t] = s/(s² + 1). Multiply by t: L[t cos t] = −d/ds [s/(s²+1)] = (s² − 1)/(s² + 1)². First shift (s → s + 2):

Answer: ((s + 2)² − 1)/((s + 2)² + 1)² = (s² + 4s + 3)/(s² + 4s + 5)²

b) L[sinh at cos t]

sinh at = (e^(at) − e^(−at))/2, so

L[sinh at cos t] = ½ { L[e^(at) cos t] − L[e^(−at) cos t] }
                 = ½ [ (s − a)/((s − a)² + 1) − (s + a)/((s + a)² + 1) ]

Combining the fractions (numerator: (s−a)((s+a)²+1) − (s+a)((s−a)²+1) = 2a(s² − a² − 1)):

Answer: a(s² − a² − 1) / [((s − a)² + 1)((s + a)² + 1)]

✓ Answer checked by Claude Opus 5.5 (high effort)

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