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Laplace transforms of functions sin2tcosh4t

Short answer

Expres cosh4t as (e^(4t)+e^(-4t))/2, apply shifting to L[sin2t]=2/(s^2+4) giving 1/(( s-4)^2+4) +1/(( s+4)^2+4), simplify to 2(s^2+20)/(s^4-24s^2+400). For t e^(2t) sin4t, L[t sin4t]=8s/(s^2+16)^2, shift s→s-2 to get 8(s-2)/(( s-2)^2+16)^2.

Find the Laplace transform of (i) sin 2t cosh 4t (ii) t e^(2t) sin 4t.

(i) L[sin 2t cosh 4t]

cosh 4t = (e^(4t) + e^(−4t))/2 and L[sin 2t] = 2/(s² + 4). By the first shifting theorem,

L[sin 2t cosh 4t] = ½ [ 2/((s − 4)² + 4) + 2/((s + 4)² + 4) ]
                  = 1/((s − 4)² + 4) + 1/((s + 4)² + 4)

Combine: the denominators multiply to (s² + 20)² − 64s² = s⁴ − 24s² + 400, and the numerator is 2(s² + 20).

Answer: 2(s² + 20)/(s⁴ − 24s² + 400)

(ii) L[t e^(2t) sin 4t]

L[sin 4t] = 4/(s² + 16). L[t sin 4t] = −d/ds [4/(s² + 16)] = 8s/(s² + 16)². Shift s → s − 2:

Answer: 8(s − 2)/((s − 2)² + 16)²

✓ Answer checked by Claude Opus 5.5 (high effort)

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