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Laplace transforms of functions t²sint

Short answer

Using L[t^2 f]=F''(s) with L[sin t]=1/(s^2+1) gives L[t^2 sin t]=2(3s^2-1)/(s^2+1)^3. Using sinh at = (e^(at)-e^(-at))/2 and first shift gives L[cos at sinh at]=a(s^2-2a^2)/(s^4+4a^4).

Find the Laplace transform of the given function: (i) t² sin t (ii) cos at sinh at.

(i) L[t² sin t]

L[sin t] = 1/(s² + 1). Use L[t² f] = F″(s).

F′(s) = −2s/(s² + 1)²
F″(s) = [−2(s²+1)² + 2s·2(s²+1)·2s]/(s²+1)⁴
      = [−2(s²+1) + 8s²]/(s²+1)³
      = (6s² − 2)/(s² + 1)³

Answer: 2(3s² − 1)/(s² + 1)³

(ii) L[cos at sinh at]

sinh at = (e^(at) − e^(−at))/2.

L[cos at sinh at] = ½ [ (s − a)/((s−a)² + a²) − (s + a)/((s+a)² + a²) ]

Denominators: (s − a)² + a² = s² − 2as + 2a², (s + a)² + a² = s² + 2as + 2a². Their product is (s² + 2a²)² − 4a²s² = s⁴ + 4a⁴. Numerator: (s−a)(s² + 2as + 2a²) − (s+a)(s² − 2as + 2a²) = 2a(s² − 2a²).

Answer: a(s² − 2a²)/(s⁴ + 4a⁴)

✓ Answer checked by Claude Opus 5.5 (high effort)

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