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Laplace transforms of functions t²cosat
Short answer
For (a), differentiate L[cos at] twice: result 2s(s²−3a²)/(s²+a²)³. For (b), use first shift on L[t²]=2/s³ to get 2/(s+3)³.
Find the Laplace transform of a) t² cos at b) t² e^(−3t).
a) L[t² cos at]
L[cos at] = s/(s² + a²) = F(s). Then L[t² cos at] = F″(s).
F′(s) = (a² − s²)/(s² + a²)²
F″(s) = (2s³ − 6a²s)/(s² + a²)³
Answer: 2s(s² − 3a²)/(s² + a²)³
b) L[t² e^(−3t)]
L[t²] = 2/s³. By the first shifting theorem (s → s + 3):
Answer: 2/(s + 3)³
✓ Answer checked by Claude Opus 5.5 (high effort)
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