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- 2076 Asoj · 2+3 marks
Laplace transforms of functions t²cosat
Short answer
i) Using L{t² f(t)} = d²F/ds² with F(s)=s/(s²+a²) yields 2s(s²−3a²)/(s²+a²)³. ii) Using L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s)=1/s − s/(s²−a²) gives ½ ln((s²−a²)/s²).
Find the Laplace transform of: i) t² cos at ii) (1 − cosh at)/t.
i) L{t² cos at}
Use L{t² f(t)} = d²F/ds² with F(s) = s/(s² + a²).
F'(s) = (a² − s²)/(s² + a²)²
F''(s) = [−2s(s² + a²)² − (a² − s²)·2(s² + a²)·2s]/(s² + a²)⁴
= [−2s(s² + a²) − 4s(a² − s²)]/(s² + a²)³
= (2s³ − 6a²s)/(s² + a²)³
= 2s(s² − 3a²)/(s² + a²)³
ii) L{(1 − cosh at)/t}
Use L{f(t)/t} = ∫ₛ^∞ F(u) du with F(s) = 1/s − s/(s² − a²).
∫ₛ^∞ [1/u − u/(u² − a²)] du = [ln u − ½ ln(u² − a²)]ₛ^∞
= 0 − ln(s/√(s² − a²))
= ln(√(s² − a²)/s) = ½ ln((s² − a²)/s²) (s > |a|)
Answers: i) 2s(s² − 3a²)/(s² + a²)³; ii) ½ ln((s² − a²)/s²).
✓ Answer checked by Claude Opus 5.5 (high effort)
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