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- 2068 Baishakh · 5 marks
Laplace transforms of functions te^(−3t)cos2t
Short answer
L[t cos 2t] = (s^2-4)/(s^2+4)^2, shift s→s+3 gives L[t e^(-3t) cos 2t] = (s^2+6s+5)/(s^2+6s+13)^2. Integration property gives L[(e^(at)-cos 6t)/t] = ln(√(s^2+36)/(s-a)).
Find the Laplace transform of the following functions: a) t e^(−3t) cos 2t b) (e^(at) − cos 6t)/t.
a) L[t e^(−3t) cos 2t]
L[cos 2t] = s/(s² + 4). L[t cos 2t] = −d/ds [s/(s²+4)] = (s² − 4)/(s² + 4)². Shift s → s + 3:
Answer: ((s + 3)² − 4)/((s + 3)² + 4)² = (s² + 6s + 5)/(s² + 6s + 13)²
b) L[(e^(at) − cos 6t)/t]
Use L[f/t] = ∫ₛ^∞ F(u) du with F(u) = 1/(u − a) − u/(u² + 36):
∫ₛ^∞ F(u) du = [ ln( (u − a)/√(u² + 36) ) ]ₛ^∞ = 0 − ln( (s − a)/√(s² + 36) )
Answer: ln( √(s² + 36)/(s − a) ), for s > a.
✓ Answer checked by Claude Opus 5.5 (high effort)
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