Skip to main content

Laplace transforms of functions te^(3t)sin3t

Short answer

For t e^(3t) sin3t, start with L[sin3t]=3/(s²+9). Multiply by t: L[t sin3t] = 6s/(s²+9)². Shift s→s-3: 6(s-3)/(( s-3)²+9)². For e^(-t) sin t/t, L[sin t/t] = cot⁻¹ s, shift s→s+1 to get cot⁻¹(s+1).

Find the Laplace transform of: i) t e^(3t) sin 3t ii) e^(−t) sin t / t.

i) L[t e^(3t) sin 3t]

L[sin 3t] = 3/(s² + 9). Multiply by t: L[t sin 3t] = −d/ds [3/(s² + 9)] = 6s/(s² + 9)². First shift (s → s − 3):

Answer: 6(s − 3)/((s − 3)² + 9)² = 6(s − 3)/(s² − 6s + 18)²

ii) L[e^(−t) sin t / t]

First, L[sin t / t] = ∫ₛ^∞ du/(u² + 1) = π/2 − tan⁻¹ s = cot⁻¹ s. Shift s → s + 1 for the factor e^(−t):

Answer: cot⁻¹(s + 1) = tan⁻¹(1/(s + 1))

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →