Laplace transforms of functions te^(−4t)sin3t

Short answer

a) t e^{-4t} sin 3t: 6(s+4)/( ( (s+4)^2+9 )^2. b) (cos at - cos bt)/t: ½ ln((s^2+b^2)/(s^2+a^2)).

Find the Laplace transforms of the following functions: a) t e^(−4t) sin 3t b) (cos at − cos bt)/t.

L[sin 3t] = 3/(s² + 9). L[t sin 3t] = −d/ds [3/(s²+9)] = 6s/(s² + 9)². Shift s → s + 4:

Answer: 6(s + 4)/((s + 4)² + 9)² = 6(s + 4)/(s² + 8s + 25)²

F(u) = u/(u² + a²) − u/(u² + b²).

∫ₛ^∞ F(u) du = ½ [ ln((u² + a²)/(u² + b²)) ]ₛ^∞
             = ½ [ 0 − ln((s² + a²)/(s² + b²)) ]

Answer: ½ ln((s² + b²)/(s² + a²))

✓ Answer checked by Claude Opus 5.5 (high effort)

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