Skip to main content

Laplace transforms of functions t²sin2t

Short answer

For (i) differentiate L{sin 2t}=2/(s²+4) twice: L{t² sin 2t}=4(3s²−4)/(s²+4)³. For (ii) integrate s/(s²+4)−s/(s²+9) from s to ∞: L{(cos 2t−cos 3t)/t}=½ ln((s²+9)/(s²+4)).

Obtain the Laplace transforms of (i) t² sin 2t (ii) (cos 2t − cos 3t)/t.

(i) L{t² sin 2t}

Use L{tⁿ f(t)} = (−1)ⁿ dⁿF/dsⁿ with F(s) = L{sin 2t} = 2/(s² + 4).

F'(s)  = −4s/(s² + 4)²
F''(s) = −4/(s² + 4)² + 16s²/(s² + 4)³
       = [−4(s² + 4) + 16s²]/(s² + 4)³
       = (12s² − 16)/(s² + 4)³
L{t² sin 2t} = F''(s) = 4(3s² − 4)/(s² + 4)³

(ii) L{(cos 2t − cos 3t)/t}

Use L{f(t)/t} = ∫ₛ^∞ F(u) du, with F(s) = s/(s² + 4) − s/(s² + 9). (f(t)/t → 0 as t → 0.)

∫ₛ^∞ [u/(u² + 4) − u/(u² + 9)] du = [½ ln((u² + 4)/(u² + 9))]ₛ^∞
 = 0 − ½ ln((s² + 4)/(s² + 9))
 = ½ ln((s² + 9)/(s² + 4))

Answers: (i) 4(3s² − 4)/(s² + 4)³; (ii) ½ ln((s² + 9)/(s² + 4)).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →