---
title: "Laplace transforms of functions t²sin2t"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-t2sin2t"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-t2sin2t
site: BCT Engineering
---

# Laplace transforms of functions t²sin2t

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2083 Baishakh, 2+3 marks
- Video: [Laplace transform of t^2 sinat, Laplace transform](https://www.youtube.com/watch?v=iUZsaYhkjMQ) — Excellence Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-t2sin2t
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

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

## 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)).

## Answer

#### (i) L{t² sin 2t}
Use L{tⁿ f(t)} = (−1)ⁿ dⁿF/dsⁿ with F(s) = L{sin 2t} = 2/(s² + 4).

```text
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.)

```text
∫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)).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which rule gives L{t^2 f(t)}?
   A. -F''(s)
   B. F'(s)
   C. integral of F from s to infinity
   D. F''(s)
   Answer: D. L{t^n f} = (-1)^n d^nF/ds^n, so for n = 2 the sign is plus. (Key point: Multiplication by t^2)

2. What is F(s) = L{sin 2t}?
   A. 2/(s^2 + 4)
   B. s/(s^2 + 4)
   C. 1/(s^2 + 4)
   D. 2/(s^2 - 4)
   Answer: A. L{sin kt} = k/(s^2 + k^2) with k = 2. (Key point: Transform of sin 2t)

3. What is L{t^2 sin 2t}?
   A. 4(3s^2 + 4)/(s^2 + 4)^3
   B. 4(3s^2 - 4)/(s^2 + 4)^3
   C. -4(3s^2 - 4)/(s^2 + 4)^3
   D. 4s/(s^2 + 4)^2
   Answer: B. F'' = (12s^2 - 16)/(s^2+4)^3 = 4(3s^2 - 4)/(s^2+4)^3. (Key point: Result for t^2 sin 2t)

4. What is L{(cos 2t - cos 3t)/t}?
   A. (1/2) ln((s^2+4)/(s^2+9))
   B. ln((s^2+9)/(s^2+4))
   C. (1/2) ln((s^2+9)/(s^2+4))
   D. (1/2)(s/(s^2+4) - s/(s^2+9))
   Answer: C. Integrating from s to infinity gives 0 - (1/2) ln((s^2+4)/(s^2+9)) = (1/2) ln((s^2+9)/(s^2+4)). (Key point: Result for (cos 2t - cos 3t)/t)

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