---
title: "Laplace transforms of functions t²cosat"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-t2cosat-three"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-t2cosat-three
site: BCT Engineering
---

# Laplace transforms of functions t²cosat

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2067 Magh
- Video: [Problem 1 Based on Multiplication By 't' Property - Laplace Transform - Engineering Mathematics 3](https://www.youtube.com/watch?v=B_f5rJ-Bues) — Ekeeda
- 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-t2cosat-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the Laplace transform of a) t² cos at  b) t² e^(−3t).

## Short answer

For (a), differentiate L[cos at] twice: result 2s(s²−3a²)/(s²+a²)³. For (b), use first shift on L[t²]=2/s³ to get 2/(s+3)³.

## Answer

#### a) L[t² cos at]
L[cos at] = s/(s² + a²) = F(s). Then L[t² cos at] = F″(s).

```text
F′(s) = (a² − s²)/(s² + a²)²
F″(s) = (2s³ − 6a²s)/(s² + a²)³
```

**Answer: 2s(s² − 3a²)/(s² + a²)³**

#### b) L[t² e^(−3t)]
L[t²] = 2/s³. By the first shifting theorem (s → s + 3):

**Answer: 2/(s + 3)³**

## Check yourself

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

1. L[t^2 cos at] is found as
   A. F''(s) where F = s/(s^2+a^2)
   B. -F'(s)
   C. F(s)/s^2
   D. F(s - a)
   Answer: A. t^n multiplication gives (-1)^n F^(n)(s). (Key point: Formula)

2. F'(s) for F = s/(s^2+a^2) is
   A. (s^2 - a^2)/(s^2 + a^2)^2
   B. (a^2 - s^2)/(s^2 + a^2)^2
   C. 1/(s^2 + a^2)
   D. -2s/(s^2 + a^2)^2
   Answer: B. Quotient rule. (Key point: First derivative)

3. L[t^2 cos at] =
   A. 2s(s^2 + 3a^2)/(s^2 + a^2)^3
   B. 2s(3s^2 - a^2)/(s^2 + a^2)^3
   C. 2s(s^2 - 3a^2)/(s^2 + a^2)^3
   D. (s^2 - 3a^2)/(s^2 + a^2)^3
   Answer: C. Differentiate again to get (2s^3 - 6a^2 s)/(s^2+a^2)^3. (Key point: Answer (a))

4. L[t^2 e^(-3t)] =
   A. 2/(s-3)^3
   B. 2/(s+3)^2
   C. 1/(s+3)^3
   D. 2/(s+3)^3
   Answer: D. L[t^2] = 2/s^3 and s -> s + 3. (Key point: Answer (b))

## Close variants set in other exams

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