---
title: "Laplace transforms of functions te^(−3t)cos2t"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:laplace-transforms-of-functions-te-3t-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/laplace-transforms-of-functions-te-3t-two
site: BCT Engineering
---

# Laplace transforms of functions te^(−3t)cos2t

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2068 Baishakh, 5 marks
- Video: [L[e^3t(tcos2t)] #Problemonmultiplicationbypowersoft #Laplacetransforms L1k,104](https://www.youtube.com/watch?v=O_zHvSir6to) — Maths Pulse - Chinnaiah Kalpana
- 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-te-3t-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the Laplace transform of the following functions: a) t e^(−3t) cos 2t  b) (e^(at) − cos 6t)/t.

## Short answer

L[t cos 2t] = (s^2-4)/(s^2+4)^2, shift s→s+3 gives L[t e^(-3t) cos 2t] = (s^2+6s+5)/(s^2+6s+13)^2. Integration property gives L[(e^(at)-cos 6t)/t] = ln(√(s^2+36)/(s-a)).

## Answer

#### a) L[t e^(−3t) cos 2t]
L[cos 2t] = s/(s² + 4).
L[t cos 2t] = −d/ds [s/(s²+4)] = (s² − 4)/(s² + 4)².
Shift s → s + 3:

**Answer: ((s + 3)² − 4)/((s + 3)² + 4)² = (s² + 6s + 5)/(s² + 6s + 13)²**

#### b) L[(e^(at) − cos 6t)/t]
Use L[f/t] = ∫ₛ^∞ F(u) du with F(u) = 1/(u − a) − u/(u² + 36):

```text
∫ₛ^∞ F(u) du = [ ln( (u − a)/√(u² + 36) ) ]ₛ^∞ = 0 − ln( (s − a)/√(s² + 36) )
```

**Answer: ln( √(s² + 36)/(s − a) )**, for s > a.

## Check yourself

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

1. How is L[t cos 2t] obtained from L[cos 2t] = s/(s^2+4)?
   A. Take -d/ds of s/(s^2+4)
   B. Take d/ds of s/(s^2+4) with no sign
   C. Integrate s/(s^2+4) from s to infinity
   D. Replace s by s + 2
   Answer: A. L[t f(t)] = -F'(s). (Key point: Multiplication by t)

2. Having L[t cos 2t] = (s^2-4)/(s^2+4)^2, how is the factor e^(-3t) handled?
   A. Replace s by s - 3
   B. Replace s by s + 3
   C. Multiply by 1/(s+3)
   D. Divide by s + 3
   Answer: B. Multiplying by e^(-3t) shifts s to s + 3. (Key point: First shifting)

3. What is L[t e^(-3t) cos 2t]?
   A. (s^2 - 6s + 5)/(s^2 - 6s + 13)^2
   B. (s^2 + 6s + 13)/(s^2 + 6s + 5)^2
   C. (s^2 + 6s + 5)/(s^2 + 6s + 13)^2
   D. (s^2 + 6s + 5)/(s^2 + 6s + 13)
   Answer: C. ((s+3)^2-4)/((s+3)^2+4)^2 expands to this. (Key point: Answer (a))

4. Which formula is used for L[(e^(at) - cos 6t)/t]?
   A. -F'(s)
   B. F(s)/s
   C. F(s - a)
   D. Integral of F(u) from s to infinity
   Answer: D. Dividing by t corresponds to integrating F from s to infinity. (Key point: Division by t)

## Close variants set in other exams

- [Laplace transforms of functions](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Alaplace-transforms-of-functions) — Mathematics III, chapter 3, asked 4×
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