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

# Laplace transforms of functions tcoshat

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2081 Bhadra, 1+2+2 marks
- Video: [Find Laplace transform of the tcosht](https://www.youtube.com/watch?v=ZUg8qtvRhZI) — Doctor of Mathematics
- 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-tcoshat
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

State the existence condition for the Laplace transform. Obtain the Laplace transform of a) t cosh at  b) sin 2t / t.

## Short answer

A sufficient condition: piecewise continuous and of exponential order. a) Differentiate s/(s²−a²) → (s²+a²)/(s²−a²)². b) Integrate 2/(s²+4) → tan⁻¹(2/s).

## Answer

**Existence condition.** The Laplace transform L{f(t)} = ∫₀^∞ e^(−st) f(t) dt exists for s > k if f(t) is piecewise continuous on every finite interval [0, T] and of exponential order, that is |f(t)| ≤ M e^(kt) for all t ≥ T. (These conditions are sufficient, not necessary.)

#### a) L{t cosh at}
Use L{t f(t)} = −dF/ds with F(s) = L{cosh at} = s/(s² − a²).

```text
−d/ds [s/(s² − a²)] = −[(s² − a²) − 2s²]/(s² − a²)²
                    = (s² + a²)/(s² − a²)²        (s > |a|)
```

#### b) L{sin 2t / t}
Use L{f(t)/t} = ∫ₛ^∞ F(u) du, with F(s) = 2/(s² + 4). (sin 2t / t → 2 as t → 0.)

```text
∫ₛ^∞ 2/(u² + 4) du = [tan⁻¹(u/2)]ₛ^∞ = π/2 − tan⁻¹(s/2) = tan⁻¹(2/s)
```

**Answers:** a) (s² + a²)/(s² − a²)²; b) tan⁻¹(2/s) = cot⁻¹(s/2).

## Check yourself

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

1. Which conditions guarantee that L{f(t)} exists for s > k?
   A. f is differentiable everywhere and bounded
   B. f is periodic and continuous at t = 0
   C. f is piecewise continuous on every finite interval and of exponential order |f(t)| <= M e^(kt)
   D. f is continuous with f(0) = 0 and f(t) -> 0 as t -> infinity
   Answer: C. The answer states these two conditions: piecewise continuity on each [0, T] and exponential order. They are sufficient, and the transform then exists for s > k. (Key point: Conditions for the Laplace transform to exist)

2. What is the logical status of the existence conditions (piecewise continuous + exponential order)?
   A. Necessary and sufficient
   B. Necessary but not sufficient
   C. Neither necessary nor sufficient
   D. Sufficient but not necessary
   Answer: D. The answer notes the conditions are sufficient, not necessary: a function can have a transform without meeting them. (Key point: Sufficient, not necessary)

3. What is L{t cosh at}?
   A. (s^2 + a^2)/(s^2 - a^2)^2
   B. (s^2 - a^2)/(s^2 + a^2)^2
   C. -(s^2 + a^2)/(s^2 - a^2)^2
   D. 2as/(s^2 - a^2)^2
   Answer: A. -d/ds [s/(s^2-a^2)] = (s^2 + a^2)/(s^2 - a^2)^2 for s > |a|. (Key point: Part (a))

4. What is L{sin 2t / t}?
   A. tan^-1(s/2)
   B. tan^-1(2/s)
   C. 2/(s^2 + 4)
   D. pi/2 + tan^-1(s/2)
   Answer: B. Integral of 2/(u^2+4) from s to infinity = pi/2 - tan^-1(s/2) = tan^-1(2/s). (Key point: Part (b))

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