---
title: "Inverse Laplace transforms 1/((s−2)(s+4))"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-1-s-2-three"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s-2-three
site: BCT Engineering
---

# Inverse Laplace transforms 1/((s−2)(s+4))

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2068 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-1-s-2-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform: (a) 1/((s − 2)(s + 4))  (b) log((s² + a²)/s²).

## Short answer

Partial fractions give L^(-1)[1/((s-2)(s+4)] = (e^(2t)-e^(-4t))/6. Using L[t f(t)] = -F'(s) yields L^(-1)[log((s^2+a^2)/s^2)] = 2(1-cos at)/t.

## Answer

#### (a) L⁻¹[1/((s − 2)(s + 4))]
```text
1/((s−2)(s+4)) = A/(s−2) + B/(s+4)
A = 1/(2 + 4) = 1/6,   B = 1/(−4 − 2) = −1/6
```

**Answer: (e^(2t) − e^(−4t))/6**

#### (b) L⁻¹[log((s² + a²)/s²)]
Let F(s) = ln(s² + a²) − 2 ln s. Then F′(s) = 2s/(s² + a²) − 2/s.
Using L[t f(t)] = −F′(s):

```text
L[t f(t)] = 2/s − 2s/(s² + a²)
t f(t)    = 2 − 2 cos at
```

**Answer: f(t) = 2(1 − cos at)/t**

## Check yourself

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

1. For 1/((s-2)(s+4)) = A/(s-2) + B/(s+4), what is A?
   A. 1/6
   B. 1/2
   C. -1/6
   D. 1/4
   Answer: A. Cover (s-2) and put s = 2: A = 1/(2+4) = 1/6. (Key point: Partial fraction constant A)

2. What is L^-1[1/((s-2)(s+4))]?
   A. (e^(2t) + e^(-4t))/6
   B. (e^(2t) - e^(-4t))/6
   C. (e^(-2t) - e^(4t))/6
   D. e^(2t) - e^(-4t)
   Answer: B. A = 1/6 and B = -1/6 give (1/6)(e^(2t) - e^(-4t)). (Key point: Result (a))

3. For F(s) = log((s^2+a^2)/s^2), which property is used to find f(t)?
   A. L[f(t)/t] = integral of F
   B. L[f'(t)] = sF - f(0)
   C. L[t f(t)] = -F'(s)
   D. L[e^(at) f(t)] = F(s - a)
   Answer: C. Differentiating the log turns it into a rational function whose inverse is t f(t). (Key point: Log transform trick)

4. What is L^-1[log((s^2+a^2)/s^2)]?
   A. 2(1 + cos at)/t
   B. 2(cos at - 1)
   C. (1 - cos at)/(2t)
   D. 2(1 - cos at)/t
   Answer: D. -F'(s) = 2/s - 2s/(s^2+a^2), so t f(t) = 2 - 2 cos at. (Key point: Final answer (b))

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