---
title: "Inverse Laplace transforms (s+2)/(s²−4s+13)"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:inverse-laplace-transforms-s-2-s2"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s-2-s2
site: BCT Engineering
---

# Inverse Laplace transforms (s+2)/(s²−4s+13)

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2073 Shrawan, 2+3 marks
- Video: [Inverse laplace transform of [(s+2)/(s^(2)-4s+13)] || B.S Grewal Example:-21.17 Question number:-02](https://www.youtube.com/watch?v=UwME_FQOCAo) — GAUTAM BUDDHA
- 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-s-2-s2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of: i) (s + 2)/(s² − 4s + 13)  ii) log((s + a)/(s − a)).

## Short answer

For (s+2)/(s²-4s+13), complete square: (s-2)²+9, rewrite as (s-2)+4. Inverse: e^(2t)[cos3t + (4/3)sin3t]. For log((s+a)/(s-a)), differentiate to get 1/(s+a)-1/(s-a); then t f(t)=e^(at)-e^(-at)=2 sinh at, so f(t)=2 sinh(at)/t.

## Answer

#### i) L⁻¹[(s + 2)/(s² − 4s + 13)]
Complete the square: s² − 4s + 13 = (s − 2)² + 9. Write the numerator as (s − 2) + 4.

```text
(s + 2)/((s−2)² + 9) = (s − 2)/((s−2)² + 9) + (4/3)·3/((s−2)² + 9)
```

By the first shifting theorem:

**Answer: e^(2t) [cos 3t + (4/3) sin 3t]**

#### ii) L⁻¹[log((s + a)/(s − a))]
Let F(s) = ln(s + a) − ln(s − a). Then F′(s) = 1/(s + a) − 1/(s − a).
Using L[t f(t)] = −F′(s):

```text
L[t f(t)] = 1/(s − a) − 1/(s + a)
t f(t)    = e^(at) − e^(−at) = 2 sinh at
```

**Answer: f(t) = 2 sinh(at) / t**

## Check yourself

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

1. s + 2 in terms of (s-2)?
   A. (s-2) + 2
   B. (s-2) - 4
   C. (s-2) + 4
   D. (s+2) + 4
   Answer: C. s + 2 = s - 2 + 4. (Key point: Numerator)

2. What is the inverse?
   A. e^(2t)[cos 3t + 4 sin 3t]
   B. e^(-2t)[cos 3t + (4/3) sin 3t]
   C. e^(2t)[cos 3t + (4/9) sin 3t]
   D. e^(2t)[cos 3t + (4/3) sin 3t]
   Answer: D. 4/((s-2)^2+9) = (4/3)*3/(...). (Key point: Result for (i))

3. What is the inverse of log((s+a)/(s-a))?
   A. 2 sinh(at)/t
   B. 2 sinh(at)
   C. 2 cosh(at)/t
   D. (e^(at)+e^(-at))/t
   Answer: A. t f = e^(at) - e^(-at). (Key point: Result for (ii))

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