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

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

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2068 Baishakh, 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-four
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Find the inverse Laplace transform of the following functions: a) 1/((s − 2)(s + 2)²)  b) 1/(s²(s + 2)).

## Short answer

Partial fractions give L^(-1)[1/((s-2)(s+2)^2] = (1/16)e^(2t) - (1/16)e^(-2t) - (t/4)e^(-2t) and L^(-1)[1/(s^2(s+2))] = t/2 - 1/4 + (1/4)e^(-2t).

## Answer

#### a) L⁻¹[1/((s − 2)(s + 2)²)]
```text
1/((s−2)(s+2)²) = A/(s−2) + B/(s+2) + C/(s+2)²
A = 1/(4)² = 1/16
C = 1/(−2 − 2) = −1/4
A + B = 0 → B = −1/16
```

Using L⁻¹[1/(s+2)²] = t e^(−2t):

**Answer: (1/16) e^(2t) − (1/16) e^(−2t) − (t/4) e^(−2t)**

#### b) L⁻¹[1/(s²(s + 2))]
```text
1/(s²(s+2)) = A/s + B/s² + C/(s+2)
B = 1/2,   C = 1/4,   A = −C = −1/4
```

**Answer: t/2 − 1/4 + (1/4) e^(−2t)**

## Check yourself

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

1. Which partial-fraction form is correct for 1/((s-2)(s+2)^2)?
   A. A/(s-2) + B/(s+2) + C/(s+2)^2
   B. A/(s-2) + B/(s+2)^2
   C. A/(s-2) + (Bs+C)/(s+2)^2
   D. A/(s-2) + B/(s+2)
   Answer: A. A repeated linear factor needs a term for each power. (Key point: Form for repeated factor)

2. What is A in 1/((s-2)(s+2)^2)?
   A. 1/4
   B. 1/16
   C. -1/16
   D. 1/8
   Answer: B. Put s = 2: 1/(4)^2 = 1/16. (Key point: Constant A)

3. L^-1[1/(s+2)^2] equals
   A. e^(-2t)
   B. t e^(2t)
   C. t e^(-2t)
   D. t^2 e^(-2t)
   Answer: C. By the first shift, 1/s^2 -> t becomes t e^(-2t). (Key point: Inverse of 1/(s+2)^2)

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

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