---
title: "Solve ODE by Laplace transform y'''+2y''−y'−2y=0"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-15"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-15
site: BCT Engineering
---

# Solve ODE by Laplace transform y'''+2y''−y'−2y=0

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2067 Asar
- Video: [18MAT31: Solution of LDE by Laplace transform y'''+2y''-y'-2y=0](https://www.youtube.com/watch?v=XOxVr2T8YE4) — Dr. Kiran Potadar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-15
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Solve the following differential equation using the Laplace transform: y''' + 2y'' − y' − 2y = 0, where y(0) = y'(0) = 0 and y''(0) = 6.

## Short answer

Take Laplace transform, factor denominator as (s+2)(s−1)(s+1), use partial fractions to get Y(s)=1/(s−1) −3/(s+1) +2/(s+2). Inverse yields y(t) = eᵗ − 3e^(−t) + 2e^(−2t).

## Answer

y‴ + 2y″ − y′ − 2y = 0, y(0) = y′(0) = 0, y″(0) = 6.

**Step 1: Transform.** With Y = L[y]:

```text
L[y‴] = s³Y − s²y(0) − s y′(0) − y″(0) = s³Y − 6
L[y″] = s²Y,   L[y′] = sY
(s³Y − 6) + 2s²Y − sY − 2Y = 0
(s³ + 2s² − s − 2) Y = 6
```

**Step 2: Factorise.** s³ + 2s² − s − 2 = s²(s + 2) − (s + 2) = (s + 2)(s − 1)(s + 1).

```text
Y = 6 / [(s − 1)(s + 1)(s + 2)] = A/(s−1) + B/(s+1) + C/(s+2)
A = 6/(2·3) = 1
B = 6/((−2)(1)) = −3
C = 6/((−3)(−1)) = 2
```

**Step 3: Invert.**

**Answer: y(t) = eᵗ − 3e^(−t) + 2e^(−2t)**

**Check:** y(0) = 1 − 3 + 2 = 0 ✓; y′(0) = 1 + 3 − 4 = 0 ✓; y″(0) = 1 − 3 + 8 = 6 ✓.

## Check yourself

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

1. With y(0) = y'(0) = 0 and y''(0) = 6, L[y'''] =
   A. s^3 Y - 6
   B. s^3 Y + 6
   C. s^3 Y
   D. s^3 Y - 6s
   Answer: A. L[y'''] = s^3 Y - s^2 y(0) - s y'(0) - y''(0). (Key point: Transform of y''')

2. s^3 + 2s^2 - s - 2 factors as
   A. (s-2)(s-1)(s+1)
   B. (s+2)(s-1)(s+1)
   C. (s+2)(s+1)^2
   D. (s-2)(s+1)(s+2)
   Answer: B. Group: s^2(s+2) - (s+2) = (s+2)(s^2-1). (Key point: Factorisation)

3. In Y = 6/[(s-1)(s+1)(s+2)], the coefficient of 1/(s-1) is
   A. 3
   B. -3
   C. 1
   D. 2
   Answer: C. A = 6/(2*3) = 1. (Key point: Coefficient A)

4. What is y(t)?
   A. e^t + 3e^(-t) + 2e^(-2t)
   B. e^t - 3e^(-t) - 2e^(-2t)
   C. 2e^t - 3e^(-t) + e^(-2t)
   D. e^t - 3e^(-t) + 2e^(-2t)
   Answer: D. Check y''(0) = 1 - 3 + 8 = 6. (Key point: Solution)

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