---
title: "Solve ODE by Laplace transform x''+4x'+4x=6e^(−t)"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-x-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-x-two
site: BCT Engineering
---

# Solve ODE by Laplace transform x''+4x'+4x=6e^(−t)

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2073 Shrawan, 5 marks
- Video: [Linear differential equation using Laplace transform example(PART-2) by easy maths easy tricks](https://www.youtube.com/watch?v=MslkxPCVygc) — EASY MATHS EASY TRICKS
- 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-x-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Solve the following initial value problem using Laplace transform: x'' + 4x' + 4x = 6e^(−t), x(0) = −2, x'(0) = −8.

## Short answer

Transform the ODE using initial conditions: (s+2)² X = 6/(s+1) - 2s - 16. Partial fractions give X = 6/(s+1) - 8/(s+2) - 18/(s+2)². Inverse Laplace yields x(t) = 6e^(-t) - 8e^(-2t) - 18 t e^(-2t).

## Answer

x″ + 4x′ + 4x = 6e^(−t), x(0) = −2, x′(0) = −8.

**Step 1: Transform.** With X = L[x]:

```text
L[x″] = s²X − s x(0) − x′(0) = s²X + 2s + 8
L[x′] = sX − x(0)            = sX + 2
(s²X + 2s + 8) + 4(sX + 2) + 4X = 6/(s + 1)
(s + 2)² X = 6/(s + 1) − 2s − 16
```

**Step 2: Solve for X and split.**

```text
X = 6/((s+1)(s+2)²) − (2s + 16)/(s + 2)²
```

```text
6/((s+1)(s+2)²) = 6/(s+1) − 6/(s+2) − 6/(s+2)²
(2s+16)/(s+2)²  = 2/(s+2) + 12/(s+2)²
X = 6/(s+1) − 8/(s+2) − 18/(s+2)²
```

**Step 3: Invert.** L⁻¹[1/(s+2)²] = t e^(−2t).

**Answer: x(t) = 6e^(−t) − 8e^(−2t) − 18t e^(−2t)**

**Check:** x(0) = 6 − 8 = −2 ✓; x′(0) = −6 + 16 − 18 = −8 ✓.

## Check yourself

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

1. x(0) = -2, x'(0) = -8. L{x''} = ?
   A. s^2X - 2s - 8
   B. s^2X + 2s + 8
   C. s^2X + 2s - 8
   D. s^2X - 2s + 8
   Answer: B. s^2X - s(-2) - (-8). (Key point: L[x''])

2. What is the coefficient of X?
   A. (s-2)^2
   B. s^2+4
   C. (s+2)^2
   D. (s+4)^2
   Answer: C. s^2 + 4s + 4. (Key point: Left coefficient)

3. 6/((s+1)(s+2)^2) takes which form?
   A. A/(s+1) + B/(s+2)^2
   B. A/(s+1) + B/(s+2)
   C. (As+B)/(s+2)^2
   D. A/(s+1) + B/(s+2) + C/(s+2)^2
   Answer: D. Repeated factor. (Key point: Partial fraction form)

4. What is x(t)?
   A. 6e^(-t) - 8e^(-2t) - 18t e^(-2t)
   B. 6e^(-t) - 8e^(-2t) + 18t e^(-2t)
   C. 6e^(-t) + 8e^(-2t) - 18t e^(-2t)
   D. 6e^(t) - 8e^(2t) - 18t e^(2t)
   Answer: A. Using L^-1{1/(s+2)^2} = t e^(-2t). (Key point: Final answer)

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