---
title: "Solve ODE by Laplace transform (D²+4D+3)x=e^(−t)"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-d2-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-d2-two
site: BCT Engineering
---

# Solve ODE by Laplace transform (D²+4D+3)x=e^(−t)

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2068 Baishakh, 5 marks
- Video: [33. Application of Laplace Transform | Complete Concept and Problem#1 | Most Important Problem](https://www.youtube.com/watch?v=ZXU6rKagW1M) — MKS TUTORIALS by Manoj Sir
- 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-d2-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Solve using Laplace transform (D² + 4D + 3)x = e^(−t), where x(0) = x'(0) = 1.

## Short answer

Apply Laplace transform, partial fractions, invert: x(t) = (7/4)e^(-t) + (t/2)e^(-t) - (3/4)e^(-3t).

## Answer

(D² + 4D + 3)x = e^(−t), x(0) = x′(0) = 1.

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

```text
(s²X − s − 1) + 4(sX − 1) + 3X = 1/(s + 1)
(s² + 4s + 3) X = 1/(s + 1) + s + 5
X = 1/[(s+1)²(s+3)] + (s + 5)/[(s+1)(s+3)]
```

**Step 2: Partial fractions.**

```text
1/[(s+1)²(s+3)] = −1/4·1/(s+1) + 1/2·1/(s+1)² + 1/4·1/(s+3)
(s+5)/[(s+1)(s+3)] = 2/(s+1) − 1/(s+3)
```

Adding:

```text
X = (7/4)/(s+1) + (1/2)/(s+1)² − (3/4)/(s+3)
```

**Step 3: Invert.**

**Answer: x(t) = (7/4) e^(−t) + (t/2) e^(−t) − (3/4) e^(−3t)**

**Check:** x(0) = 7/4 − 3/4 = 1 ✓; x′(0) = −7/4 + 1/2 + 9/4 = 1 ✓.

## Check yourself

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

1. With x(0) = x'(0) = 1, L[x''] equals
   A. s^2 X - s - 1
   B. s^2 X - 1
   C. s^2 X + s + 1
   D. s^2 X - s
   Answer: A. L[x''] = s^2 X - s x(0) - x'(0). (Key point: Transform with initial values)

2. After moving initial-condition terms across, (s^2+4s+3)X equals
   A. 1/(s+1) + s + 1
   B. 1/(s+1) + s + 5
   C. 1/(s+1) - s - 5
   D. 1/(s+1)
   Answer: B. s + 1 from x'' and 4 from 4x' give s + 5. (Key point: Right-hand side)

3. Why does the answer contain t e^(-t)?
   A. Because of the factor s + 3
   B. Because x(0) is not zero
   C. The term 1/(s+1)^2 has a repeated pole at s = -1
   D. Because the right side is exponential
   Answer: C. L^-1[1/(s+1)^2] = t e^(-t). (Key point: Repeated pole)

4. What is x(t)?
   A. (7/4)e^(-t) - (t/2)e^(-t) - (3/4)e^(-3t)
   B. (3/4)e^(-t) + (t/2)e^(-t) - (7/4)e^(-3t)
   C. (7/4)e^(-t) + (t/2)e^(-t) + (3/4)e^(-3t)
   D. (7/4)e^(-t) + (t/2)e^(-t) - (3/4)e^(-3t)
   Answer: D. Check: x(0) = 7/4 - 3/4 = 1. (Key point: Final solution)

## Related in Mathematics III

- [Solve ODE by Laplace transform y''+3y'+2y=e^(−t)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-y-two) — Mathematics III, chapter 3, asked 18×
- [Solve ODE by Laplace transform y''+4y'+3y=0](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-y-four) — Mathematics III, chapter 3, asked 18×
- [Solve ODE by Laplace transform x''+2x'+5x=e^(−t)sint](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-x) — Mathematics III, chapter 3, asked 17×
- [Solve ODE by Laplace transform y''+2y'+2y=5sinx](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-y-17) — Mathematics III, chapter 3, asked 17×
- [Solve ODE by Laplace transform y''+4y'+3y=eᵗ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-y-five) — Mathematics III, chapter 3, asked 16×
- [Solve ODE by Laplace transform x''+4x'+4x=6e^(−t)](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-x-two) — Mathematics III, chapter 3, asked 16×