---
title: "Solve ODE by Laplace transform"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform
site: BCT Engineering
---

# Solve ODE by Laplace transform

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2067 Magh
- Video: [Using Laplace Transforms to solve Differential Equations ***full example***](https://www.youtube.com/watch?v=fuxFrpaMLtw) — Dr. Trefor Bazett
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Use the Laplace transform to solve d²y/dt² + 4 dy/dt + 3y = e^(−t), y(0) = y'(0) = 1.

## Short answer

Transform, solve for Y(s), partial fractions yield Y(s) = (7/4)/(s+1) + (1/2)/(s+1)² − (3/4)/(s+3). Inverse gives y(t) = (7/4 + t/2)e^(−t) − (3/4)e^(−3t).

## Answer

d²y/dt² + 4 dy/dt + 3y = e^(−t), y(0) = y′(0) = 1.

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

```text
(s²Y − s·1 − 1) + 4(sY − 1) + 3Y = 1/(s + 1)
(s² + 4s + 3) Y = 1/(s + 1) + s + 5
Y = 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)/(s+1) + (1/2)/(s+1)² + (1/4)/(s+3)
(s+5)/[(s+1)(s+3)] = 2/(s+1) − 1/(s+3)
Y = (7/4)/(s+1) + (1/2)/(s+1)² − (3/4)/(s+3)
```

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

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

**Check:** y(0) = 1 ✓ and y′(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. For y(0) = y'(0) = 1, the transformed equation gives (s^2+4s+3)Y =
   A. 1/(s+1) + s + 5
   B. 1/(s+1) + s + 1
   C. 1/(s+1) - s - 5
   D. 1/(s+1)
   Answer: A. Initial-condition terms add s + 1 + 4. (Key point: Transform)

2. s^2 + 4s + 3 factors as
   A. (s-1)(s-3)
   B. (s+1)(s+3)
   C. (s+1)(s-3)
   D. (s+2)^2
   Answer: B. Roots -1 and -3. (Key point: Factor)

3. The 1/(s+1)^2 term inverts to
   A. e^(-t)
   B. t e^(t)
   C. t e^(-t)
   D. t^2 e^(-t)
   Answer: C. First shifting of 1/s^2. (Key point: Repeated pole)

4. y(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 y(0) = 1. (Key point: Solution)

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