---
title: "Solve ODE by Laplace transform y''+3y'+2y=e^(−t)"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-two
site: BCT Engineering
---

# Solve ODE by Laplace transform y''+3y'+2y=e^(−t)

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2083 Baishakh, 5 marks; 2076 Asoj, 5 marks (asked 2 times)
- Video: [Solving a Second Order IVP By Laplace Transform | y^''-3y^'+2y=e^(-4t),       y(0)=1    y^' (0)=5](https://www.youtube.com/watch?v=y9WPraSR9WI) — math experts
- 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-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Using Laplace transform, solve the initial value problem y'' + 3y' + 2y = e^(−t), y(0) = 0, y'(0) = 0.

## Short answer

Take Laplace transform, solve Y = 1/[(s+1)^2(s+2)], use partial fractions and invert. The solution is y(t) = (t-1)e^(-t) + e^(-2t).

## Answer

Take the Laplace transform of y'' + 3y' + 2y = e^(−t), with y(0) = y'(0) = 0 and Y = L{y}.

```text
s²Y + 3sY + 2Y = 1/(s + 1)
(s + 1)(s + 2) Y = 1/(s + 1)
Y = 1/[(s + 1)²(s + 2)]
```

Partial fractions: Y = A/(s + 1) + B/(s + 1)² + C/(s + 2).

```text
B = 1/(s + 2) at s = −1  = 1
C = 1/(s + 1)² at s = −2 = 1
Compare s² terms: A + C = 0 → A = −1
Y = −1/(s + 1) + 1/(s + 1)² + 1/(s + 2)
```

Inverse transform (L⁻¹{1/(s + a)²} = t e^(−at)):

```text
y(t) = −e^(−t) + t e^(−t) + e^(−2t)
```

**Answer: y(t) = (t − 1) e^(−t) + e^(−2t).** Check: y(0) = −1 + 1 = 0; y'(0) = 1 + 1 − 2 = 0 ✓.

## Check yourself

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

1. Y=
   A. 1/[(s+1)(s+2)]
   B. 1/[(s-1)(s+2)]
   C. 1/[(s+1)^2(s+2)]
   D. s/[(s+1)(s+2)]
   Answer: C. Factor and divide. (Key point: Y)

2. Partial fractions needed:
   A. A/(s+1)+B/(s+2)
   B. A/(s+1)^2 only
   C. Cs/(s+1)^2
   D. A/(s+1)+B/(s+1)^2+C/(s+2)
   Answer: D. Repeated root. (Key point: Repeated)

3. L^-1{1/(s+1)^2}=
   A. t e^(-t)
   B. e^(-t)
   C. t^2 e^(-t)
   D. t
   Answer: A. Shift. (Key point: Inverse)

4. Solution:
   A. (t+1)e^(-t)+e^(-2t)
   B. (t-1)e^(-t) + e^(-2t)
   C. e^(-t)-e^(-2t)
   D. t e^(-t)
   Answer: B. A=-1,B=1,C=1. (Key point: y)

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