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

# Solve ODE by Laplace transform y''+2y'−3y=sint

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2079 Bhadra, 5 marks; 2076 Chaitra, 5 marks; 2074 Asoj, 5 marks (asked 3 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Solve the following initial value problem by using the Laplace transform: y'' + 2y' − 3y = sin t, y(0) = y'(0) = 0.

## Short answer

Taking Laplace transforms gives Y(s) = 1/[(s−1)(s+3)(s²+1)]. Partial fractions and inverse transformation yield y(t) = e^t/8 − e^{−3t}/40 − (cos t)/10 − (sin t)/5.

## Answer

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

```text
L{y''} = s²Y − s y(0) − y'(0) = s²Y
L{y'}  = sY − y(0) = sY
L{sin t} = 1/(s² + 1)

(s² + 2s − 3) Y = 1/(s² + 1)
Y = 1/[(s − 1)(s + 3)(s² + 1)]
```

Partial fractions: Y = A/(s − 1) + B/(s + 3) + (Cs + D)/(s² + 1).

```text
A = 1/[(1 + 3)(1 + 1)] = 1/8           (put s = 1)
B = 1/[(−3 − 1)(9 + 1)] = −1/40        (put s = −3)
Coefficient of s³: A + B + C = 0  → C = −1/10
Put s = 0: −1/3 = −A + B/3 + D        → D = −1/5
Y = 1/(8(s − 1)) − 1/(40(s + 3)) − (s + 2)/(10(s² + 1))
```

Inverse transform:

```text
y(t) = (1/8) eᵗ − (1/40) e^(−3t) − (1/10) cos t − (1/5) sin t
```

**Answer: y(t) = eᵗ/8 − e^(−3t)/40 − (cos t)/10 − (sin t)/5.** Check: y(0) = 1/8 − 1/40 − 1/10 = 0 ✓.

## Check yourself

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

1. L{y''}=
   A. sY
   B. s^2 Y - 1
   C. s^2 Y
   D. Y/s^2
   Answer: C. Zero initial conditions. (Key point: Transform)

2. Y(s)=
   A. 1/[(s+1)(s-3)(s^2+1)]
   B. 1/[(s-1)(s+3)]
   C. s/[(s-1)(s+3)(s^2+1)]
   D. 1/[(s-1)(s+3)(s^2+1)]
   Answer: D. s^2+2s-3 factors. (Key point: Y)

3. A=
   A. 1/8
   B. 1/4
   C. 1/40
   D. -1/8
   Answer: A. Residue at s=1. (Key point: A)

4. Solution:
   A. e^t/8 + e^(-3t)/40
   B. e^t/8 - e^(-3t)/40 - cos t/10 - sin t/5
   C. e^t/8 - sin t
   D. e^t - e^(-3t)
   Answer: B. Inverse transform. (Key point: y(t))

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