---
title: "Solve ODE by Laplace transform y''+y=sin3t"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-nine"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-nine
site: BCT Engineering
---

# Solve ODE by Laplace transform y''+y=sin3t

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2080 Bhadra, 5 marks
- Video: [Application of Laplace Transformation in Differential equations](https://www.youtube.com/watch?v=THqzPHS8cDE) — Anshuman Purohit
- 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-nine
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Solve the initial value problem by the Laplace transform method: y'' + y = sin 3t; y(0) = 0, y'(0) = 0.

## Short answer

y(t) = (3/8) sin t - (1/8) sin 3t.

## Answer

Transform y'' + y = sin 3t with y(0) = y'(0) = 0. Let Y = L{y}.

```text
s²Y + Y = 3/(s² + 9)
Y = 3/[(s² + 1)(s² + 9)]
```

Since 1/[(s² + 1)(s² + 9)] = (1/8)[1/(s² + 1) − 1/(s² + 9)]:

```text
Y = (3/8) · 1/(s² + 1) − (3/8) · 1/(s² + 9)
y(t) = (3/8) sin t − (3/8)(sin 3t)/3
```

**Answer: y(t) = (3/8) sin t − (1/8) sin 3t.** Check: y(0) = 0; y'(0) = 3/8 − 3/8 = 0 ✓.

## Check yourself

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

1. y'' + y = sin 3t with zero initial conditions. What is Y?
   A. 3/(s^2+9)
   B. 3/[(s^2+1)(s^2+9)]
   C. 1/[(s^2+1)(s^2+9)]
   D. 3s/[(s^2+1)(s^2+9)]
   Answer: B. (s^2+1)Y = L{sin 3t} = 3/(s^2+9). (Key point: Y(s))

2. 1/[(s^2+1)(s^2+9)] equals what?
   A. (1/9)[1/(s^2+1) - 1/(s^2+9)]
   B. (1/8)[1/(s^2+1) + 1/(s^2+9)]
   C. (1/8)[1/(s^2+1) - 1/(s^2+9)]
   D. (1/10)[1/(s^2+1) - 1/(s^2+9)]
   Answer: C. The difference of the two fractions is 8/[(s^2+1)(s^2+9)]. (Key point: Partial fractions)

3. In y(t), what is the coefficient of sin 3t?
   A. -3/8
   B. -1/24
   C. 1/8
   D. -1/8
   Answer: D. -(3/8)/(s^2+9) inverts to -(3/8)(sin 3t)/3 = -(1/8) sin 3t. (Key point: Coefficient of sin 3t)

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