---
title: "Solve ODE by Laplace transform y''+4y=sint"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-16"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-16
site: BCT Engineering
---

# Solve ODE by Laplace transform y''+4y=sint

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2070 Asar, 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-16
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Using Laplace transform, solve: y'' + 4y = sin t; y(0) = 0 = y'(0).

## Short answer

Apply Laplace transform, use partial fractions to write Y = 1/3 [1/(s^2+1) - 1/(s^2+4)], then invert to get y(t) = (1/3) sin t - (1/6) sin 2t.

## Answer

y″ + 4y = sin t, y(0) = y′(0) = 0.

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

```text
s²Y + 4Y = 1/(s² + 1)
Y = 1/[(s² + 1)(s² + 4)]
```

**Step 2: Partial fractions.** Since (s² + 4) − (s² + 1) = 3:

```text
Y = (1/3) [ 1/(s² + 1) − 1/(s² + 4) ]
```

**Step 3: Invert.** L⁻¹[1/(s² + 4)] = (1/2) sin 2t.

**Answer: y(t) = (1/3) sin t − (1/6) sin 2t**

**Check:** y(0) = 0; y′(0) = 1/3 − 1/3 = 0; y″ + 4y = −(1/3) sin t + (2/3) sin 2t + (4/3) sin t − (2/3) sin 2t = sin t ✓.

## Check yourself

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

1. For y'' + 4y = sin t with zero initial values, Y =
   A. 1/(s^2+4)
   B. s/((s^2+1)(s^2+4))
   C. 1/((s^2+1)(s^2+4))
   D. 1/((s^2-1)(s^2+4))
   Answer: C. (s^2+4)Y = 1/(s^2+1). (Key point: Transform)

2. Y =
   A. (1/3)[1/(s^2+1) + 1/(s^2+4)]
   B. [1/(s^2+1) - 1/(s^2+4)]
   C. (1/3)[1/(s^2+4) - 1/(s^2+1)]
   D. (1/3)[1/(s^2+1) - 1/(s^2+4)]
   Answer: D. (s^2+4)-(s^2+1) = 3. (Key point: Partial fractions)

3. L^-1[1/(s^2+4)] =
   A. (1/2) sin 2t
   B. sin 2t
   C. 2 sin 2t
   D. (1/2) cos 2t
   Answer: A. Divide by 2. (Key point: Inverse)

4. y(t) =
   A. (1/3) sin t - (1/3) sin 2t
   B. (1/3) sin t - (1/6) sin 2t
   C. (1/3) sin t + (1/6) sin 2t
   D. (1/3) cos t - (1/6) cos 2t
   Answer: B. y'(0) = 1/3 - 1/3 = 0. (Key point: Solution)

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