---
title: "Solve ODE by Laplace transform y''+2y'+5y=e^(−t)sint"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-six"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-six
site: BCT Engineering
---

# Solve ODE by Laplace transform y''+2y'+5y=e^(−t)sint

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2072 Chaitra, 5 marks; 2071 Chaitra, 5 marks (asked 2 times)
- Video: [Solve initial value problem y"+2y'+2y=0, y(0)=1, y'(0) =-3 using Laplace Transform Engineering math](https://www.youtube.com/watch?v=l6uJ1JoteRY) — Swati Theng Mathematics
- 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-six
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Apply Laplace transform to solve the differential equation y'' + 2y' + 5y = e^(−t) sin t, x(0) = 0, x'(0) = 1.

## Short answer

Set p = s+1; the transformed equation yields Y = 1/(p^2+4) + 1/[(p^2+1)(p^2+4)]. Simplifying and applying the first shift theorem gives y(t) = (1/3)e^(-t)(sin t + sin 2t).

## Answer

Transform y'' + 2y' + 5y = e^(−t) sin t with y(0) = 0, y'(0) = 1. (The question writes the unknown as both y and x; it is the same function of t, so y is used here.) Let Y = L{y}.

```text
(s²Y − 1) + 2sY + 5Y = L{e^(−t) sin t} = 1/((s + 1)² + 1)
(s² + 2s + 5) Y = 1 + 1/((s + 1)² + 1)
```

Put p = s + 1, so s² + 2s + 5 = p² + 4:

```text
Y = 1/(p² + 4) + 1/[(p² + 1)(p² + 4)]
  = 1/(p² + 4) + (1/3)[1/(p² + 1) − 1/(p² + 4)]
  = (1/3) · 1/(p² + 1) + (2/3) · 1/(p² + 4)
```

Inverse transform (first shifting theorem, s → p = s + 1 gives factor e^(−t)):

```text
y(t) = e^(−t) [ (1/3) sin t + (2/3)(sin 2t)/2 ]
```

**Answer: y(t) = (1/3) e^(−t) (sin t + sin 2t).** Check: y(0) = 0; y'(0) = (1/3)(1 + 2) = 1 ✓.

## Check yourself

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

1. With y(0) = 0 and y'(0) = 1, what is L{y''}?
   A. s^2 Y - 1
   B. s^2 Y - s
   C. s^2 Y - s - 1
   D. s^2 Y + 1
   Answer: A. L{y''} = s^2 Y - s y(0) - y'(0) = s^2 Y - 0 - 1. (Key point: Transform of y'')

2. What is L{e^(-t) sin t}?
   A. 1/(s^2 + 1)
   B. 1/((s+1)^2 + 1)
   C. 1/((s-1)^2 + 1)
   D. (s+1)/((s+1)^2 + 1)
   Answer: B. First shifting replaces s by s+1 in L{sin t} = 1/(s^2+1). (Key point: Transform of the forcing term)

3. How can s^2 + 2s + 5 be written to use shifting?
   A. (s+1)^2 + 5
   B. (s-1)^2 + 4
   C. (s+1)^2 + 4
   D. (s+1)^2 - 4
   Answer: C. s^2 + 2s + 5 = (s+1)^2 + 4, so with p = s+1 it becomes p^2 + 4. (Key point: Completing the square)

4. The term (2/3) / (p^2+4) with p = s+1 inverts to what?
   A. (2/3) e^(-t) sin 2t
   B. (2/3) e^(-t) cos 2t
   C. (1/3) e^(t) sin 2t
   D. (1/3) e^(-t) sin 2t
   Answer: D. L^-1{1/(p^2+4)} = (sin 2t)/2, so the factor 2/3 becomes 1/3, and s -> s+1 gives the factor e^(-t). (Key point: Inverting the 1/(p^2+4) part)

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