---
title: "Solve ODE by Laplace transform y''+2y=r(t)"
subject: "Mathematics III"
chapter: 3
key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-13"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-13
site: BCT Engineering
---

# Solve ODE by Laplace transform y''+2y=r(t)

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2074 Chaitra, 5 marks
- 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-13
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

By using Laplace transform, solve the initial value problem y'' + 2y = r(t), y(0) = y'(0) = 0, where r(t) = 1 for 0 < t < 1 and r(t) = 0 otherwise.

## Short answer

Write r(t) = 1 - u(t-1). Transform gives Y = (1 - e^(-s)) / (s(s²+2)). Partial fractions yield g(t) = ½(1 - cos√2 t). By second shift, y(t) = g(t) - g(t-1)u(t-1). For 0<t<1: y = ½(1 - cos√2 t); for t>1: y = ½[cos√2(t-1) - cos√2 t].

## Answer

The right side is a square pulse: r(t) = 1 − u(t − 1), where u is the unit step function.

**Step 1: Transform both sides** (y(0) = y′(0) = 0).

```text
s²Y + 2Y = 1/s − e^(−s)/s = (1 − e^(−s))/s
Y = (1 − e^(−s)) / [ s(s² + 2) ]
```

**Step 2: Partial fractions.**

```text
1/[s(s² + 2)] = (1/2) [ 1/s − s/(s² + 2) ]
L⁻¹ = g(t) = (1/2)(1 − cos √2 t)
```

**Step 3: Second shifting theorem.** L⁻¹[e^(−s) G(s)] = g(t − 1) u(t − 1). So

```text
y(t) = g(t) − g(t − 1) u(t − 1)
```

**Answer:**

```text
y(t) = ½ (1 − cos √2 t)                     for 0 < t < 1
y(t) = ½ [cos √2 (t − 1) − cos √2 t]        for t > 1
```

The solution and its first derivative are continuous at t = 1 (since y″ has only a jump there).

## Check yourself

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

1. How is the pulse written?
   A. r(t) = 1 - u(t-1)
   B. r(t) = 1 + u(t-1)
   C. r(t) = u(t-1)
   D. r(t) = u(t) + 1
   Answer: A. 1 for 0<t<1, 0 afterwards. (Key point: Writing r(t))

2. L{r} equals what?
   A. (1 + e^(-s))/s
   B. (1 - e^(-s))/s
   C. e^(-s)/s
   D. 1/s
   Answer: B. L{1} = 1/s, L{u(t-1)} = e^(-s)/s. (Key point: Transform of r)

3. 1/[s(s^2+2)] inverts to?
   A. (1/2)(1 - cos 2t)
   B. 1 - cos sqrt2 t
   C. (1/2)(1 - cos sqrt2 t)
   D. (1/2)(1 - sin sqrt2 t)
   Answer: C. 1/[s(s^2+2)] = (1/2)[1/s - s/(s^2+2)]. (Key point: Inverse g(t))

4. What is y for t>1?
   A. (1/2)(1 - cos sqrt2 t)
   B. (1/2)[cos sqrt2 t - cos sqrt2(t-1)]
   C. (1/2)[1 - cos sqrt2(t-1)]
   D. (1/2)[cos sqrt2(t-1) - cos sqrt2 t]
   Answer: D. y = g(t) - g(t-1). (Key point: Solution for t>1)

## Close variants set in other exams

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