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

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

Mathematics III · Chapter 3: Laplace Transform

- Asked in: 2070 Chaitra, 5 marks; 2069 Chaitra, 5 marks (asked 2 times)
- 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-seven
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3

## Question

Solve the following differential equation by using Laplace transform: y'' + y' − 2y = x, y(0) = 1, y'(0) = 0.

## Short answer

Rewrite independent variable as t. Y = (s^3+s^2+1)/[s^2(s-1)(s+2)]; partial fractions: -1/(4s) -1/(2s^2) +1/(s-1) +1/(4(s+2)). Inverse yields y(x) = e^x + e^(-2x)/4 - x/2 - 1/4.

## Answer

The independent variable is x here; write it as t for the Laplace transform. Solve y'' + y' − 2y = t with y(0) = 1, y'(0) = 0. Let Y = L{y}.

```text
(s²Y − s) + (sY − 1) − 2Y = 1/s²
(s² + s − 2) Y = s + 1 + 1/s² = (s³ + s² + 1)/s²
Y = (s³ + s² + 1)/[s² (s − 1)(s + 2)]
```

Partial fractions: Y = A/s + B/s² + C/(s − 1) + D/(s + 2).

```text
B = 1/[(−1)(2)] = −1/2                    (s = 0 in the other factors)
C = (1 + 1 + 1)/[1 · 3] = 1               (s = 1)
D = (−8 + 4 + 1)/[4 · (−3)] = 1/4         (s = −2)
As s → ∞, Y ≈ 1/s, so A + C + D = 1 → A = 1 − 1 − 1/4 = −1/4
Y = −1/(4s) − 1/(2s²) + 1/(s − 1) + 1/(4(s + 2))
```

Inverse transform:

```text
y = −1/4 − x/2 + eˣ + (1/4) e^(−2x)
```

**Answer: y(x) = eˣ + e^(−2x)/4 − x/2 − 1/4.** Check: y(0) = 1 + 1/4 − 1/4 = 1 ✓; y'(0) = 1 − 1/2 − 1/2 = 0 ✓.

## Check yourself

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

1. With y(0) = 1, y'(0) = 0, what does (s^2 + s - 2)Y equal?
   A. s + 1 + 1/s^2
   B. s - 1 + 1/s^2
   C. s + 1 + 1/s
   D. 1/s^2
   Answer: A. L{y''} = s^2Y - s and L{y'} = sY - 1, so moving the constants across gives s + 1, plus L{x} = 1/s^2. (Key point: Transforming the derivatives)

2. Y = (s^3 + s^2 + 1)/[s^2 (s-1)(s+2)]. Which form is correct?
   A. A/s^2 + C/(s-1) + D/(s+2)
   B. A/s + B/s^2 + C/(s-1) + D/(s+2)
   C. A/s + B/(s-1) + C/(s+2)
   D. (As+B)/s^2 + C/(s-1) + D/(s+2)
   Answer: B. A repeated factor s^2 needs both A/s and B/s^2, plus one term for each simple factor. (Key point: Partial fraction form)

3. Cover (s-1) and put s = 1 to find C. What is C?
   A. 1/3
   B. 3
   C. 1
   D. -1
   Answer: C. C = (1 + 1 + 1)/[1^2 * (1 + 2)] = 3/3 = 1. (Key point: Finding C)

## Close variants set in other exams

- [Solve ODE by Laplace transform y'''+2y''−y'−2y=0](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-y-15) — Mathematics III, chapter 3, asked 3×
- [Solve ODE by Laplace transform y''+3y'+2y=eᵗ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-ode-by-laplace-transform-y-three) — Mathematics III, chapter 3, asked 4×

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