---
title: "Initial value problem y''+2y'+5y=0"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:initial-value-problem-y-2y"
source: https://www.bctengineer.com/q/semester_2/mathematics/initial-value-problem-y-2y
site: BCT Engineering
---

# Initial value problem y''+2y'+5y=0

Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions

- Asked in: 2069 Bhadra, 5 marks
- 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_2/mathematics/initial-value-problem-y-2y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Solve the initial value problem y'' + 2y' + 5y = 0, given y(0) = 1, y'(0) = 5.

## Short answer

Roots of m^2+2m+5=0 are -1±2i, so y = e^{-x}(A cos2x + B sin2x). y(0)=1 gives A=1; y'(0)=5 gives -1+2B=5, B=3. Answer: y = e^{-x}(cos2x + 3 sin2x).

## Answer

Equation: y″ + 2y′ + 5y = 0, y(0) = 1, y′(0) = 5.

#### General solution
Auxiliary equation: m² + 2m + 5 = 0, so m = (−2 ± √(4 − 20))/2 = −1 ± 2i.

The roots are complex (α ± iβ with α = −1, β = 2), so

y = e⁻ˣ (A cos 2x + B sin 2x).

#### Apply the conditions
y(0) = 1: A = 1.

y′ = −e⁻ˣ(A cos 2x + B sin 2x) + e⁻ˣ(−2A sin 2x + 2B cos 2x).

y′(0) = −A + 2B = 5, so −1 + 2B = 5 and B = 3.

**Answer: y = e⁻ˣ (cos 2x + 3 sin 2x).**

Check: y(0) = 1 ✓; y′(0) = −1 + 6 = 5 ✓.

## Check yourself

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

1. What are the roots of m^2 + 2m + 5 = 0?
   A. 1 ± 2i
   B. −1 ± 2i
   C. −1 ± 4i
   D. −2 ± 5i
   Answer: B. m = (−2 ± sqrt(4 − 20))/2 = (−2 ± 4i)/2 = −1 ± 2i. (Key point: Auxiliary equation roots)

2. For complex roots α ± iβ, the general solution is:
   A. e^(αx)(A cos βx + B sin βx)
   B. e^(βx)(A cos αx + B sin αx)
   C. A e^(αx) + B e^(βx)
   D. (A + Bx) e^(αx)
   Answer: A. Complex roots give an exponential e^(αx) times cos and sin of βx; here α = −1, β = 2. (Key point: Form for complex roots)

3. With y = e^(−x)(A cos 2x + B sin 2x) and y(0) = 1, y'(0) = 5, what is B?
   A. 2
   B. 5
   C. 1
   D. 3
   Answer: D. A = 1; y'(0) = −A + 2B = 5 gives −1 + 2B = 5, so B = 3. (Key point: Using y'(0))

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