---
title: "Power series solution of y''+xy'+y=0"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:power-series-solution-y-xy"
source: https://www.bctengineer.com/q/semester_2/mathematics/power-series-solution-y-xy
site: BCT Engineering
---

# Power series solution of y''+xy'+y=0

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

- Asked in: 2074 Bhadra, 5 marks; 2070 Magh, 5 marks; 2069 Poush, 5 marks (asked 3 times)
- Video: [Power Series Method | Numerical | Series Solution of Differential Equation | Maths](https://www.youtube.com/watch?v=p8hvZGFTq7I) — Gautam Varde
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/power-series-solution-y-xy
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Solve by power series method the differential equation y'' + xy' + y = 0.

## Short answer

Assume y = Σ a_n x^n. Recurrence a_{n+2} = -a_n/(n+2) yields even terms a_{2m}=(-1)^m a_0/(2^m m!), forming a_0 e^(-x²/2), and odd terms a_{2m+1}=(-1)^m a_1/(1·3·…·(2m+1)). The general solution is y = a_0 e^(-x²/2) + a_1 (x - x³/3 + x⁵/15 - …).

## Answer

Equation: y″ + xy′ + y = 0. x = 0 is an ordinary point, so take y = Σ aₙxⁿ.

#### Substitution
y″ = Σ (n+2)(n+1) aₙ₊₂ xⁿ, xy′ = Σ n aₙ xⁿ, y = Σ aₙ xⁿ.

Coefficient of xⁿ: (n+2)(n+1)aₙ₊₂ + (n + 1)aₙ = 0, so

**aₙ₊₂ = − aₙ / (n + 2)**

#### Coefficients
a₀ and a₁ arbitrary.

```text
Even:  a₂ = −a₀/2,  a₄ = −a₂/4 = a₀/(2·4),  a₆ = −a₀/(2·4·6) …
       a₂ₘ = (−1)^m a₀ / (2^m m!)
Odd:   a₃ = −a₁/3,  a₅ = −a₃/5 = a₁/(3·5),  a₇ = −a₁/(3·5·7) …
       a₂ₘ₊₁ = (−1)^m a₁ / (1·3·5···(2m+1))
```

#### Solution
```text
y = a₀(1 − x²/2 + x⁴/8 − x⁶/48 + …) + a₁(x − x³/3 + x⁵/15 − x⁷/105 + …)
```
The first series is e^(−x²/2), because Σ (−x²/2)^m / m! = e^(−x²/2). So

**y = a₀ e^(−x²/2) + a₁ (x − x³/3 + x⁵/15 − …)**

Check: y = e^(−x²/2) gives y′ = −x e^(−x²/2), y″ = (x² − 1)e^(−x²/2); then y″ + xy′ + y = (x² − 1 − x² + 1)e^(−x²/2) = 0.

## Check yourself

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

1. For y'' + xy' + y = 0 the recurrence is:
   A. a_(n+2) = -a_n/(n+2)
   B. a_(n+2) = a_n/(n+2)
   C. a_(n+2) = -a_n/(n+1)
   D. a_(n+2) = -(n+1) a_n/(n+2)
   Answer: A. (n+2)(n+1)a_(n+2) + (n+1)a_n = 0. (Key point: Recurrence)

2. The even series is a0 times:
   A. e^(x^2/2)
   B. e^(-x^2/2)
   C. cos x
   D. e^(-x^2)
   Answer: B. a_2m = (-1)^m a0/(2^m m!). (Key point: Even series)

3. a3 in terms of a1 is:
   A. a1/3
   B. -a1/6
   C. -a1/3
   D. -a1/2
   Answer: C. a3 = -a1/3 from n = 1. (Key point: Odd coefficients)

## Close variants set in other exams

- [Power series: y''−4xy'+(4x²−2)y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apower-series-y-4xy-4x2) — Mathematics, chapter 4, asked 12×
- [Power series solution of y''−y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apower-series-solution-y-y) — Mathematics, chapter 4, asked 12×
- [Define power series; solve y'+2xy=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Adefine-power-series-solve-y) — Mathematics, chapter 4, asked 12×
- [Power series solution of y''+x²y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apower-series-solution-y-x2y) — Mathematics, chapter 4, asked 12×
- [Power series solution of (1−x²)y''−2xy'+2y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apower-series-solution-1-x2) — Mathematics, chapter 4, asked 12×

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