---
title: "Series solution of (1+x²)y''+xy'−y=0"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:series-solution-1-x2-y"
source: https://www.bctengineer.com/q/semester_2/mathematics/series-solution-1-x2-y
site: BCT Engineering
---

# Series solution of (1+x²)y''+xy'−y=0

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

- Asked in: 2067 Mangsir, 5 marks
- 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/series-solution-1-x2-y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Solve in series: (1+x²) d²y/dx² + x dy/dx − y = 0.

## Short answer

Recurrence a_{n+2}=(1−n)a_n/(n+2). Solution: y=a₀(1+x²/2−x⁴/8+x⁶/16−…)+a₁x, where a₁x is a polynomial solution.

## Answer

Equation: (1 + x²)y″ + xy′ − y = 0. x = 0 is an ordinary point; let y = Σ aₙxⁿ.

#### Substitution
```text
(1+x²)y″ = Σ (n+2)(n+1)aₙ₊₂ xⁿ + Σ n(n−1)aₙ xⁿ
xy′ = Σ n aₙ xⁿ,    −y = −Σ aₙ xⁿ
```
Coefficient of xⁿ: (n+2)(n+1)aₙ₊₂ + [n(n−1) + n − 1]aₙ = 0. Since n(n−1) + n − 1 = n² − 1,

**aₙ₊₂ = − (n² − 1) aₙ / ((n+1)(n+2)) = (1 − n)aₙ/(n+2)**

(because n² − 1 = (n − 1)(n + 1)).

#### Coefficients (a₀, a₁ arbitrary)
```text
n=0:  a₂ = a₀/2
n=1:  a₃ = 0   (so a₅ = a₇ = … = 0)
n=2:  a₄ = −a₂/4 = −a₀/8
n=4:  a₆ = −3a₄/6 = a₀/16
```

#### Solution
```text
y = a₀(1 + x²/2 − x⁴/8 + x⁶/16 − …) + a₁x
```
The odd solution terminates: y = x satisfies the equation exactly (0 + x − x = 0). The even series is the expansion of √(1 + x²).

## Check yourself

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

1. The recurrence relation is:
   A. a(n+2) = (1 − n)aₙ/(n + 2)
   B. a(n+2) = (n − 1)aₙ/(n + 2)
   C. a(n+2) = −aₙ/(n + 2)
   D. a(n+2) = (1 − n)a(n+1)/(n + 2)
   Answer: A. (n+2)(n+1)a(n+2) + (n^2 − 1)aₙ = 0, so a(n+2) = −(n−1)aₙ/(n+2). (Key point: Recurrence)

2. Why do all odd coefficients beyond a1 vanish?
   A. a1 is zero
   B. Odd n makes the denominator infinite
   C. x = 0 is singular
   D. At n = 1 the factor (1 − n) is zero, so a3 = 0
   Answer: D. a3 = 0 and every later odd term is built from a3. (Key point: Odd terms vanish)

3. What is a2?
   A. −a0/2
   B. a0/4
   C. a0/2
   D. −a0/8
   Answer: C. n = 0: a2 = (1 − 0)a0/2 = a0/2. (Key point: First coefficient)

4. Which simple function solves the equation exactly?
   A. y = x^2
   B. y = x
   C. y = 1
   D. y = 1/x
   Answer: B. Check: (1+x^2)(0) + x(1) − x = 0. (Key point: Closed solution)

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