---
title: "Power series solution of y''+x²y=0"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:power-series-solution-y-x2y"
source: https://www.bctengineer.com/q/semester_2/mathematics/power-series-solution-y-x2y
site: BCT Engineering
---

# Power series solution of y''+x²y=0

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

- Asked in: 2068 Magh, 5 marks
- Video: [Power Series Solution of Differential Equation: y'' + x^2y = 0](https://www.youtube.com/watch?v=X8vSBRO60Mw) — Academic Videos (Solved Examples)
- 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-x2y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Apply the power series method and solve the differential equation y'' + x²y = 0.

## Short answer

Power series yields a₂=a₃=0 and recurrence a_{n+2}=−a_{n−2}/((n+1)(n+2)). Solution: y=a₀(1−x⁴/12+x⁸/672−…)+a₁(x−x⁵/20+x⁹/1440−…).

## Answer

Equation: y″ + x²y = 0. x = 0 is an ordinary point; take y = Σ aₙxⁿ.

#### Substitution
y″ = Σ (n+2)(n+1)aₙ₊₂ xⁿ and x²y = Σ aₙ₋₂ xⁿ (n ≥ 2).

- x⁰: 2a₂ = 0, so a₂ = 0.
- x¹: 6a₃ = 0, so a₃ = 0.
- xⁿ (n ≥ 2): (n+2)(n+1)aₙ₊₂ + aₙ₋₂ = 0, so

**aₙ₊₂ = − aₙ₋₂ / ((n+1)(n+2))**

#### Coefficients (a₀, a₁ arbitrary)
```text
n=2:  a₄ = −a₀/(3·4)  = −a₀/12
n=3:  a₅ = −a₁/(4·5)  = −a₁/20
n=4:  a₆ = −a₂/30 = 0
n=5:  a₇ = −a₃/42 = 0
n=6:  a₈ = −a₄/(7·8) = a₀/672
n=7:  a₉ = −a₅/(8·9) = a₁/1440
```
Only powers x⁴ᵐ and x⁴ᵐ⁺¹ survive.

#### Solution
```text
y = a₀(1 − x⁴/12 + x⁸/672 − …) + a₁(x − x⁵/20 + x⁹/1440 − …)
```
a₀ and a₁ are arbitrary constants.

## Check yourself

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

1. Why are a2 = 0 and a3 = 0?
   A. Because a0 = a1 = 0
   B. Because x = 0 is singular
   C. Because the equation is first order
   D. y'' gives 2a2 and 6a3 at x^0 and x^1, while x^2 y has no such terms
   Answer: D. x^2 y starts at x^2, so the x^0 and x^1 coefficients give 2a2 = 0 and 6a3 = 0. (Key point: Low coefficients)

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

3. Which powers of x survive in the solution?
   A. Only even powers
   B. x^(4m) and x^(4m+1)
   C. All powers
   D. x^(2m) and x^(2m+1)
   Answer: B. The recurrence links coefficients 4 apart and a2 = a3 = 0. (Key point: Surviving powers)

4. What is a4?
   A. −a0/12
   B. a0/12
   C. −a0/6
   D. −a0/4
   Answer: A. For n = 2: a4 = −a0/(3·4) = −a0/12. (Key point: Coefficient)

## Close variants set in other exams

- [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×
- [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''+xy'+y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apower-series-solution-y-xy) — 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×

## Related in Mathematics

- [Series solution of Legendre's equation](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aseries-solution-legendre-s-equation) — Mathematics, chapter 4, asked 4×
- [Power series solution of y''−y=x](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apower-series-solution-y-y-2) — Mathematics, chapter 4
- [Series solution of x²y''+xy'+(x²−4)y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aseries-solution-x2y-xy-x2) — Mathematics, chapter 4
- [Series solution of (1+x²)y''+xy'−y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aseries-solution-1-x2-y) — Mathematics, chapter 4

## Related in other subjects

- [Solve a partial differential equation](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-a-partial-differential-equation) — Mathematics III, chapter 4
- [Solve a partial differential equation](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asolve-a-partial-differential-equation-two) — Mathematics III, chapter 4
- [LC oscillation differential equation](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Alc-oscillation-differential-equation) — Engineering Physics, chapter 1
- [Three moment equation: derive and apply to fixed beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Athree-moment-derive-fixed-beam-udl) — Theory of Structures II, chapter 2