---
title: "Series solution of Legendre's equation"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:series-solution-legendre-s-equation"
source: https://www.bctengineer.com/q/semester_2/mathematics/series-solution-legendre-s-equation
site: BCT Engineering
---

# Series solution of Legendre's equation

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

- Asked in: 2075 Baisakh, 5 marks; 2070 Magh, 5 marks; 2069 Poush, 5 marks; 2068 Magh, 5 marks (asked 4 times)
- Video: [How to solve Series Solution of Legendre's Differential Equation by using simple steps](https://www.youtube.com/watch?v=IOoXumI6_lE) — EASY MATHS EASY TRICKS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/series-solution-legendre-s-equation
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Solve in series, the Legendre's equation (1−x²)y'' − 2xy' + n(n+1)y = 0.

## Short answer

Substituting y = Σ a_k x^k yields recurrence a_{k+2} = -(n-k)(n+k+1)/((k+1)(k+2)) a_k. The general solution is a0 y1(x) + a1 y2(x) with even and odd series; for integer n, one series terminates to the Legendre polynomial P_n(x).

## Answer

Legendre's equation: (1 − x²)y″ − 2xy′ + n(n + 1)y = 0. x = 0 is an ordinary point, so a power-series solution y = Σ aₖxᵏ exists.

#### Substitution
y′ = Σ k aₖ x^(k−1), y″ = Σ k(k−1) aₖ x^(k−2).

```text
Σ k(k−1)aₖ x^(k−2) − Σ k(k−1)aₖ xᵏ − 2Σ k aₖ xᵏ + n(n+1)Σ aₖ xᵏ = 0
```
Shift the first sum (k → k + 2) and collect the coefficient of xᵏ:

(k + 2)(k + 1)aₖ₊₂ − [k(k − 1) + 2k − n(n + 1)]aₖ = 0.

Since k(k − 1) + 2k − n(n + 1) = k(k + 1) − n(n + 1) = (k − n)(k + n + 1),

**aₖ₊₂ = − (n − k)(n + k + 1) / ((k + 1)(k + 2)) · aₖ**

#### Coefficients
a₀ and a₁ are arbitrary.

```text
a₂ = − n(n+1)/2! · a₀
a₃ = − (n−1)(n+2)/3! · a₁
a₄ = (n−2)n(n+1)(n+3)/4! · a₀
a₅ = (n−3)(n−1)(n+2)(n+4)/5! · a₁
```

#### Solution
y = a₀y₁(x) + a₁y₂(x), where

```text
y₁ = 1 − n(n+1)/2! x² + (n−2)n(n+1)(n+3)/4! x⁴ − …
y₂ = x − (n−1)(n+2)/3! x³ + (n−3)(n−1)(n+2)(n+4)/5! x⁵ − …
```
Both series converge for |x| < 1.

When n is a non-negative integer, one of the series terminates (y₁ for even n, y₂ for odd n), giving a polynomial, the Legendre polynomial Pₙ(x) after suitable choice of the constant. For example n = 2 gives y₁ = 1 − 3x², proportional to P₂ = (3x² − 1)/2.

## Check yourself

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

1. Why can y = sum a_k x^k be assumed?
   A. x = 0 is an ordinary point
   B. x = 0 is a regular singular point
   C. n is an integer
   D. The equation is linear with constant coefficients
   Answer: A. At an ordinary point power series solutions exist. (Key point: Series validity)

2. The recurrence relation is:
   A. a_(k+2) = (n-k)(n+k+1) a_k / ((k+1)(k+2))
   B. a_(k+2) = -(n-k)(n+k+1) a_k / ((k+1)(k+2))
   C. a_(k+2) = -(n+k)(n-k+1) a_k / (k(k+1))
   D. a_(k+1) = -(n-k)(n+k+1) a_k / ((k+1)(k+2))
   Answer: B. k(k+1) - n(n+1) = (k-n)(k+n+1). (Key point: Recurrence)

3. Which coefficients are arbitrary?
   A. a0 only
   B. a1 and a2
   C. a0 and a1
   D. All of them
   Answer: C. The recurrence skips by 2, giving two independent series. (Key point: Arbitrary constants)

4. When n is a non-negative integer, what happens?
   A. One series terminates, giving a Legendre polynomial
   B. Both series diverge
   C. Both terminate
   D. The solution is zero
   Answer: A. The factor (n-k) vanishes at k = n. (Key point: Polynomial case)

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