---
title: "Power series solution of (1−x²)y''−2xy'+2y=0"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:power-series-solution-1-x2"
source: https://www.bctengineer.com/q/semester_2/mathematics/power-series-solution-1-x2
site: BCT Engineering
---

# Power series solution of (1−x²)y''−2xy'+2y=0

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

- Asked in: 2067 Mangsir, 5 marks
- Video: [5. (1-x2)y''-2xy'+2y=0 Shorter Method - ODE Solution || Safayat Munna,BUET'19](https://www.youtube.com/watch?v=ysiFN-lMjtw) — Safayat Munna
- 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-1-x2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Apply the power series method to solve the following differential equation (1−x²) d²y/dx² − 2x dy/dx + 2y = 0.

## Short answer

Substituting series gives a_{k+2}=(k−1)a_k/(k+1). Solution: y=a₀(1−x²−x⁴/3−x⁶/5−…)+a₁x, with y=x the Legendre polynomial P₁(x).

## Answer

Equation: (1 − x²)y″ − 2xy′ + 2y = 0. This is Legendre's equation with n(n + 1) = 2, i.e. n = 1. x = 0 is an ordinary point; take y = Σ aₖxᵏ.

#### Recurrence
Coefficient of xᵏ: (k+2)(k+1)aₖ₊₂ − k(k−1)aₖ − 2k aₖ + 2aₖ = 0.

Since k(k−1) + 2k − 2 = k² + k − 2 = (k − 1)(k + 2),

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

#### Coefficients (a₀, a₁ arbitrary)
```text
k=0:  a₂ = −a₀
k=1:  a₃ = 0       (so all further odd terms vanish)
k=2:  a₄ = a₂/3 = −a₀/3
k=4:  a₆ = 3a₄/5 = −a₀/5
k=6:  a₈ = 5a₆/7 = −a₀/7
```
Even part: 1 − x² − x⁴/3 − x⁶/5 − x⁸/7 − …

#### Solution
```text
y = a₀(1 − x² − x⁴/3 − x⁶/5 − …) + a₁x
```
The odd solution y = x (= P₁) is a polynomial; the even solution is an infinite series (convergent for |x| < 1) = 1 − x·tanh⁻¹x.

## Check yourself

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

1. (1 − x^2)y'' − 2xy' + 2y = 0 is Legendre's equation with:
   A. n = 1, since n(n+1) = 2
   B. n = 2, since n(n+1) = 6
   C. n = 2, since n^2 = 2
   D. n = 0
   Answer: A. n(n+1) = 2 gives n = 1. (Key point: Identify equation)

2. The recurrence relation is:
   A. a(k+2) = (k + 1)aₖ/(k − 1)
   B. a(k+2) = (k − 1)aₖ/(k + 2)
   C. a(k+2) = −aₖ/(k + 1)
   D. a(k+2) = (k − 1)aₖ/(k + 1)
   Answer: D. (k+2)(k+1)a(k+2) = (k−1)(k+2)aₖ. (Key point: Recurrence)

3. Why does the odd part terminate as y = x?
   A. a1 = 0
   B. a3 = a1
   C. a3 = 0 because the factor (k − 1) vanishes at k = 1
   D. The odd series diverges
   Answer: C. k = 1 gives a3 = 0, so only a1 x remains, which is P₁. (Key point: Odd solution)

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

## Close variants set in other exams

- [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: 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×
- [Polar coordinates: ∫∫(x²+y²) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apolar-coordinates-x2-y2-dy) — Mathematics, chapter 2, asked 5×
- [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
- [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 (1+x²)y''+xy'−y=0](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aseries-solution-1-x2-y) — Mathematics, chapter 4
- [Evaluate ∫∫ e^(y/x) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-y-x-dy) — Mathematics, chapter 2

## Related in other subjects

- [Solve dy/dx = (x+2y−3)/(2x+y−3)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-homogeneous-linear-x-2y-3) — Mathematics I, chapter 4
- [Solve (x²−y²)dx+2xy dy=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x2-minus-y2-dx-2xy-dy) — Mathematics I, chapter 4
- [Solve dy/dx − 2y tan x = y² tan x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-bernoulli-2y-tan-x) — Mathematics I, chapter 4, asked 6×
- [Solve x²y''−2xy'+2y=1/x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-cauchy-x2y2-minus-2xy1-plus-2y-1-over-x) — Mathematics I, chapter 4