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

# Power series solution of y''−y=0

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

- Asked in: 2075 Baisakh, 5 marks; 2072 Magh, 5 marks (asked 2 times)
- Video: [Power Series Method -  Solve y''+y=0](https://www.youtube.com/watch?v=L93PLTrMGYI) — Polar Pi
- 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-y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Solve the differential equation by power series method: y'' − y = 0.

## Short answer

Assume y = Σ a_n x^n. Recurrence a_{n+2}=a_n/((n+1)(n+2)) gives a_{2n}=a_0/(2n)!, a_{2n+1}=a_1/(2n+1)!. Summing yields y = a_0 cosh x + a_1 sinh x = C_1 e^x + C_2 e^{-x}.

## Answer

Equation: y″ − y = 0. x = 0 is an ordinary point, so assume y = Σ aₙxⁿ (n = 0, 1, 2, …).

#### Substitution
y″ = Σ (n+2)(n+1)aₙ₊₂ xⁿ. Putting in the equation and equating the coefficient of xⁿ to zero:

(n+2)(n+1)aₙ₊₂ − aₙ = 0, so **aₙ₊₂ = aₙ / ((n+1)(n+2))**

#### Coefficients
a₀ and a₁ are arbitrary.

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

#### Solution
```text
y = a₀(1 + x²/2! + x⁴/4! + …) + a₁(x + x³/3! + x⁵/5! + …)
  = a₀ cosh x + a₁ sinh x
```
Writing cosh x = (eˣ + e⁻ˣ)/2 and sinh x = (eˣ − e⁻ˣ)/2, this is the same as

**y = C₁eˣ + C₂e⁻ˣ**,

which agrees with the direct solution of the auxiliary equation m² − 1 = 0.

## Check yourself

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

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

2. The even series is:
   A. a0 cos x
   B. a0 e^x
   C. a0 cosh x
   D. a0 sinh x
   Answer: C. It is 1 + x^2/2! + x^4/4! + ... (Key point: Series)

3. This is equal to:
   A. C1 e^x + C2 e^(-x)
   B. C1 cos x + C2 sin x
   C. C1 e^x + C2 x e^x
   D. C1 + C2 x
   Answer: A. cosh and sinh combine exponentials; matches m^2 = 1. (Key point: Equivalent form)

## 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×
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- [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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