---
title: "Bessel function second-derivative identity"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:bessel-function-second-derivative-identity"
source: https://www.bctengineer.com/q/semester_2/mathematics/bessel-function-second-derivative-identity
site: BCT Engineering
---

# Bessel function second-derivative identity

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

- Asked in: 2070 Bhadra, 5 marks; 2069 Poush, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/bessel-function-second-derivative-identity
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Show that 4J_n''(x) = J_(n−2)(x) − 2J_n(x) + J_(n+2)(x).

## Short answer

Using the identity 2J_n' = J_{n-1} - J_{n+1}, differentiate to obtain 2J_n'' = J_{n-1}' - J_{n+1}', then substitute the same identity for J_{n-1}' and J_{n+1}'. This gives 4J_n'' = J_{n-2} - 2J_n + J_{n+2}.

## Answer

Use the standard relation obtained from the series of Jₙ:

2Jₙ′(x) = Jₙ₋₁(x) − Jₙ₊₁(x)    ... (1)

This holds for every order n, so we can apply it again to each term.

#### Differentiate (1)
2Jₙ″ = Jₙ₋₁′ − Jₙ₊₁′

Apply (1) with n replaced by n − 1 and by n + 1:

```text
2Jₙ₋₁′ = Jₙ₋₂ − Jₙ
2Jₙ₊₁′ = Jₙ − Jₙ₊₂
```

#### Combine
Multiply 2Jₙ″ = Jₙ₋₁′ − Jₙ₊₁′ by 2:

```text
4Jₙ″ = 2Jₙ₋₁′ − 2Jₙ₊₁′
     = (Jₙ₋₂ − Jₙ) − (Jₙ − Jₙ₊₂)
     = Jₙ₋₂ − 2Jₙ + Jₙ₊₂
```

**Hence 4Jₙ″(x) = Jₙ₋₂(x) − 2Jₙ(x) + Jₙ₊₂(x).**

#### Proof of relation (1) (for completeness)
From the two standard results xJₙ′ = nJₙ − xJₙ₊₁ and xJₙ′ = −nJₙ + xJₙ₋₁ (obtained from d/dx[x⁻ⁿJₙ] = −x⁻ⁿJₙ₊₁ and d/dx[xⁿJₙ] = xⁿJₙ₋₁), adding gives 2xJₙ′ = x(Jₙ₋₁ − Jₙ₊₁).

## Check yourself

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

1. The relation applied repeatedly is:
   A. 2 J_n' = J_(n-1) + J_(n+1)
   B. J_n' = J_(n-1) J_(n+1)
   C. 2 J_n' = J_(n-1) - J_(n+1)
   D. 2 J_n' = J_(n+1) - J_(n-1)
   Answer: C. It holds for every n. (Key point: Starting relation)

2. Differentiating once gives:
   A. 2 J_n'' = J_(n-1)' - J_(n+1)'
   B. 2 J_n'' = J_(n-1) - J_(n+1)
   C. J_n'' = J_(n-1)' + J_(n+1)'
   D. 2 J_n'' = J_(n-2) - J_(n+2)
   Answer: A. Differentiating both sides. (Key point: Differentiating)

3. 2 J_(n-1)' and 2 J_(n+1)' equal:
   A. J_(n-2) + J_n and J_n + J_(n+2)
   B. J_(n-2) - J_n and J_n - J_(n+2)
   C. J_(n-1) - J_(n+1) twice
   D. J_n - J_(n-2) and J_(n+2) - J_n
   Answer: B. Apply the relation with n replaced. (Key point: n-1 and n+1)

4. Subtracting gives:
   A. J_(n-2) - 2J_n + J_(n+2)
   B. J_(n-2) + 2J_n + J_(n+2)
   C. J_(n-2) - J_(n+2)
   D. 2J_n - J_(n-2) - J_(n+2)
   Answer: A. (J_(n-2) - J_n) - (J_n - J_(n+2)). (Key point: Combination)

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