---
title: "Solve (x²D²+xD+1)y=sin(log x²)"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-cauchy-x2d2-xd-1-sin-log-x2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-cauchy-x2d2-xd-1-sin-log-x2
site: BCT Engineering
---

# Solve (x²D²+xD+1)y=sin(log x²)

Mathematics I · Chapter 4: Ordinary Differential Equations and their Applications

- Asked in: 2071 Chaitra
- Video: [x^2d^2y/dx^2-3xdy/dx+5y=x^2Sin(logx) #CauchyEulerLinearEquation L912](https://www.youtube.com/watch?v=H5B0u1hGW54) — Maths Pulse - Chinnaiah Kalpana
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-cauchy-x2d2-xd-1-sin-log-x2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve (x²D² + xD + 1)y = sin(log x²).

## Short answer

Equation is Cauchy–Euler; substitution z = log x gives (θ²+1)y = sin 2z. General solution: y = c₁ cos(log x) + c₂ sin(log x) − (1/3) sin(2 log x).

## Answer

(x²D² + xD + 1)y = sin(log x²), where D = d/dx. This is a Cauchy–Euler equation.

#### Transformation
Put z = log x. With θ = d/dz: x²D² = θ(θ − 1) and xD = θ. Also log x² = 2z. So

[θ(θ − 1) + θ + 1]y = sin 2z
(θ² + 1)y = sin 2z

#### Complementary function
m² + 1 = 0, m = ±i: CF = c₁ cos z + c₂ sin z = c₁ cos(log x) + c₂ sin(log x)

#### Particular integral
Put θ² = −4:

PI = 1/(θ² + 1) · sin 2z = sin 2z/(−4 + 1) = −(1/3) sin 2z

#### General solution
**y = c₁ cos(log x) + c₂ sin(log x) − (1/3) sin(2 log x)**

## Check yourself

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

1. sin(log x^2) becomes, with z = log x:
   A. 2 sin z
   B. sin 2z
   C. sin z^2
   D. sin z
   Answer: B. log x^2 = 2 log x. (Key point: RHS)

2. The transformed equation is:
   A. (theta^2 + 1)y = sin z
   B. (theta^2 + 2theta + 1)y = sin 2z
   C. (theta^2 - 1)y = sin 2z
   D. (theta^2 + 1)y = sin 2z
   Answer: D. theta(theta - 1) + theta + 1. (Key point: Transformed equation)

3. PI for sin 2z uses:
   A. theta^2 = 4
   B. theta = 2
   C. theta^2 = -1
   D. theta^2 = -4
   Answer: D. For sin az, theta^2 = -a^2. (Key point: PI rule)

4. The solution is:
   A. c1 cos(log x) + c2 sin(log x) - (1/3) sin(2 log x)
   B. ... + (1/3) sin(2 log x)
   C. ... - sin(2 log x)
   D. ... - (1/5) sin(2 log x)
   Answer: A. 1/(-4 + 1) = -1/3. (Key point: Final solution)

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