---
title: "Solve x²y''−xy'+2y=x log x"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-cauchy-x2y2-minus-xy1-plus-2y-x-log-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-cauchy-x2y2-minus-xy1-plus-2y-x-log-x
site: BCT Engineering
---

# Solve x²y''−xy'+2y=x log x

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

- Asked in: 2070 Asar
- Video: [x^2d^2y/dx^2+4xdy/dx+2y=xlogx , CauchyEulerLinearEquation  L901](https://www.youtube.com/watch?v=mmTzNcDhAvk) — 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-x2y2-minus-xy1-plus-2y-x-log-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve the differential equation x² d²y/dx² − x dy/dx + 2y = x log x.

## Short answer

Cauchy–Euler equation, transform with z = log x. General solution: y = x[c₁ cos(log x) + c₂ sin(log x)] + x log x.

## Answer

x²y'' − xy' + 2y = x log x is a Cauchy–Euler equation.

#### Transformation
Put z = log x, D = d/dz. The right side becomes x log x = z eᶻ.

[D(D − 1) − D + 2]y = z eᶻ
(D² − 2D + 2)y = z eᶻ

#### Complementary function
m² − 2m + 2 = 0, so m = 1 ± i.

CF = eᶻ(c₁ cos z + c₂ sin z) = x[c₁ cos(log x) + c₂ sin(log x)]

#### Particular integral
Replace D by D + 1:

PI = eᶻ · 1/[(D + 1)² − 2(D + 1) + 2] · z = eᶻ · 1/(D² + 1) · z
 = eᶻ(1 − D² + …) z = z eᶻ

#### General solution
**y = x[c₁ cos(log x) + c₂ sin(log x)] + x log x**

## Check yourself

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

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

2. The roots of m^2 - 2m + 2 = 0 give CF:
   A. c1 x + c2 x^2
   B. x[c1 cos(log x) + c2 sin(log x)]
   C. x^2[c1 cos(log x) + ...]
   D. (c1 + c2 log x)x
   Answer: B. m = 1 ± i. (Key point: CF)

3. Replacing D by D + 1 in D^2 - 2D + 2 gives:
   A. D^2 - 1
   B. D^2 + 2D + 1
   C. D^2 + 2
   D. D^2 + 1
   Answer: D. (D+1)^2 - 2(D+1) + 2 = D^2 + 1. (Key point: Shifted operator)

4. The PI is:
   A. z e^z = x log x
   B. x
   C. e^z/2
   D. x log x/2
   Answer: A. (1 - D^2 + ...) z = z. (Key point: PI)

## Close variants set in other exams

- [Solve x dy/dx + y log y = xy eˣ](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-dy-dx-y-log-y) — Mathematics I, chapter 4, asked 4×
- [Solve x log x dy/dx + y = 2 log x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-x-log-x-dy-dx) — Mathematics I, chapter 4, asked 4×

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