---
title: "Solve x dy/dx + y log y = xy eˣ"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-x-dy-dx-y-log-y"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-x-dy-dx-y-log-y
site: BCT Engineering
---

# Solve x dy/dx + y log y = xy eˣ

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

- Asked in: 2072 Kartik; 2069 Chaitra (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-x-dy-dx-y-log-y
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve the differential equation x dy/dx + y log y = xy eˣ.

## Short answer

Divide by xy, set z=log y to obtain dz/dx + z/x = eˣ. Integrating factor x gives xz = (x−1)eˣ + c. Hence x log y = (x−1)eˣ + c.

## Answer

x dy/dx + y log y = xy eˣ

#### Reduce to a linear equation
Divide by xy:

(1/y) dy/dx + (log y)/x = eˣ

Put z = log y, so dz/dx = (1/y) dy/dx:

dz/dx + z/x = eˣ

This is linear in z with P = 1/x, Q = eˣ.

#### Integrating factor
IF = e^(∫dx/x) = x

#### Solution
z · x = ∫ x eˣ dx = x eˣ − eˣ + c = (x − 1)eˣ + c

Replace z = log y:

**x log y = (x − 1)eˣ + c**, i.e. log y = (1 − 1/x)eˣ + c/x.

## Check yourself

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

1. Which substitution makes x dy/dx + y log y = xy e^x linear?
   A. z = y/x
   B. z = log y
   C. z = e^y
   D. z = 1/y
   Answer: B. Dividing by xy and setting z = log y gives dz/dx + z/x = e^x. (Key point: Reduction)

2. The resulting equation is:
   A. dz/dx - z/x = e^x
   B. dz/dx + z/x = e^x
   C. dz/dx + z = x e^x
   D. dz/dx + xz = e^x
   Answer: B. (1/y)dy/dx = dz/dx and (log y)/x = z/x. (Key point: Linear form)

3. The integrating factor is:
   A. x
   B. 1/x
   C. log x
   D. e^x
   Answer: A. IF = e^(integral dx/x) = x. (Key point: Integrating factor)

4. The solution is:
   A. x log y = (x - 1)e^x + c
   B. x log y = x e^x + c
   C. y = (x - 1)e^x + c
   D. log y = (x - 1)e^x + c
   Answer: A. z x = integral of x e^x dx = (x - 1)e^x + c, with z = log y. (Key point: Final solution)

## Close variants set in other exams

- [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×
- [Solve x²y''−xy'+2y=x log x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-cauchy-x2y2-minus-xy1-plus-2y-x-log-x) — Mathematics I, chapter 4, asked 4×

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