---
title: "Solve dy/dx = (x+2y−3)/(2x+y−3)"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-homogeneous-linear-x-2y-3"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-homogeneous-linear-x-2y-3
site: BCT Engineering
---

# Solve dy/dx = (x+2y−3)/(2x+y−3)

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

- Asked in: 2073 Chaitra
- Video: [Solve: dy/dx = x+2y-3/2x+y-3 | Reducible to homogeneous differential equations](https://www.youtube.com/watch?v=ObXbOt05Fhw) — RK Eduworld (Classic)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-homogeneous-linear-x-2y-3
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve: dy/dx = (x + 2y − 3)/(2x + y − 3).

## Short answer

Shift the origin to (1,1) by setting x=X+1, y=Y+1 to make it homogeneous. Substitute Y=vX, separate variables, integrate, and back-substitute to obtain x + y - 2 = C(x - y)³.

## Answer

dy/dx = (x + 2y − 3)/(2x + y − 3)

The numerator and denominator are linear but not homogeneous. The lines x + 2y − 3 = 0 and 2x + y − 3 = 0 meet at a point; solving them gives x = 1, y = 1.

#### Shift the origin to (1, 1)
Put x = X + 1, y = Y + 1. Then dy/dx = dY/dX and

dY/dX = (X + 2Y)/(2X + Y)

#### Homogeneous substitution
Put Y = vX, so dY/dX = v + X dv/dX:

v + X dv/dX = (1 + 2v)/(2 + v)

X dv/dX = (1 + 2v − 2v − v²)/(2 + v) = (1 − v²)/(2 + v)

(2 + v)dv/[(1 − v)(1 + v)] = dX/X

#### Partial fractions
(2 + v)/[(1 − v)(1 + v)] = (3/2)/(1 − v) + (1/2)/(1 + v)

Integrating:

−(3/2) ln|1 − v| + (1/2) ln|1 + v| = ln|X| + const

(1 + v)/(1 − v)³ = C X²

Put v = Y/X: (X + Y)/X · X³/(X − Y)³ = CX², so X + Y = C(X − Y)³.

#### Back to x and y
X + Y = x + y − 2 and X − Y = x − y:

**x + y − 2 = C(x − y)³**

## Check yourself

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

1. Why is the origin shifted to (1, 1)?
   A. It makes the equation exact
   B. It makes y = 0
   C. It is the intersection of the two lines, removing constants
   D. It is the equation's singular point
   Answer: C. x + 2y - 3 = 0 and 2x + y - 3 = 0 meet at (1, 1). (Key point: Reducing to homogeneous)

2. After x = X+1, y = Y+1, we get:
   A. dY/dX = (X + 2Y)/(2X + Y)
   B. dY/dX = (X + 2Y - 3)/(2X + Y - 3)
   C. dY/dX = (X + Y)/(2X + 2Y)
   D. dY/dX = (2X + Y)/(X + 2Y)
   Answer: A. The constants -3 vanish. (Key point: Transformed equation)

3. (2 + v)/[(1-v)(1+v)] equals:
   A. (1/2)/(1 - v) + (3/2)/(1 + v)
   B. (3/2)/(1 - v) - (1/2)/(1 + v)
   C. 1/(1 - v) + 1/(1 + v)
   D. (3/2)/(1 - v) + (1/2)/(1 + v)
   Answer: D. At v = 1: 3/2; at v = -1: 1/2. (Key point: Partial fractions)

4. The solution is:
   A. x - y = C(x + y - 2)^3
   B. x + y = C(x - y)^3
   C. x + y - 2 = C(x - y)
   D. x + y - 2 = C(x - y)^3
   Answer: D. X + Y = C(X - Y)^3 with X + Y = x + y - 2 and X - Y = x - y. (Key point: Final solution)

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