---
title: "Solve (x²−y²)dx+2xy dy=0"
subject: "Mathematics I"
chapter: 4
key: "semester_1/mathematics_i:de-x2-minus-y2-dx-2xy-dy"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/de-x2-minus-y2-dx-2xy-dy
site: BCT Engineering
---

# Solve (x²−y²)dx+2xy dy=0

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

- Asked in: 2074 Chaitra, 5 marks
- Video: [(x^2-y^2)dx+2xydy=0 || Solve the differential equation (x^2-y^2)dx+2xydy=0](https://www.youtube.com/watch?v=S88nr3zaVCI) — concepts & Methods
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/de-x2-minus-y2-dx-2xy-dy
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-4

## Question

Solve the differential equation (x² − y²)dx + 2xy dy = 0.

## Short answer

The equation is homogeneous. Substituting y=vx, separating variables, and integrating gives the general solution x² + y² = cx, representing circles through the origin with centers on the x-axis.

## Answer

(x² − y²)dx + 2xy dy = 0

The equation is homogeneous of degree 2.

#### Substitution
Put y = vx, dy = v dx + x dv:

(x² − v²x²)dx + 2x(vx)(v dx + x dv) = 0

x²[(1 − v²) + 2v²]dx + 2x³v dv = 0

(1 + v²)dx + 2xv dv = 0

#### Separate variables
2v dv/(1 + v²) = −dx/x

ln(1 + v²) = −ln x + ln c

(1 + v²)x = c

Replace v = y/x:

x(1 + y²/x²) = c → x² + y² = cx

**x² + y² = cx**

This is a family of circles through the origin with centres on the x-axis.

## Check yourself

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

1. (x^2 - y^2)dx + 2xy dy = 0 is:
   A. Linear
   B. Homogeneous of degree 2
   C. Bernoulli
   D. Clairaut
   Answer: B. Both coefficients are degree 2 in x, y. (Key point: Type)

2. After y = vx, the equation becomes:
   A. dx + v dv = 0
   B. (1 + v^2)dx - 2xv dv = 0
   C. (1 + v^2)dx + 2xv dv = 0
   D. (1 - v^2)dx + 2xv dv = 0
   Answer: C. (1 - v^2) + 2v^2 = 1 + v^2. (Key point: Reduced equation)

3. Integrating 2v dv/(1+v^2) = -dx/x gives:
   A. (1 + v^2)/x = c
   B. (1 + v^2)x = c
   C. v^2 x = c
   D. (1 + v^2) = cx^2
   Answer: B. ln(1+v^2) = -ln x + ln c. (Key point: Separation)

4. The solution x^2 + y^2 = cx represents:
   A. Straight lines
   B. Circles centred at origin
   C. Parabolas
   D. Circles through the origin with centres on the x-axis
   Answer: D. Completing the square shows centre (c/2, 0). (Key point: Geometry)

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