---
title: "Homogeneous function: sin u=(√x−√y)/(√x+√y)"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:homogeneous-function-sin-u-sqrt"
source: https://www.bctengineer.com/q/semester_2/mathematics/homogeneous-function-sin-u-sqrt
site: BCT Engineering
---

# Homogeneous function: sin u=(√x−√y)/(√x+√y)

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2069 Bhadra, 5 marks
- Video: [solve u= tan^-1 [(x^3+y^3)/(x-y)] EULERS PARTIAL DERIVATIVE](https://www.youtube.com/watch?v=Coga9fXOgMM) — m-easy maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/homogeneous-function-sin-u-sqrt
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

If sin u = (√x − √y)/(√x + √y), show that x ∂u/∂x + y ∂u/∂y = 0.

## Short answer

Let z = sin u = (√x−√y)/(√x+√y). Replacing x,y by tx,ty shows z is homogeneous of degree 0. By Euler’s theorem, x ∂z/∂x + y ∂z/∂y = 0. Substituting ∂z/∂x = cos u ∂u/∂x etc. gives cos u (x u_x + y u_y)=0, hence x ∂u/∂x + y ∂u/∂y = 0.

## Answer

Given sin u = (√x − √y)/(√x + √y). Let

z = sin u = (x^(1/2) − y^(1/2))/(x^(1/2) + y^(1/2)).

#### Homogeneity
Replace x, y by tx, ty. The factor t^(1/2) cancels between numerator and denominator:

z(tx, ty) = t^(1/2)(√x − √y)/(t^(1/2)(√x + √y)) = t⁰ z.

So z is a homogeneous function of degree n = 0.

#### Euler's theorem
For a homogeneous function of degree *n*: x ∂z/∂x + y ∂z/∂y = n z. Here n = 0:

x ∂z/∂x + y ∂z/∂y = 0

Since z = sin u, ∂z/∂x = cos u ∂u/∂x and ∂z/∂y = cos u ∂u/∂y:

```text
cos u ( x ∂u/∂x + y ∂u/∂y ) = 0
x ∂u/∂x + y ∂u/∂y = 0      (cos u ≠ 0)
```

Hence proved. ∎

## Check yourself

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

1. z = sin u = (√x−√y)/(√x+√y) is homogeneous of degree:
   A. 0
   B. −1
   C. 1
   D. 1/2
   Answer: A. t^(1/2) cancels. (Key point: Homogeneity)

2. For degree n = 0, Euler's theorem gives:
   A. z_x = z_y
   B. x z_x = y z_y
   C. x z_x + y z_y = z
   D. x z_x + y z_y = 0
   Answer: D. n z = 0. (Key point: Euler)

3. Why is cos u ≠ 0 needed?
   A. To divide the equation cos u (x u_x + y u_y) = 0 by cos u
   B. Because u = 0
   C. To cancel √x
   D. To make z homogeneous
   Answer: A. z_x = cos u u_x, so dividing by cos u leaves the result. (Key point: Passing to u)

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