---
title: "Euler's theorem verified for (x^¼+y^¼)/(x^½+y^⅕)"
subject: "Mathematics"
chapter: 1
key: "semester_2/mathematics:euler-s-theorem-verified-x-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/euler-s-theorem-verified-x-2
site: BCT Engineering
---

# Euler's theorem verified for (x^¼+y^¼)/(x^½+y^⅕)

Mathematics · Chapter 1: Calculus of Two or more variables

- Asked in: 2067 Mangsir, 1+4 marks
- Video: [Mathematics: Illustration on Euler's Theorem on Homogeneous Function](https://www.youtube.com/watch?v=palSdK9P-ns) — Edredo for Learners
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/euler-s-theorem-verified-x-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-1

## Question

State Euler's Theorem for a homogeneous function of two independent variables and verify it for the function u = (x^(1/4) + y^(1/4))/(x^(1/2) + y^(1/5)). [Exponent of y in the denominator hard to read]

## Short answer

Euler: If f(tx,ty)=tⁿf, then x f_x+y f_y=n f. For u=(x^(1/4)+y^(1/4))/(x^k+y^k) (denominator exponents equal), it is homogeneous of degree n=1/4−k. Direct partial differentiation gives x ∂u/∂x+y ∂u/∂y=(1/4−k)u, verifying the theorem; with k=1/5, n=1/20; with k=1/2, n=−1/4.

## Answer

**Reading of the question.** As printed, u = (x^(1/4) + y^(1/4))/(x^(1/2) + y^(1/5)) is not homogeneous (the two denominator terms have degrees 1/2 and 1/5), so Euler's theorem cannot be verified for it. The two exponents in the denominator must be equal. The standard textbook version of this problem is u = (x^(1/4) + y^(1/4))/(x^(1/5) + y^(1/5)); another possible reading is x^(1/2) + y^(1/2) in the denominator. Both are covered below by writing the denominator as x^k + y^k.

#### Euler's theorem (1 mark)
If f(x, y) is homogeneous of degree *n*, then x ∂f/∂x + y ∂f/∂y = n f.

#### Degree of u
Let N = x^(1/4) + y^(1/4) and D = x^k + y^k, so u = N/D. Replacing x, y by tx, ty turns N into t^(1/4) N and D into t^k D, so u(tx, ty) = t^(1/4 − k) u. Thus u is homogeneous of degree n = 1/4 − k, and Euler's theorem predicts x u_x + y u_y = (1/4 − k) u.

#### Direct verification (4 marks)
```text
∂u/∂x = [ (1/4)x^(−3/4) D − N k x^(k−1) ] / D²
x ∂u/∂x = [ (1/4)x^(1/4) D − k x^k N ] / D²

y ∂u/∂y = [ (1/4)y^(1/4) D − k y^k N ] / D²
```

Add them, using x^(1/4) + y^(1/4) = N and x^k + y^k = D:

```text
x u_x + y u_y = [ (1/4) N D − k D N ] / D²
              = (1/4 − k) N D / D²
              = (1/4 − k) N/D = (1/4 − k) u
```

This equals n u, so Euler's theorem is verified. ∎

#### The two readings
- Denominator x^(1/5) + y^(1/5) (k = 1/5): n = 1/4 − 1/5 = 1/20, so x ∂u/∂x + y ∂u/∂y = u/20.
- Denominator x^(1/2) + y^(1/2) (k = 1/2): n = 1/4 − 1/2 = −1/4, so x ∂u/∂x + y ∂u/∂y = −u/4.

## Check yourself

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

1. For u = (x^(1/4)+y^(1/4))/(x^(1/2)+y^(1/2)), degree n is:
   A. 3/4
   B. −1/2
   C. 1/4
   D. −1/4
   Answer: D. t^(1/4)/t^(1/2) = t^(−1/4). (Key point: Degree)

2. Direct computation gives x u_x + y u_y equal to:
   A. 0
   B. −(1/4) u
   C. −u
   D. (1/4) u
   Answer: B. Equals n u with n = −1/4. (Key point: Verification)

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