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

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)

∂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:

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.