Euler's theorem with log function

Use Euler's theorem on a suitable function of u.

Euler's theorem

If f(x, y) is homogeneous of degree n, then x ∂f/∂x + y ∂f/∂y = n f.

Proof of the result

u = log((x² + y²)/(x + y)) is not homogeneous itself, so put

z = eᵘ = (x² + y²)/(x + y).

Replacing x by tx and y by ty: z(tx, ty) = t²(x² + y²)/(t(x + y)) = t z. So z is homogeneous of degree n = 1.

By Euler's theorem:

x ∂z/∂x + y ∂z/∂y = 1 · z

Since z = eᵘ, ∂z/∂x = eᵘ ∂u/∂x and ∂z/∂y = eᵘ ∂u/∂y. Therefore

eᵘ ( x ∂u/∂x + y ∂u/∂y ) = eᵘ
x ∂u/∂x + y ∂u/∂y = 1

Hence proved. ∎