If \(u=\log\left(\dfrac{x^2+y^2}{x+y}\right)\), find the value of \(x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}\).

Answer: \(x\,u_x+y\,u_y=1\).

Here \(u\) itself is not homogeneous, but \(e^u=\dfrac{x^2+y^2}{x+y}=f(x,y)\) is homogeneous of degree \(n=2-1=1\) (replace \(x\to tx,\ y\to ty\): \(f\to t\,f\)).

By Euler's theorem on \(f=e^u\): \(x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=1\cdot f\).

\(\dfrac{\partial f}{\partial x}=e^u\dfrac{\partial u}{\partial x}\), \(\dfrac{\partial f}{\partial y}=e^u\dfrac{\partial u}{\partial y}\), so \(e^u\left(x u_x+y u_y\right)=e^u\).

\[\therefore\ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=1.\]