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Statement: If \(u=f(x,y)\) is a homogeneous function of degree n in x and y, then \(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nu\).

Proof: Since u is homogeneous of degree n, it can be written as \(u=x^n\phi\left(\frac yx\right)\).

\(\frac{\partial u}{\partial x}=nx^{n-1}\phi\left(\frac yx\right)+x^n\phi'\left(\frac yx\right)\left(-\frac y{x^2}\right)=nx^{n-1}\phi-yx^{n-2}\phi'\)

\(\Rightarrow x\frac{\partial u}{\partial x}=nx^n\phi-yx^{n-1}\phi'\) ...(1)

\(\frac{\partial u}{\partial y}=x^n\phi'\left(\frac yx\right)\cdot\frac1x=x^{n-1}\phi'\Rightarrow y\frac{\partial u}{\partial y}=yx^{n-1}\phi'\) ...(2)

Adding (1) and (2): \(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nx^n\phi\left(\frac yx\right)=nu\).

Hence \(x u_x+y u_y=nu\) (proved). Example: \(u=x^2+xy\) (n=2): \(x(2x+y)+y(x)=2(x^2+xy)=2u\).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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