In one line
If \(u\) is a homogeneous function of degree \(n\) in \(x\) and \(y\), then \(x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=nu\).
Explained simply
Go back to the rectangle: area \(A=xy\) has degree 2. Euler's theorem says \(xA_x+yA_y=x(y)+y(x)=2xy=2A\). In words: if you weight each partial derivative by its own variable and add, you get the function back multiplied by its degree. It is a neat shortcut that saves a lot of differentiation in exam problems.
Statement: if \(u=f(x,y)\) is homogeneous of degree \(n\) and has continuous first partial derivatives, then \(xu_x+yu_y=nu\). The syllabus asks for this theorem with proof.
The key idea of the proof is that any homogeneous function of degree \(n\) can be written as \(x^n\) times a function of the single ratio \(y/x\). Once you write it that way, the two partial derivatives contain matching terms that cancel, leaving only \(n\) times the original function. Remember this idea and the proof will follow naturally in the exam.
How it works
Proof.
- Since \(u\) is homogeneous of degree \(n\), write \(u=x^n\phi(v)\) where \(v=\dfrac yx\).
- Note \(v_x=-\dfrac{y}{x^2}\) and \(v_y=\dfrac1x\).
- \(u_x=nx^{n-1}\phi(v)+x^n\phi'(v)\left(-\dfrac y{x^2}\right)=nx^{n-1}\phi(v)-yx^{n-2}\phi'(v)\).
- \(u_y=x^n\phi'(v)\cdot\dfrac1x=x^{n-1}\phi'(v)\).
- \(xu_x+yu_y=nx^n\phi(v)-yx^{n-1}\phi'(v)+yx^{n-1}\phi'(v)=nx^n\phi(v)=nu\). Hence proved.
Key formula
\(x\dfrac{\partial u}{\partial x}+y\dfrac{\partial u}{\partial y}=nu\), where \(n\) is the degree of homogeneity of \(u(x,y)\).
Quick example
Verify for \(u=x^3+y^3\) (degree 3): \(u_x=3x^2\), \(u_y=3y^2\). Then \(xu_x+yu_y=3x^3+3y^3=3u\). Verified.
Use as a shortcut: if \(u=\dfrac{x^2y^2}{x+y}\) (degree 4 − 1 = 3), then without differentiating, \(xu_x+yu_y=3u=\dfrac{3x^2y^2}{x+y}\).
Degree-zero case: \(u=\tan^{-1}\dfrac yx\) gives \(xu_x+yu_y=0\). Check: \(u_x=\dfrac{-y}{x^2+y^2}\), \(u_y=\dfrac{x}{x^2+y^2}\), so \(-\dfrac{xy}{x^2+y^2}+\dfrac{xy}{x^2+y^2}=0\).
Common mistakes
- Applying the theorem to a non-homogeneous function such as \(x^2+y\).
- In the proof, forgetting the chain-rule factor \(v_x=-y/x^2\).
- Using the wrong degree for a quotient: degree of numerator minus degree of denominator.
How MU asks it
"State and prove Euler's theorem for a homogeneous function of two variables" is a classic 5/6-mark question (proof is in the syllabus). Write the statement, the form \(u=x^n\phi(y/x)\), both derivatives in full, and the cancellation. "Verify Euler's theorem for \(u=\dfrac{x^{1/4}+y^{1/4}}{x^{1/5}+y^{1/5}}\)" (degree \(\frac14-\frac15=\frac1{20}\)) is another common form: compute both sides and show they are equal.