How do I apply Euler's theorem when f(u), not u, is homogeneous, like u = sin⁻¹((x+y)/(√x+√y))?

In one line

If \(f(u)\) is homogeneous of degree \(n\), then \(xu_x+yu_y=n\dfrac{f(u)}{f'(u)}=G(u)\) and \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=G(u)[G'(u)-1]\).

Explained simply

Sometimes the function is wrapped in a coat, like \(\sin^{-1}\), \(\log\) or \(\tan^{-1}\), and the coat spoils homogeneity. Take the coat off: \(\sin u\) or \(e^u\) or \(\tan u\) may be homogeneous. Apply Euler's theorem to that, then use the chain rule to come back to \(u\).

Technically, let \(z=f(u)\) be homogeneous of degree \(n\). Euler gives \(xz_x+yz_y=nz\). Since \(z_x=f'(u)u_x\) and \(z_y=f'(u)u_y\), we get \(f'(u)(xu_x+yu_y)=nf(u)\), so \(xu_x+yu_y=n\dfrac{f(u)}{f'(u)}=G(u)\). Differentiating again gives the second deduction (stated without proof in the syllabus).

Common choices of \(f\): for \(u=\sin^{-1}(\cdot)\) take \(f(u)=\sin u\), so \(G=n\tan u\); for \(u=\tan^{-1}(\cdot)\) take \(\tan u\), so \(G=n\sin u\cos u=\frac n2\sin2u\); for \(u=\log(\cdot)\) take \(e^u\), so \(G=n\); for \(u=\cos^{-1}(\cdot)\) take \(\cos u\), so \(G=-n\cot u\). Knowing these four saves time.

How it works

1. Identify the outer function and form \(z=f(u)\) (e.g. \(\sin u\), \(\tan u\), \(e^u\)) that is homogeneous.
2. Find its degree \(n\) by the \(t\)-test.
3. Compute \(G(u)=n\dfrac{f(u)}{f'(u)}\).
4. For the second-order result compute \(G'(u)\) and then \(G(u)[G'(u)-1]\).
5. Simplify using trig identities.

Key formula

\(xu_x+yu_y=G(u)=n\dfrac{f(u)}{f'(u)}\); \(\ x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=G(u)\left[G'(u)-1\right]\).

Quick example

\(u=\sin^{-1}\dfrac{x+y}{\sqrt x+\sqrt y}\). Then \(\sin u=\dfrac{x+y}{\sqrt x+\sqrt y}\) has degree \(1-\tfrac12=\tfrac12\).

\(G(u)=\dfrac12\cdot\dfrac{\sin u}{\cos u}=\dfrac12\tan u\). So \(xu_x+yu_y=\dfrac12\tan u\).

\(G'(u)=\dfrac12\sec^2u\). Second order: \(\dfrac12\tan u\left(\dfrac12\sec^2u-1\right)=\dfrac{\tan u(\sec^2u-2)}{4}=\dfrac{\sin u(1-2\cos^2u)}{4\cos^3u}=-\dfrac{\sin u\cos2u}{4\cos^3u}\).

Common mistakes

• Applying \(xu_x+yu_y=nu\) directly to \(u=\sin^{-1}(\dots)\), which is not homogeneous.
• Computing \(G'(u)\) as derivative with respect to \(x\); it is with respect to \(u\).
• Forgetting the "−1" in \(G(u)[G'(u)-1]\).

How MU asks it

Very common 6-mark question: "If \(u=\tan^{-1}\dfrac{x^3+y^3}{x-y}\), prove \(xu_x+yu_y=\sin2u\)" (here \(\tan u\) has degree 2, so \(G=2\tan u\cos^2u=\sin2u\)), and then "hence prove \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=2\cos3u\sin u\)". Write the homogeneous \(f(u)\), its degree with the \(t\)-test, \(G(u)\), \(G'(u)\) and simplify step by step.