If \(u=\sin^{-1}\left(\dfrac{x+y}{\sqrt x+\sqrt y}\right)\), prove that \(x^2\dfrac{\partial^2u}{\partial x^2}+2xy\dfrac{\partial^2u}{\partial x\partial y}+y^2\dfrac{\partial^2u}{\partial y^2}=-\dfrac{\sin u\cos2u}{4\cos^3u}\). (asked in MU Dec 2024…

Answer: \(\sin u\) is homogeneous of degree \(\frac12\); with \(g(u)=\frac12\tan u\), \(g(g'-1)=-\dfrac{\sin u\cos2u}{4\cos^3u}\).

Let \(f=\sin u=\dfrac{x+y}{\sqrt x+\sqrt y}\). Degree \(n=1-\frac12=\frac12\).

Step 1: By Euler's theorem \(x f_x+y f_y=\frac12 f\Rightarrow \cos u\,(x u_x+y u_y)=\frac12\sin u\)

\[x u_x+y u_y=\frac12\tan u=g(u).\]

Step 2: \(x^2u_{xx}+2xy\,u_{xy}+y^2u_{yy}=g(u)[g'(u)-1]\), with \(g'(u)=\frac12\sec^2u\).

\[=\frac12\tan u\left(\frac{1}{2\cos^2u}-1\right)=\frac12\cdot\frac{\sin u}{\cos u}\cdot\frac{1-2\cos^2u}{2\cos^2u}=\frac{\sin u\,(1-2\cos^2u)}{4\cos^3u}.\]

Since \(1-2\cos^2u=-\cos2u\):

\[x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=-\frac{\sin u\cos2u}{4\cos^3u}.\]

Note: the Dec 2024 paper printed the last term as \(y^2\partial^2u/\partial x^2\); the intended term is \(y^2\partial^2u/\partial y^2\).