If \(u=\sin^{-1}\left(\frac{x+y}{\sqrt x+\sqrt y}\right)\), then prove that \(x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial x^2}=-\frac{\sin u\cos2u}{4\cos^3u}\).

(Note: the last term on the left is \(y^2\frac{\partial^2u}{\partial y^2}\); it is misprinted as \(\partial x^2\) in the paper.)

Given: \(u=\sin^{-1}\left(\frac{x+y}{\sqrt x+\sqrt y}\right)\Rightarrow z=\sin u=\frac{x+y}{\sqrt x+\sqrt y}\).

Step 1 (homogeneity): \(z(tx,ty)=\frac{t(x+y)}{\sqrt t(\sqrt x+\sqrt y)}=t^{1/2}z\), so \(z=\sin u\) is homogeneous of degree \(n=\tfrac12\).

Step 2 (Euler's theorem): \(x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz\Rightarrow \cos u\left(x u_x+y u_y\right)=\tfrac12\sin u\Rightarrow x u_x+y u_y=\tfrac12\tan u=g(u)\).

Step 3 (deduction): If \(xu_x+yu_y=g(u)\) then \(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=g(u)\,[g'(u)-1]\).

Here \(g(u)=\tfrac12\tan u,\ g'(u)=\tfrac12\sec^2u\). So

\(x^2u_{xx}+2xyu_{xy}+y^2u_{yy}=\tfrac12\tan u\left(\tfrac12\sec^2u-1\right)=\frac{\tan u\,(1-2\cos^2u)}{4\cos^2u}=\frac{\sin u}{\cos u}\cdot\frac{-\cos 2u}{4\cos^2u}\)

\(\therefore\ x^2\frac{\partial^2u}{\partial x^2}+2xy\frac{\partial^2u}{\partial x\partial y}+y^2\frac{\partial^2u}{\partial y^2}=-\frac{\sin u\cos 2u}{4\cos^3u}\) (proved).