If $u= \arcsin\left(\dfrac{x+y}{\sqrt x+\sqrt y}\right)$ then show that :
$$x^2\frac{\partial^2 u}{\partial x^2}+ y^2\frac{\partial^2 u}{\partial y^2}+ 2xy\frac{\partial^2 u}{\partial x \partial y}=-\dfrac{\sin u\cos 2u}{4\cos^3 u}$$
I tried using euler's formula : Where f=sinu $$x^2\frac{\partial^2 f}{\partial x^2}+y^2\frac{\partial^2 f}{\partial x^2}+2xy\frac{\partial^2 f}{\partial x \partial y}=n(n-1) f$$ where n is the degree of homogenous function f which comes out to be 1/2.
Now I get: $\dfrac{\partial^2 f}{\partial x^2}=-\sin u\dfrac{\partial u}{\partial x}+\cos u\dfrac{\partial^2 u}{\partial x^2}$
$\dfrac{\partial^2 f}{\partial y^2}=-\sin u\dfrac{\partial u}{\partial y}+\cos u\dfrac{\partial^2 u}{\partial y^2}$
$\dfrac{\partial^2 f}{\partial x \partial y}=-\sin u\cos u\dfrac{\partial u}{\partial x}$$\dfrac {\partial u}{\partial y}+\cos u\dfrac{\partial u}{\partial x \partial y}$
Substituting this in my euler formula is not giving me answer..... Any help please.
Since $$ \frac{{\partial f}} {{\partial x}} = \left[ {\cos u} \right]\frac{{\partial u}} {{\partial x}} $$ you have that $$ \frac{{\partial ^2 f}} {{\partial x^2 }} = \left[ { - \sin \left( u \right)\frac{{\partial u}} {{\partial x}}} \right]\frac{{\partial u}} {{\partial x}} + \cos \left( u \right)\frac{{\partial ^2 u}} {{\partial x^2 }} $$ so that $$ \frac{{\partial ^2 f}} {{\partial x^2 }} = - \sin \left( u \right)\left( {\frac{{\partial u}} {{\partial x}}} \right)^2 + \cos \left( u \right)\frac{{\partial ^2 u}} {{\partial x^2 }} $$ and so on...