Question

$\rm If \: u=sin^{-1} \: \dfrac{x}{y} , \: prove \: that \\ \rm x \frac{ \partial u}{ \partial x} + y\frac{ \partial u}{ \partial y} = 0$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: u=sin^{-1} \: \dfrac{x}{y}$

can be rewritten as

$\rm \: sinu= \: \dfrac{x}{y}$

can be further rewritten as

$\rm \: sinu = {x}^{0} \times \dfrac{1}{ \: \: \: \: \dfrac{y}{x} \: \: \: \: }$

$\rm\implies \:sinu \: is \: a \: homogenous \: function \: of \: degree \: 0$

We know, Euler's Theorem,

This theorem states that if u is a homogeneous function of x and y of degree n, then

$\boxed{ \rm{ \: x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu \: }}$

So, using this, we get

$\rm \: \: x\frac{\partial }{\partial x} sinu+ y\frac{\partial }{\partial y}sinu = 0 \times sinu\:$

$\rm \: \:x(cosu )\frac{\partial u}{\partial x}+ y( cosu)\frac{\partial u}{\partial y} = 0 \:$

$\rm \: \:cosu \bigg(x\frac{\partial u}{\partial x}+ y\frac{\partial u}{\partial y}\bigg) = 0 \:$

$\rm\implies \:x\frac{\partial u}{\partial x}+ y\frac{\partial u}{\partial y} = 0 \:$

Hence, Proved

$\rule{190pt}{2pt}$

Additional Information :-

If u is a homogeneous function of x and y of degree n, then

$\boxed{ \rm{ \: x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = nu \: }}$

$\boxed{ \rm{ \: x\frac{\partial^{2} u}{\partial {x}^{2} } + y\frac{\partial^2u}{\partial x\partial y} = (n - 1)\frac{\partial u}{\partial x}\: }}$

$\boxed{ \rm{ \: y\frac{\partial^{2} u}{\partial {y}^{2} } + x\frac{\partial^2u}{\partial x\partial y} = (n - 1)\frac{\partial u}{\partial y}\: }}$

$\boxed{ \rm{ \: {x}^{2} \frac{ {\partial}^{2} u}{\partial {x}^{2} } + {y}^{2} \frac{\partial^{2} u}{\partial {y}^{2} } + 2xy\frac{\partial^2u}{\partial x\partial y} = n(n - 1)u\: }}$