Question

$engineering \: math$
​

Answer 1

Given :-

$\rm :\longmapsto\:u \: = \: log_{e}\bigg|\dfrac{ {x}^{4} + {y}^{4} }{x + y} \bigg|$

To Prove :-

$\rm :\longmapsto\:x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = 3$

Basic Concept Used :-

Eulers theorem :-

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

$\rm :\longmapsto\:x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = nu$

Solution :-

Given that

$\rm :\longmapsto\:u \: = \: log_{e}\bigg|\dfrac{ {x}^{4} + {y}^{4} }{x + y} \bigg|$

We know that

$\green{\bf :\longmapsto\: log_{a}(b) = c \implies \: b = {a}^{c}}$

Thus,

$\rm :\longmapsto\: {e}^{u} \: = \: \dfrac{ {x}^{4} + {y}^{4} }{x + y}$

$\rm :\longmapsto\: {e}^{u} \: = \: \dfrac{\bigg(1 + \dfrac{ {y}^{4} }{{x}^{4}}\bigg) {x}^{4}}{x\bigg(1 + \dfrac{y}{x}\bigg) }$

$\rm :\longmapsto\: {e}^{u} \: = \: \dfrac{\bigg(1 + \dfrac{ {y}^{4} }{{x}^{4}}\bigg) {x}^{3}}{\bigg(1 + \dfrac{y}{x}\bigg) }$

$\rm :\longmapsto\: {e}^{u} \: = {x}^{3}f\bigg(\dfrac{y}{x} \bigg)$

$\bf\implies \: {e}^{u} \: is \: a \: homogeneous \: funtion \: of \: degree \: 3$

Therefore, By Euler's Theorem,

$\rm :\longmapsto\:x\dfrac{\partial}{\partial x} {e}^{u} + y\dfrac{\partial}{\partial y} {e}^{u} = 3{e}^{u}$

$\rm :\longmapsto\:x{e}^{u}\dfrac{\partial \: u}{\partial x} + y{e}^{u}\dfrac{\partial \: u}{\partial y}= 3 {e}^{u}$

$\bf\implies \:\:x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = 3$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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

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

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

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