Question

$\rm{If \: f=a_0x^n+a_1x^{n-1}y\dots+a_{n-1}xy^{n-1}+a_ny^n} \\ \rm then \: x \frac{ \partial f}{ \partial x} + \frac{ \partial f}{ \partial y} \: is$​

Answer 1

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

Given that,

$\rm \: f=a_0x^n+a_1x^{n-1}y\dots+a_{n-1}xy^{n-1}+a_ny^n$

can be rewritten as

$\rm \: f= {x}^{n}\bigg( a_0+a_1 \frac{y}{x} \dots+a_{n-1}\dfrac{ {y}^{n - 1} }{ {x}^{n - 1} } +a_n \frac{ {y}^{n} }{ {x}^{n} }\bigg)$

$\rm\implies \:\rm \: f \:is \: a \: homogenous \: function \: of \: degree \: n$

We know, Euler's Theorem :- This theorem states that if u is a homogeneous function of x and y of degree n then

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

So, by using Euler's theorem, we have

$\rm\implies \:\rm \: x\dfrac{\partial f}{\partial x} \: + \: y\dfrac{\partial f}{\partial y} \: = \: nf$

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

Additional Information :-

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

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

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

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

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