Question

State and prove Euler's Theorem of partial differentiation for two variables​

Answer 1

Answer:

Hope it helps you

Step-by-step explanation:

Refer to the attachment.

Answer 2

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

Statement :- If u is a homogeneous function in x and y of degree n then $\rm \: x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = nu$

Proof :-

As u is a homogeneous function in x and y. So, let assume that

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

On differentiating partially w. r. t. x, we get

$\rm \: \dfrac{\partial u}{\partial x} = {x}^{n} \dfrac{\partial }{\partial x}f\bigg(\dfrac{y}{x} \bigg) + f\bigg(\dfrac{y}{x} \bigg)\dfrac{\partial }{\partial x} {x}^{n}$

$\rm \: \dfrac{\partial u}{\partial x} = {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \dfrac{\partial }{\partial x}\bigg(\dfrac{y}{x} \bigg) + f\bigg(\dfrac{y}{x} \bigg) \times {nx}^{n - 1}$

$\rm \: \dfrac{\partial u}{\partial x} = {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \bigg(\dfrac{ - y}{ {x}^{2} } \bigg) + {nx}^{n - 1} f\bigg(\dfrac{y}{x} \bigg)$

On multiply both sides by x, we get

$\rm \: x\dfrac{\partial u}{\partial x} = {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \bigg(\dfrac{ - y}{ {x}} \bigg) + {nx}^{n} f\bigg(\dfrac{y}{x} \bigg)$

$\rm\implies \:\rm \: x\dfrac{\partial u}{\partial x} = \: - \: {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \bigg(\dfrac{y}{ {x}} \bigg) + nu - - - (1)$

Now, Consider again

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

On differentiating partially w. r. t. y, we get

$\rm \: \dfrac{\partial u}{\partial y} = {x}^{n}\dfrac{\partial }{\partial y} f\bigg(\dfrac{y}{x} \bigg)$

$\rm \: \dfrac{\partial u}{\partial y} = {x}^{n}f'\bigg(\dfrac{y}{x} \bigg)\dfrac{\partial }{\partial y} \bigg(\dfrac{y}{x} \bigg)$

$\rm \: \dfrac{\partial u}{\partial y} = {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \bigg(\dfrac{1}{x} \bigg)$

On multiply both sides by y, we get

$\rm\implies \:\rm \: y\dfrac{\partial u}{\partial y} = {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \bigg(\dfrac{y}{x} \bigg) - - - (2)$

So, on adding equation (1) and (2), we get

$\rm \: x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} = - {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \bigg(\dfrac{y}{x} \bigg) + nu + {x}^{n}f'\bigg(\dfrac{y}{x} \bigg) \bigg(\dfrac{y}{x} \bigg)$

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

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

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{d}{dx}(u.v) = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

$\boxed{ \rm{ \:\dfrac{d}{dx} \frac{1}{ {x}^{n} } = \frac{ - n}{ {x}^{n + 1} } \: }}$