Question

If z = log [(x ^ 2 - y ^ 2) / (x ^ 2 + y ^ 2)] then . x (partial x /partial x )+y (partial z/ partial y) =0​

Answer 1

Appropriate Question :-

$\rm \: If \: z = log\bigg[\dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} } \bigg], \: prove \: that \: x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} = 0$

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

Given function is

$\rm \: z = log\bigg[\dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} } \bigg]$

can be rewritten as

$\rm \: {e}^{z} = \dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} }$

$\rm \: {e}^{z} = \dfrac{ {x}^{2}\bigg(1 - \dfrac{ {y}^{2} }{ {x}^{2} } \bigg) }{{x}^{2}\bigg(1 + \dfrac{ {y}^{2} }{ {x}^{2} } \bigg)}$

$\rm \: {e}^{z} = \dfrac{1 - \dfrac{ {y}^{2} }{ {x}^{2} }}{1 + \dfrac{ {y}^{2} }{ {x}^{2} }}$

$\rm\implies \:{e}^{z} \: is \: a \: homogenous \: function \: of \: degree \: 0$

We know, Euler's Theorem

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

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

So, using Euler's theorem, we have

$\rm \: x\dfrac{\partial }{\partial x}{e}^{z} + y\dfrac{\partial }{\partial y} {e}^{z}= 0 \times {e}^{z}$

$\rm \: x{e}^{z}\dfrac{\partial z}{\partial x} + y{e}^{z}\dfrac{\partial z}{\partial y} = 0$

$\rm \: {e}^{z}\bigg(x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} \bigg)= 0$

$\bf\implies \:x\dfrac{\partial z}{\partial x} + y\dfrac{\partial z}{\partial y} = 0$

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

Additional Information :-

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

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

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

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

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

Answer 2

Answer:

Given function is

\begin{gathered}\rm \: z = log\bigg[\dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} } \bigg] \\ \end{gathered}

z=log[

x

2

+y

2

x

2

−y

2

]

can be rewritten as

\begin{gathered}\rm \: {e}^{z} = \dfrac{ {x}^{2} - {y}^{2} }{ {x}^{2} + {y}^{2} } \\ \end{gathered}