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If $u=\log\begin{pmatrix}\large\frac{x^{2}+y^{2}}{xy}\end{pmatrix}$ then $x\large\frac{\partial u}{\partial x}+$$y\large\frac{\partial u}{\partial y}$ is

\[\begin{array}{1 1}(1)0&(2)u\\(3)2u&(4)u^{-1}\end{array}\]

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1 Answer

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$u= \log \bigg( \large\frac{x^2+y^2} {xy}\bigg)$
$e^u=-e^{\log} \bigg( \large\frac{x^2+y^2} {xy}\bigg)$
$e= \large\frac{x^2+y^2} {xy} $
$e^u $ is a homogeneous function of degree 0.
by Eulers theorem,
$x.\large\frac{\partial}{\partial x} (e^u)+ y . \large\frac{\partial }{\partial y} (e^u)=0.e^u$
$x.e^u \large\frac{\partial u}{\partial x} + y .e^u.\large\frac{\partial u}{\partial y} =0$
$e^u \bigg( x \large\frac{\partial u}{\partial x} + y .e^u.\large\frac{\partial u}{\partial y} \bigg)=0$
$ x \large\frac{\partial u}{\partial x} + \large\frac{\partial u}{\partial y} =0$
Hence 1 is the correct answer.
answered May 20, 2014 by meena.p
 

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