If u = x y^2 tan^{-1}left(frac{y}{x}right),th | vedclass

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(C) The given function is $u(x, y) = x y^2 	an^{-1}\left(\frac{y}{x}ight)$.This is a homogeneous function of degree $n = 1 + 2 = 3$ because $u(tx,

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$3 u$

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(C) The given function is $u(x, y) = x y^2 	an^{-1}\left(\frac{y}{x}ight)$.This is a homogeneous function of degree $n = 1 + 2 = 3$ because $u(tx, ty) = (tx)(ty)^2 	an^{-1}\left(\frac{ty}{tx}ight) = t^3 x y^2 	an^{-1}\left(\frac{y}{x}ight) = t^3 u(x, y)$.According to Euler's Theorem on homogeneous functions,if $u$ is a homogeneous function of degree $n$ in $x$ and $y$,then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n u$.Here,$n = 3$.Therefore,$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 3 u$.

If $u = x y^2 \tan^{-1}\left(\frac{y}{x}\right)$,then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$ is equal to

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