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

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Question

(C) Given the function $u(x, y) = x{y^2}{	an ^{ - 1}}\left( {\frac{y}{x}} ight)$.To check for homogeneity,we replace $x$ with $tx$ and $y$ with $ty

Vedclass · Accepted Answer

$3u$

Vedclass · Answer

(C) Given the function $u(x, y) = x{y^2}{	an ^{ - 1}}\left( {\frac{y}{x}} ight)$.To check for homogeneity,we replace $x$ with $tx$ and $y$ with $ty$:$u(tx, ty) = (tx)(ty)^2 	an^{-1}\left(\frac{ty}{tx}ight)$$u(tx, ty) = t^3 x y^2 	an^{-1}\left(\frac{y}{x}ight)$$u(tx, ty) = t^3 u(x, y)$.Since the function is homogeneous of degree $n = 3$,by Euler's Theorem on homogeneous functions,we have:$x{u_x} + y{u_y} = n u$$x{u_x} + y{u_y} = 3u$.

If $u = x{y^2}{\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$,then $x{u_x} + y{u_y} = $

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