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Question Number 121361 by rs4089 last updated on 07/Nov/20

	Answered by TANMAY PANACEA last updated on 07/Nov/20
	$u=x^3 sin^(−1) ((y/x))−y^3 sin^(−1) ((x/y)) =x^3 {sin^(−1) ((y/x))−((y/x))^3 sin^(−1) ((1/(((y/x)))))} so u is homogeneous function of degree 3 so using Euler theorem on homogeneous function x^2 u_(xx) +2xyu_(xy) +y^2 u_(yy) =n(n−1)u here n=3 x^2 u_(xx) +2xyu_(xy ) +y^2 u_(yy) =3(3−1)u=6u$
	$u = x^{3} {s i n}^{- 1} (\frac{y}{x}) - y^{3} {s i n}^{- 1} (\frac{x}{y})$ $= x^{3} {{s i n}^{- 1} (\frac{y}{x}) - {(\frac{y}{x})}^{3} {s i n}^{- 1} (\frac{1}{(\frac{y}{x})})}$ $s o u i s h o m o g e n e o u s f u n c t i o n o f d e g r e e 3$ $s o u s i n g E u l e r t h e o r e m o n h o m o g e n e o u s$ $f u n c t i o n$ $x^{2} u_{x x} + 2 {x y u}_{x y} + y^{2} u_{y y} = n (n - 1) u$ $h e r e n = 3$ $x^{2} u_{x x} + 2 {x y u}_{x y} + y^{2} u_{y y} = 3 (3 - 1) u = 6 u$
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