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Question-121361




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
$${u}={x}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)−{y}^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\frac{{x}}{{y}}\right) \\ $$$$={x}^{\mathrm{3}} \left\{{sin}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)−\left(\frac{{y}}{{x}}\right)^{\mathrm{3}} {sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\left(\frac{{y}}{{x}}\right)}\right)\right\} \\ $$$${so}\:{u}\:{is}\:{homogeneous}\:{function}\:{of}\:{degree}\:\mathrm{3} \\ $$$${so}\:{using}\:{Euler}\:{theorem}\:{on}\:{homogeneous} \\ $$$${function} \\ $$$${x}^{\mathrm{2}} {u}_{{xx}} +\mathrm{2}{xyu}_{{xy}} +{y}^{\mathrm{2}} {u}_{{yy}} ={n}\left({n}−\mathrm{1}\right){u} \\ $$$${here}\:{n}=\mathrm{3} \\ $$$${x}^{\mathrm{2}} {u}_{{xx}} +\mathrm{2}{xyu}_{{xy}\:} +{y}^{\mathrm{2}} {u}_{{yy}} =\mathrm{3}\left(\mathrm{3}−\mathrm{1}\right){u}=\mathrm{6}{u} \\ $$

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