Question-9686

Question Number 9686 by tawakalitu last updated on 24/Dec/16

Answered by mrW last updated on 24/Dec/16

u=log (x^2 +y^2 +z^2 ) (du/dy)=((2y)/(x^2 +y^2 +z^2 ))×log e (d^2 u/(dydz))=−((2y×2z)/((x^2 +y^2 +z^2 )^2 ))×log e x(d^2 u/(dydz))=−((4xyz)/((x^2 +y^2 +z^2 )^2 ))×log e similarly: y(d^2 u/(dxdz))=−((4xyz)/((x^2 +y^2 +z^2 )^2 ))×log e z(d^2 u/(dxdy))=−((4xyz)/((x^2 +y^2 +z^2 )^2 ))×log e hence: x(d^2 u/(dydz))=y(d^2 u/(dxdz))=z(d^2 u/(dxdy))

$$\mathrm{u}=\mathrm{log}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{du}}{\mathrm{dy}}=\frac{\mathrm{2y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} }×\mathrm{log}\:\mathrm{e} \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{u}}{\mathrm{dydz}}=−\frac{\mathrm{2y}×\mathrm{2z}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \right)^{\mathrm{2}} }×\mathrm{log}\:\mathrm{e} \\ $$$$\mathrm{x}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{u}}{\mathrm{dydz}}=−\frac{\mathrm{4xyz}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \right)^{\mathrm{2}} }×\mathrm{log}\:\mathrm{e} \\ $$$$\mathrm{similarly}: \\ $$$$\mathrm{y}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{u}}{\mathrm{dxdz}}=−\frac{\mathrm{4xyz}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \right)^{\mathrm{2}} }×\mathrm{log}\:\mathrm{e} \\ $$$$\mathrm{z}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{u}}{\mathrm{dxdy}}=−\frac{\mathrm{4xyz}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \right)^{\mathrm{2}} }×\mathrm{log}\:\mathrm{e} \\ $$$$\:\mathrm{hence}: \\ $$$$\mathrm{x}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{u}}{\mathrm{dydz}}=\mathrm{y}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{u}}{\mathrm{dxdz}}=\mathrm{z}\frac{\mathrm{d}^{\mathrm{2}} \mathrm{u}}{\mathrm{dxdy}} \\ $$

Commented by tawakalitu last updated on 24/Dec/16

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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