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If $ u \;=\; f \bigg( \dfrac{x-y}{xy} , \dfrac{z-x}{xz} \bigg) $ prove the following

prove that

$ x^2 \dfrac{\partial u}{\partial x} + y^2 \dfrac{\partial u}{\partial y} + z^2 \dfrac{\partial u}{\partial z} \;=\; 0 $

partial differentiation • 680  views
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Let, $ l = \dfrac{x-y}{xy} \; and \; m = \dfrac{z-x}{xz} \\ \; \\ \; \\ \therefore u=f(l,m) \; where \; l \; \& \; m \; are \; functions \; of \; x \& y \; , \; x \& z \; respectively \\ \; \\ \; \\ \therefore \dfrac{\partial l}{\partial x} \;=\; \dfrac{xy(1)-(x-y)(y)}{(xy)^2} \;=\; \dfrac{xy-(xy-y^2)}{(x^2y^2)} \;=\; \dfrac{1}{x^2} \\ \; \\ \; \\ \dfrac{\partial l}{\partial y} \;=\; \dfrac{xy(-1)-(x-y)(x)}{(xy)^2} \;=\; \dfrac{-xy-(x^2-xy)}{(x^2y^2)} \;=\; \dfrac{-1}{y^2} \\ \; \\ \; \\ m \;=\; \dfrac{z-x}{xz} \; \; \therefore \dfrac{\partial m}{\partial x} \;=\; \dfrac{xz(-1)-(z-x)(z)}{(xz)^2} \;=\; \dfrac{-xz-(z^2-xz)}{(x^2z^2)} \;=\; \dfrac{-1}{x^2} $

Similarly, $ \dfrac{\partial m}{\partial z} \;=\; \dfrac{xz(1)-(z-x)(x)}{(xz)^2} \;=\; \dfrac{xz-(xz-x^2)}{(x^2z^2)} \;=\; \dfrac{1}{z^2} $

$ \dfrac{\partial m}{\partial y} \;=\; 0 $

Now, by Chain Rule,

$ \dfrac{\partial u}{\partial x} \;=\; \dfrac{\partial u}{\partial l} . \dfrac{\partial l}{\partial x} + \dfrac{\partial u}{\partial m} . \dfrac{\partial m}{\partial x} \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial x} \;=\; \dfrac{\partial u}{\partial l} . \dfrac{1}{x^2} + \dfrac{\partial u}{\partial m} . \dfrac{-1}{x^2} \\ \; \\ \; \\ \therefore x^2 \dfrac{\partial u}{\partial x} \;=\; \dfrac{\partial u}{\partial l} - \dfrac{\partial u}{\partial m} \; \; \; \; \ldots (i) \\ \; \\ \; \\ \dfrac{\partial u}{\partial y} \;=\; \dfrac{\partial u}{\partial l} . \dfrac{\partial l}{\partial y} + \dfrac{\partial u}{\partial m} . \dfrac{\partial m}{\partial y} \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial y} \;=\; \dfrac{\partial u}{\partial l} . \dfrac{-1}{y^2} + \dfrac{\partial u}{\partial m} . (0) \\ \; \\ \; \\ \therefore y^2 \dfrac{\partial u}{\partial y} \;=\; \dfrac{-\partial u}{\partial l} \; \; \; \; \ldots (ii) \\ \; \\ \; \\ \; \\ \dfrac{\partial u}{\partial z} \;=\; \dfrac{\partial u}{\partial l} . \dfrac{\partial l}{\partial z} + \dfrac{\partial u}{\partial m} . \dfrac{\partial m}{\partial z} \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial z} \;=\; \dfrac{\partial u}{\partial l} (0) + \dfrac{\partial u}{\partial m} . \dfrac{1}{z^2} \\ \; \\ \; \\ \therefore z^2 \dfrac{\partial u}{\partial z} \;=\; \dfrac{\partial u}{\partial m} \; \; \; \; \ldots (iii) $

Adding (i) , (ii) , (iii) , we get

$ x^2 \dfrac{\partial u}{\partial x} + y^2 \dfrac{\partial u}{\partial y} + z^2 \dfrac{\partial u}{\partial z} \;=\; \dfrac{\partial u}{\partial l} - \dfrac{\partial u}{\partial m} - \dfrac{\partial u}{\partial l} + \dfrac{\partial u}{\partial m} \;=\; 0 $

Hence Proved.

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