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If $u \; = \; tan^{-1} (\frac{y}{x}), \; find \; \frac{\partial^2 u}{\partial x^2} \; + \frac{\partial^2 u}{\partial y^2}$
partial differentiation • 582  views
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u$ \; = \; tan^{-1} (\frac{y}{x}) \\ \; \\ Differentiating \; partially \; w. r. t. x, \\ \frac{\partial u}{\partial x} \; = \; \frac{1}{1 + \frac{y^2}{x^2}} \bigg( \frac{-y}{x^2} \bigg) \; = \; \frac{-y}{x^2 \; + \; y^2} \\ \; \\ Again \; differentiating \; partially \; w. r. t. \; x, \\ \frac{\partial^2 u}{\partial x^2} \; = \; \frac{y}{(x^2 \; + \; y^2)^2} (2x) \; = \; \frac{2xy}{(x^2 \; + \; y^2)^2} \; \; \; \; \ldots (i)\\ \; \\ Now \; differentiating \; u \; partially \; w. r. t. \; y, \\ \frac{\partial u}{\partial y} \; = \; \frac{1}{1 + \frac{y^2}{x^2}} \bigg( \frac{1}{x} \bigg) \; = \; \frac{x}{x^2 \; + \; y^2} \\ \; \\ Again \; differentiating \; partially \; w. r. t. \; y, \\ \frac{\partial^2 u}{\partial x^2} \; = \; \frac{-2xy}{(x^2 \; + \; y^2)^2} \; \; \; \; \ldots (ii)\\ Adding \; (i) \; and \; (ii), \; \\ \; \\ \therefore \frac{\partial^2 u}{\partial x^2} \; + \; \frac{\partial^2 u}{\partial y^2} \; = \; \frac{2xy}{(x^2 \; + \; y^2)^2} \; - \; \frac{2xy}{(x^2 \; + \; y^2)^2} \; = \; 0 $

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