The solution of the differential equation $\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$ is :

Question

The solution of the differential equation

$\left( {1 + {y^2}} \right) + \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right){{dy} \over {dx}} = 0,$ is :

$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$

$\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}$

$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$

$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$

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