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Differential Equations question

2003 · Shift 0 · Q86

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JEE Main
Mathematics
Differential Equations
2003 · Shift 0 · Q86

Differential Equations question

2003 · Shift 0 · Q86

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

The solution of the differential equation

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

is :

A
$x{e^{2{{\tan }^{ - 1}}y}} = {e^{{{\tan }^{ - 1}}y}} + k$
B
$\left( {x - 2} \right) = k{e^{2{{\tan }^{ - 1}}y}}$
C
$2x{e^{{{\tan }^{ - 1}}y}} = {e^{2{{\tan }^{ - 1}}y}} + k$
D
$x{e^{{{\tan }^{ - 1}}y}} = {\tan ^{ - 1}}y + k$

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