Find the general solution of the differential equation

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

Given, differential equation is

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

$\Rightarrow$ $\left( {1 + {y^2}} \right) = - \left( {x - {e^{{{\tan }^{ - 1}}y}}} \right)\frac{{dy}}{{dx}}$

$\left( {1 + {y^2}} \right)\frac{{dx}}{{dy}} = - x + {e^{{{\tan }^{ - 1}}y}}$

$\Rightarrow$ $\left( {1 + {y^2}} \right)\frac{{dx}}{{dy}} + x = {e^{{{\tan }^{ - 1}}y}}$

$\Rightarrow$ $\frac{{dx}}{{dy}} + \frac{x}{{1 + {y^2}}} = \frac{{{e^{{{\tan }^{ - 1}}y}}}}{{1 + {y^2}}}$

[dividing throughout by $\left( {1 + {y^2}} \right)$]

which is a linear differential equation.
On comparing it with

$\frac{{dx}}{{dy}} + Px = Q,$

we get
$P = \frac{1}{{1 + {y^2}}},Q = \frac{{{e^{{{\tan }^{ - 1}}y}}}}{{1 + {y^2}}}$

${\rm{IF}} = {e^{\int P dy}} = {e^{\int {\frac{1}{{1 + {y^2}}}} dy}} = {e^{{{\tan }^{ - 1}}y}}$

The general solution is
$x \cdot {e^{{{\tan }^{ - 1}}y}} = \int {\frac{{{e^{{{\tan }^{ - 1}}y}}}}{{1 + {y^2}}}} \cdot {e^{{{\tan }^{ - 1}}y}}dy + C$

$\Rightarrow$ $x \cdot {e^{{{\tan }^{ - 1}}y}} = \int {\frac{{{{\left( {{e^{{{\tan }^{ - 1}}y}}} \right)}^2}}}{{1 + {y^2}}}} \cdot dy + C$

Put ${\tan ^{ - 1}}y = t \Rightarrow \frac{1}{{1 + {y^2}}}dy = dt$

$\therefore$ $x \cdot {e^{{{\tan }^{ - 1}}y}} = \int {{e^{2t}}} dt + C$

$\Rightarrow$ $x \cdot {e^{{{\tan }^{ - 1}}y}} = \frac{1}{2}{e^{2{{\tan }^{ - 1}}y}} + C$

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

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