The Solution of differential equation $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{1 + {x^2}}}$ is

Given that, $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{1 + {x^2}}}$

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

On integrating both sides,

we get${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + C$

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

$\Rightarrow$ ${\tan ^{ - 1}}\left( {\frac{{y - x}}{{1 + xy}}} \right) = C$

$\Rightarrow$ $\frac{{y - x}}{{1 + xy}} = \tan C$

$\Rightarrow$ $y - x = \tan c(1 + xy)$

$\Rightarrow$ $y - x = K(1 + xy)$

where, $k = \tan C$