$x + y = {\tan ^{ - 1}}y:{y^2}y' + {y^2} + 1 = 0$

.: We have, . $x + y = {\tan ^{ - 1}}y$

…(1)
Differentiating (1) w.r.t. $x$,

we get
$1 + y' = \cfrac{1}{{1 + {y^2}}}(y') \Rightarrow \;(1 + y')(1 + {y^2}) = y'$

$\Rightarrow 1 + f + y' + {y^2}y' = y' \Rightarrow 1 + y + {y^2}y' = 0$

…(2)
$\therefore$ $x + y = {\tan ^{ - 1}}y$ is a
solution of the given differential equation.