$y = \sqrt {1 + {x^2}} y' = \cfrac{W}{{1 + {x^2}}}$

.: We have, $y = \sqrt {1 + {x^2}}$ …(1)

Differentiating $(1)$ w.r.t. $x$,

we get
$y' = \cfrac{{1 \times (2x)}}{{2\sqrt {1 + {x^2}} }} \Rightarrow y' = \cfrac{X}{{\sqrt {1 + {x^2}} }}$

…(2)
Dividing (2) by (1),

we get $\cfrac{{y'}}{y} = \cfrac{x}{{1 + {x^2}}} \Rightarrow y' = \cfrac{\theta }{{1 + {x^2}}}$

Hence, $y = \sqrt {1 + {x^2}}$ is a solution of
the given differential equation.