$y = \sqrt {{a^2} - {x^2}} ,x \in ( - a,a):x + y\cfrac{{dy}}{{dx}} = 0(y \ne 0)$

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

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

we get
$y' = \cfrac{{1 \times ( - 2x)}}{{2\sqrt {{a^2} - {x^2}} }}\; \Rightarrow y' = \cfrac{{ - x}}{{\sqrt {{a^2} - {x^2}} }}$

$\Rightarrow y' = \cfrac{{ - x}}{y}$

(using (i))
$\Rightarrow yy' = - x \Rightarrow x + yy' = 0$

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