Verify that the given function $y = \sqrt{a^{2} - x^{2}}$,where $x \in (-a, a)$,is a solution of the differential equation $x + y \frac{dy}{dx} = 0$ (where $y \neq 0$).

(A) Given function: $y = \sqrt{a^{2} - x^{2}}$
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{a^{2} - x^{2}})$
Using the chain rule:
$\frac{dy}{dx} = \frac{1}{2\sqrt{a^{2} - x^{2}}} \cdot \frac{d}{dx}(a^{2} - x^{2})$
$\frac{dy}{dx} = \frac{1}{2\sqrt{a^{2} - x^{2}}} \cdot (-2x)$
$\frac{dy}{dx} = \frac{-x}{\sqrt{a^{2} - x^{2}}}$
Now,substitute the value of $\frac{dy}{dx}$ into the given differential equation $x + y \frac{dy}{dx} = 0$:
$L.H.S = x + y \left( \frac{-x}{\sqrt{a^{2} - x^{2}}} \right)$
Since $y = \sqrt{a^{2} - x^{2}}$,we substitute this into the expression:
$L.H.S = x + \sqrt{a^{2} - x^{2}} \cdot \left( \frac{-x}{\sqrt{a^{2} - x^{2}}} \right)$
$L.H.S = x - x = 0$
$L.H.S = R.H.S$
Hence,the given function is a solution of the differential equation.