Verify that the given function $y = x \sin x$ is a solution of the differential equation $x y^{\prime} = y + x \sqrt{x^2 - y^2}$ (where $x \neq 0$ and $x > y$ or $x < -y$).

(N/A) Given function: $y = x \sin x$
Differentiating both sides with respect to $x$ using the product rule:
$y^{\prime} = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x)$
$y^{\prime} = \sin x + x \cos x$
Now,consider the $L.H.S.$ of the differential equation:
$L.H.S. = x y^{\prime} = x(\sin x + x \cos x) = x \sin x + x^2 \cos x$
Substitute $y = x \sin x$ into the $R.H.S.$:
$R.H.S. = y + x \sqrt{x^2 - y^2}$
$= x \sin x + x \sqrt{x^2 - (x \sin x)^2}$
$= x \sin x + x \sqrt{x^2(1 - \sin^2 x)}$
$= x \sin x + x \sqrt{x^2 \cos^2 x}$
$= x \sin x + x(x \cos x)$
$= x \sin x + x^2 \cos x$
Since $L.H.S. = R.H.S.$,the given function is a solution of the differential equation.