$y=x \sin x \quad: x y^{\prime}=y+x \sqrt{x^{2}-y^{2}}(x \neq 0$ and $x>y$ or $x<-y)$
$y=x \sin x$
Differentiating both sides of this equation with respect to x, we get:
$y^{\prime}=\frac{d}{d x}(x \sin x)$
$\Rightarrow y^{\prime}=\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)$
$\Rightarrow y^{\prime}=\sin x+x \cos x$
Substituting the value of
in the given differential equation, we get:
L.H.S. $=x y^{\prime}=x(\sin x+x \cos x)$
$=x \sin x+x^{2} \cos x$
$=y+x^{2} \cdot \sqrt{1-\sin ^{2}} x$
$=y+x^{2} \sqrt{1-\left(\frac{y}{x}\right)^{2}}$
$=y+x \sqrt{y^{2}-x^{2}}$
$=$ R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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