Verify that the given function y = x sin x is | vedclass

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(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 +

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(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.

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 $x < -y$).

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