If the solve the problem


If $y=\sin (\log x)$, prove that $x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0$


Formula: -

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$

(ii) $\frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}=\frac{1}{\mathrm{x}}$

(iii) $\frac{d}{d x} \cos x=\sin x$

(iv) $\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=-\cos \mathrm{x}$

(v) $\frac{d}{d x} x^{n}=n x^{n-1}$

(vi) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wou })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

Given: -

$y=\sin (\log x)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\cos (\log \mathrm{x}) \frac{1}{\mathrm{x}}$

$\Rightarrow x \frac{d y}{d x}=\cos (\log x)$

$\Rightarrow x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\sin (\log x) \frac{1}{x}$

$\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}=-y$

$\Rightarrow x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0$

Hence proved.

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