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.