Question
$\text{If y = 3} \cos (\log\text{x}) + 4\sin (\log \text{x}), \text{then show that x}^{2} .\frac{\text{d}^{2}{\text{y}}}{\text{dx}^{2}} + \text{y} = 0$

Answer

$\text{y} = 3\cos(\log\text{x}) + 4\sin (\log\text{x}) \Rightarrow \frac{\text{dy}}{\text{dx}} = - \frac{3\sin(\log\text{x})}{\text{x}} +\frac{4\cos(\log\text{x})}{\text{x}}$$\text{x} \frac{\text{dy}}{\text{dx}} = -3\sin (\log\text{x}) + 4\cos(\log\text{x})$
$\Rightarrow \text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} = - \frac{3\cos(\log\text{x})}{\text{x}} -\frac{4\sin(\log\text{x})}{\text{x}}$
$\Rightarrow \text{x} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} + \text{x} \frac{\text{dy}}{\text{dx}} = -[3\cos (\log\text{x}) + 4\sin (\log{\text{x}})] = \text{-y}$
$\text{or x}^{2} \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} +\text{y = o} $

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