If  y = 3 cos ( log x ) + 4 sin ( log x ), show that

${\mathrm{x}}^2 \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} + \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} + \mathrm{y} = 0$

It is given that,  y = 3 cos ( log x ) + 4 sin ( log  x )

Then,

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}} = 3 \times \frac{\mathrm{d}}{\mathrm{dx}} \left[ \cos \log \mathrm{x} \right] + 4 \times \frac{{\displaystyle \mathrm{d}}}{{\displaystyle \mathrm{dx}}} \left[ \sin \log \mathrm{x} \right] \\ = 3 \times \left[ - \sin \log \mathrm{x} \times \frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x} \right] + 4 \times \left[ \cos \log \mathrm{x} \times \frac{{\displaystyle \mathrm{d}}}{{\displaystyle \mathrm{dx}}} \log \mathrm{x} \right] \\ = \frac{ - 3 \sin \log \mathrm{x}}{\mathrm{x}} + \frac{4 \cos \log \mathrm{x}}{\mathrm{x}} = \frac{4 \cos \log \mathrm{x} - 3 \sin \log \mathrm{x}}{\mathrm{x}} \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} = \frac{\mathrm{d}}{\mathrm{dx}} \left( \frac{4 \cos ( \log \mathrm{x} ) - 3 \sin ( \log \mathrm{x} )}{\mathrm{x}}\right) \end{gathered}$$

$$\begin{gathered} = \frac{\mathrm{x} \left\{ 4 \cos ( \log \mathrm{x} ) - 3 \sin ( \log \mathrm{x} ) \right\}\text{'} - \left\{ 4 \cos ( \log \mathrm{x} ) - 3 \sin ( \log \mathrm{x} ) \right\} \left( \mathrm{x} \right)\text{'}}{{\mathrm{x}}^2} \\ = \frac{\mathrm{x} \left[ - 4 \sin ( \log \mathrm{x} ) \times ( \log \mathrm{x} )\text{'} - 3 \cos ( \log \mathrm{x} ) \times ( \log \mathrm{x} )\text{'} \right] - 4 \cos ( \log \mathrm{x} ) + 3 \sin ( \log \mathrm{x} ) }{{\mathrm{x}}^2} \\ = \frac{- 4 \sin ( \log \mathrm{x} ) - 3 \cos ( \log \mathrm{x} ) - 4 \cos ( \log \mathrm{x} ) + 3 \sin ( \log \mathrm{x} ) }{{\mathrm{x}}^2} \\ = \frac{- \sin ( \log \mathrm{x} ) - 7 \cos ( \log \mathrm{x} )}{{\mathrm{x}}^2} \\ \therefore {\mathrm{x}}^2 \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2} + \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} + \mathrm{y} \end{gathered}$$

$$\begin{gathered} = {\mathrm{x}}^{2 } \left( \frac{ - \sin ( \log \mathrm{x} ) - 7 \cos ( \log \mathrm{x} ) }{{\mathrm{x}}^2}\right) + \mathrm{x} \left( \frac{ 4 \cos ( \log \mathrm{x} ) - 3 \sin ( \log \mathrm{x} ) }{\mathrm{x}}\right) + 3 \cos ( \log \mathrm{x} ) + 4 \sin ( \log \mathrm{x} ) \\ = - \sin ( \log \mathrm{x} ) - 7 \cos ( \log \mathrm{x} ) + 4 \cos ( \log \mathrm{x} ) - 3 \sin ( \log \mathrm{x} ) + 3 \cos ( \log \mathrm{x} ) + 4 \sin ( \log \mathrm{x} ) \\ = 0 \end{gathered}$$

Hence proved.