If $y = 3\cos (\log x) + 4\sin (\log x),$ then show that ${x^2}{y_2} + x{y_1} + y = 0$

We have, $y = 3\cos (\log x) + 4\sin (\log x),$ ….(i)
Differentiating (i) w.r.t. x, we get

$\cfrac{{dy}}{{dx}} = 3[ - \sin (\log x)]\cfrac{1}{x} + 4[\cos (\log x)]\cfrac{1}{x}$ …(ii)

Differentiating (ii) w.r.t. x, we get

$\cfrac{{{d^2}y}}{{{d^2}x}} = 3\left[ {\left( { - \sin (\log x)\left( { - \cfrac{1}{{{x^2}}}} \right)} \right) + \cfrac{1}{x}( - \cos (\log x))\cfrac{1}{x}} \right]$ $+ 4\left[ {\cos (\log x) \cdot \left( {\cfrac{{ - 1}}{{{x^2}}}} \right) + \cfrac{1}{x}\{ - \sin (\log x)\} \cfrac{1}{x}} \right]$

$= \cfrac{1}{{{x^2}}}[3\sin (\log x) - 3\cos (\log x) - 4\cos (\log x) - 4\sin (\log x)]$

$\Rightarrow$ ${x^2}\cfrac{{{d^2}y}}{{d{x^2}}} = [3\sin (\log x) - 4\cos (\log x) - \{ 3\cos (\log x) + 4\sin (\log x)\}$

$\Rightarrow$ ${x^2}\cfrac{{{d^2}y}}{{d{x^2}}} = - x\cfrac{{dy}}{{dx}} - y$

$\Rightarrow$ ${x^2}\cfrac{{{d^2}y}}{{d{x^2}}} + x\cfrac{{dy}}{{dx}} + y = 0$
$\Rightarrow$ ${x^2}{y_2} + x{y_1} + y = 0$

Hence proved.