If $y=3 \cos (\log x)+4 \sin (\log x),$ show that $x^{2} y_{2}+x y_{1}+y=0$.

Given $y=3 \cos (\log x)+4 \sin (\log x)$.
Differentiating with respect to $x$:
$y_{1} = 3 \cdot \frac{d}{dx}[\cos (\log x)] + 4 \cdot \frac{d}{dx}[\sin (\log x)]$
$y_{1} = 3 \cdot [-\sin (\log x) \cdot \frac{1}{x}] + 4 \cdot [\cos (\log x) \cdot \frac{1}{x}]$
$x y_{1} = -3 \sin (\log x) + 4 \cos (\log x)$
Differentiating again with respect to $x$:
$\frac{d}{dx}(x y_{1}) = \frac{d}{dx}[-3 \sin (\log x) + 4 \cos (\log x)]$
$x y_{2} + y_{1} \cdot 1 = -3 \cos (\log x) \cdot \frac{1}{x} - 4 \sin (\log x) \cdot \frac{1}{x}$
$x^{2} y_{2} + x y_{1} = -3 \cos (\log x) - 4 \sin (\log x)$
$x^{2} y_{2} + x y_{1} = -(3 \cos (\log x) + 4 \sin (\log x))$
$x^{2} y_{2} + x y_{1} = -y$
$x^{2} y_{2} + x y_{1} + y = 0$
Hence,proved.