If y=3 cos (log x)+4 sin (log x), then choose correct option :

The correct answer is Option (3) → $x^2y"+xy'+y=0$

$y=3\cos \log x+4 \sin \log x$

$y'=\frac{-3\sin\log x+4\cos\log x}{x}$

$y''=\frac{1(-3\sin\log x+4\cos\log x)}{x^2}-\frac{(3\cos\log x+4\sin\log x)}{x^2}$

$x^2y''=-xy'-y$

$x^2y''+xy'+y=0$