**Given,**

y = 3 cos (log x) + 4 sin (log x)

**Differentiating with respect to x, we get**

\(\frac{dy}{dx}=-\frac{3sin(log\,x)}{x}+\)\(\frac{4cos(log\,x)}{x}\)

⇒ y_{1} = \(\frac{1}{x}\)[- 3 sin(log x) +4 cos (log x)]

**Again differentiating with respect to x, we get**

**Now,**

LHS = x^{2}y_{2} + xy_{1} +y

= (\(x^2(\frac{-sin(logx)-7\,cos(logx)}{x^2})\)+ \(x\times\frac{1}x\)[-3 sin(log x) + 4 cos(log x)] +3 cos (log x) + 4 sin(log x)

= - sin (log x) - 7 cos (log x) - 3 sin (log x) + 4 sin (log x)

= 0 **= RHS**