It is given that, y = 3cos(log x) + 4sin(log x) Then, $\frac{dy}{dx}$ = 3 × $\frac{d}{dx}$ [cos log x] + 4 × $\frac{d}{dx}$ [sin log x] = 3 × $-\sin(\log x) \times \frac{d}{dx} \log x]$ + 4 × $\left[\cos(\log x) \times \frac{d}{dx} \log x\right]$ = $\frac{-3\sin(\log x)}{x}$ + $\frac{4\cos(\log x)}{x}$ = $\frac{4\cos(\log x)-3\sin(\log x)}{x}$ $\frac{d^2y}{dx^2}$ = $\frac{d}{dx} \left( \frac{4\cos(\log x)-3\sin(\log x)}{x} \right)$ = = = = ∴ $x^2 \frac{d^2y}{dx^2}$ + x $\frac{dy}{dx}$ + y = $x^2 \left( \frac{-\sin(\log x)-7\cos(\log x)}{x^2} \right)$ + x $\left( \frac{4\cos(\log x)-3\sin(\log x)}{x} \right)$ + 3 cos (log x) + 4 sin (log x) = - sin (log x) - 7 cos (log x) + 4 cos (log x) - 3 sin (log x) + 3 cos (log x) + 4 sin (log x) = 0 Hence proved.