# If $$y = 3\cos (\log\: x)+4\sin (\log\: x),$$ show that $x^2\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}$$+y=0$

Toolbox:
• $y=f(x)$
• $\large\frac{dy}{dx}$$=f'(x) • \large\frac{d^2y}{dx^2}=\frac{d}{dx}\big(\frac{dy}{dx}\big) • \large\frac{d}{dx}$$(\sin x)=\cos x$
• $\large\frac{d}{dx}$$(\cos x)=-\sin x • \large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
Step 1:
$y=3\cos(\log x)+4\sin(\log x)$
$\large\frac{dy}{dx}$$=y_1=3\sin(\log x).\large\frac{1}{x}$$+4\cos(\log x).\large\frac{1}{x}$
$x\large\frac{dy}{dx}$$=3\sin(\log x)+4\cos(\log x) Step 2: Differentiating with respect to x we get x\large\frac{d^2y}{dx^2}+\frac{dy}{dx}$$.1=\large\frac{3\cos(\log x)}{x}-\large\frac{4\sin(\log x)}{x}$
$\qquad\qquad\quad\;\;=\large\frac{3\cos(\log x)}{x}-\large\frac{4\sin(\log x)}{x}$
$\qquad\qquad\quad\;\;=\large\frac{-1}{x}$$[3\cos (\log x)+4\sin(\log x)] \qquad\qquad\quad\;\;=\large\frac{-y}{x} x\big[x\large\frac{d^2y}{dx^2}+\frac{dy}{dx}$$\big]=-y$
$x^2\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}$$=-y$
$x^2\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}$$+y=0$