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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$

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  • $y=f(x)$
  • $\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}$$[3\cos (\log x)+4\sin(\log x)]$
$ x^2\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}$$=-y$
$ x^2\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}$$+y=0$
answered Sep 26, 2013 by sreemathi.v

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