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If \( y = 3 \cos \: (\log x ) + 4 \sin ( \log x )\), show that \(x^2y_2 + xy_1 \)

$\begin{array}{1 1} -y \\ y\\1+y \\ 1-y \end{array} $

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1 Answer

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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}$$=y_1=3\sin(\log x).\frac{1}{x}$$+4\cos(\log x).\large\frac{1}{x}$
$xy_1=3\sin(\log x)+4\cos(\log x)$
Step 2:
Differentiating with respect to $x$ we get
$xy_2+y_1.1=\large\frac{3\cos(\log x)}{x}-\large\frac{4\sin(\log x)}{x}$
$xy_2+y_1=\large\frac{-1}{x}$$[3\cos (\log x)+4\sin(\log x)]$
$xy_2+y_1=\large\frac{-y}{x}$
$x^2y_2+xy_1=-y$
answered Apr 10, 2014 by balaji.thirumalai
 

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