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Integrating factor of $\large\frac{dy}{dx}+\frac{1}{x\log x}$$.y=\large\frac{2}{x^{2}}$ is

\[\begin{array}{1 1}(1)e^{x}&(2)\log x\\(3)\frac{1}{x}&(4)e^{-x}\end{array}\]

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$\large\frac{dy}{dx}+ \frac{1}{x \log x }$$y =\large\frac{2}{x^2}$
$P= \large\frac{1}{x \log x }$
$\int Pdx =\int \large\frac{1}{\log x} . \frac{1}{x} $$dx$
$\qquad= \log (\log x)$
I.F $e^{\int Pdx}=e^{\log(\log x)} $
$\qquad=\log x$
Hence 2 is the correct answer.
answered May 21, 2014 by meena.p
 

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