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The integrating factor of $dx+xdy=e^{-y}\sec^{2}y dy$ is

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

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$dx= e^{-y} \sec^2 y dy- xdy$
$dx= (e^{-y} \sec^2 y-x)dy$
$\large\frac{dx}{dy}$$=e^{-y} \sec^2 y-x$
$\large\frac{dx}{dy}$$+x=e^{-y} \sec^2 y$
Comparing with $\large\frac{dx}{dy} $$+Px =Q$
$P=1$
$I.F=e^{\int Pdy}=e ^{\int dy}$
$\qquad= e^y$
Hence 2 is the correct answer.
answered May 21, 2014 by meena.p
 
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