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The integrating factor of the differential equation $\large\frac{dy}{dx}$$-y\tan x=\cos x $ is

\[\begin{array}{1 1}(1)\sec x&(2)\cos x\\(3)e^{\tan x}&(4)x^{2}/2e^{2x}\end{array}\]

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$\large\frac{dy}{dx}$$-y \tan x =\cos x$
$P_1= -\tan x$
Integrating factor $=e^{\int Pdx}$
$\qquad= e^{- \int \tan x .dx}$
$\qquad= e^{-\int \large\frac{\sin x}{\cos x } dx}$
$\qquad= e^{\log \cos x}$
$\qquad= \cos x$
Hence 2 is the correct answer.
answered May 22, 2014 by meena.p
 
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