The integrating factor for the differential equation $({cos}^{2} x) \frac{d y}{d x} + y = t a n x$ , is _______

e^{t a n x}

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cos 2x

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e^{- t a n x}

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sin2x

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Solution

The correct option is A $e^{t a n x}$
(a)

$F o r t h e e q u a t i o n \frac{d y}{d x} + P y = Q t h e i n t e r g r a t i n g f a c t o r i s e^{\int P d x}$ ,
$c o s^{2} x \frac{d y}{d x} + y = t a n x$
$\frac{d y}{d x} + \frac{y}{c o s^{2} x} = \frac{t a n x}{c o s^{2} x}$
$P = \frac{1}{c o s^{2} x} = s e c^{2} x$
$I n t e g r a t i n g f a c t o r = e^{\int s e c^{2} x d x} = e^{t a n x}$

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