Solution of the differential equation
$c o s x d y = y (s i n x - y) d x, 0 < x < \frac{π}{2}$ is

y s e c x = t a n x + c

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y t a n x = s e c x + c

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t a n x = (s e c x + c) y

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s e c x = (t a n x + c) y

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Solution

The correct option is D $s e c x = (t a n x + c) y$
$c o s x d y = y (s i n x - y) d x \frac{d y}{d x} = y t a n x - y^{2} s e c x \frac{1}{y^{2}} \frac{d y}{d x} - \frac{1}{y} t a n x = - s e c x$ ....(i)
Let $\frac{1}{y} = t \Rightarrow - \frac{1}{y^{2}} \frac{d y}{d x} = \frac{d t}{d x}$
From equations (i)
$- \frac{d t}{d x} - t (t a n x) = - s e c x \Rightarrow \frac{d t}{d x} + (t a n x) t = s e c x$
I.F. $= e^{\int t a n x d x} = (e)^{l o g | s e c x |} s e c x$
Solution: $t (I . F) = \int (I . F) s e c x d x \Rightarrow \frac{1}{y} s e c x = t a n x + c$

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