Solve: $(x + tan y) d y = sin 2 y d x$

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Solution

$(x + t a n y) d y = s i n 2 y d x \Rightarrow \frac{d x}{d y} = \frac{x + t a n y}{s i n 2 y} \Rightarrow \frac{d x}{d y} = \frac{x}{s i n 2 y} + \frac{t a n y}{s i n 2 y} \Rightarrow \frac{d x}{d y} - c o s e c 2 y . x = \frac{1}{2} {s e c}^{2} y (∵ s i n x = \frac{1}{c o s e c x}; t a n x = \frac{s i n x}{c o s x}; s i n 2 x = 2 s i n x c o s x) s i n c e t h i s i s i n t h e f o r m o f \frac{d x}{d y} + R x = S w h e r e, R = - c o s e c 2 y a n d S = \frac{1}{2} {s e c}^{2} y ∴ I . F = e^{\int R d y} = e^{\int (- c o s e c 2 y) d y} = e^{- l o g | c o s e c 2 y - c o t 2 y |} = e^{- l o g (t a n y)} = e^{l o g (c o t y)} = c o t y s o t h e s o l u t i o n o f t h e g i v e n d i f f e r e n t i a l e q u a t i o n i s g i v e n b y x . (I . F) = \int S . (I . F) d y + C \Rightarrow x . (c o t y) = \int \frac{1}{2} {s e c}^{2} y . (c o t y) d y + C \Rightarrow x . (c o t y) = \int \frac{1}{s i n 2 y} d y + C \Rightarrow x . (c o t y) = \int c o s e c 2 y d y + C \Rightarrow x . (c o t y) = \frac{1}{2} l o g | c o s 2 y - c o t 2 y | + C$

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