Differential Equations of first order and first Degree

Q. Cos(x+y) dy =dx

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Solution

Dear student
$\cos (x + y) d y = d x \frac{d y}{d x} = \frac{1}{\cos (x + y)} L e t x + y = v . T h e n 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1 P u t t i n g x + y = v a n d \frac{d y}{d x} = \frac{d v}{d x} - 1 t h e g i v e n d i f f e r e n t i a l e q, w e g e t \Rightarrow \frac{d v}{d x} - 1 = \frac{1}{\cos v} \Rightarrow \frac{d v}{d x} = \frac{1 + \cos v}{\cos v} \Rightarrow \frac{\cos v}{1 + \cos v} d v = d x \Rightarrow \frac{\cos v (1 - \cos v)}{1 - \cos^{2} v} d v = d x \Rightarrow (c o t v \cos e c v - c o t^{2} v) d v = d x \Rightarrow (c o t v \cos e c v - \cos e c^{2} v + 1) d v = d x \Rightarrow - \cos e c v + c o t v + v = x + C [O n i n t e g r a t i n g] \Rightarrow - \cos e c (x + y) + c o t (x + y) + y = C \Rightarrow \frac{- 1 - \cos (x + y)}{\sin (x + y)} + y = C \Rightarrow - \tan (\frac{x + y}{2}) + y = C \Rightarrow y = \tan (\frac{x + y}{2}) + C$
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