NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of d y = cos x 2 - y c o s e c x d x , where y = 2 when x = π / 4 , is
Options
- Ay = sin x + 1 2 c o s e c x
- By = tan ( x / 2 ) + cot x / 2
- Cy = 1 / 2 sec x / 2 + 2 cos x / 2
- DNone of the above
Correct answer
A. y = sin x + 1 2 c o s e c x
Step-by-step solution
Given, d y d x = 2 cos x - y cos x c o s e c x ⇒ d y d x + y cot x = 2 cos x ∴ I F = e ∫ cot x d x = e ln sin x = sin x ∴ Solution is y sin x = ∫ 2 cos x sin x d x + c ⇒ y sin x = ∫ sin 2 x d x + c ⇒ y sin x = - cos 2 x 2 + c At x = π 4 , y = 2 ∴ 2 sin π 4 = - cos 2 ( π / 4 ) 2 + c ⇒ c = 1 ∴ y sin x = - 1 2 cos 2 x + 1 ⇒ y = - 1 2 . cos 2 x sin x + c o s e c x ⇒ y = - 1 2 sin x 1 - 2 sin 2 x + c o s e c x ⇒ y = 1 2 c o s e c x + sin x