NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation cos x d y d x + y sin x = 1 is (where, c is an arbitrary constant)
Options
- Ay = c sin x + cos x
- By = sin x + c cos x
- Cy = tan x + c
- Dy sin x = sin x + c
Correct answer
B. y = sin x + c cos x
Step-by-step solution
d y d x + y tan ⁡ x = sec ⁡ x Ι . F . = e ∫ tan ⁡ x d x = e l o g s e c x = sec ⁡ x Hence, the solution is, y ⋅ sec ⁡ x = ∫ s e c 2 x d x ⇒ y sec ⁡ x = tan ⁡ x + c ⇒ y = sin ⁡ x + c cos ⁡ x