NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation x d x + y sin 2 ⁡ x d y = y d y + x sin 2 ⁡ y d x is (where, c is an arbitrary constant)
Options
- Ax tan x = sec y + c
- Bx tan y = sec x + c
- Cx tan x - ln sec x = y tan y - ln sec y + c
- Dx tan x = ln sec y + c
Correct answer
C. x tan x - ln sec x = y tan y - ln sec y + c
Step-by-step solution
The given equation is x cos 2 ⁡ y d x = y cos 2 ⁡ x d y ∫ x sec 2 ⁡ x d x = ∫ y sec 2 ⁡ y d y Using integration by parts, we get, x tan ⁡ x - ln ⁡ sec ⁡ x = y tan ⁡ y - ln ⁡ sec ⁡ y + c