NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The general solution of the differential equation d y d x = 2 y tan x + tan 2 x , ∀ x ∈ 0 , π 2 is yf x = x 2 - sin 2 x 4 + C , (where, C is an arbitrary constant). If f π 4 = 1 2 , then the value of f π 3 is equal to
Options
- A1 2
- B1 4
- C2
- D4
Correct answer
B. 1 4
Step-by-step solution
Given equation is d y d x + y - 2 tan x = tan 2 x IF = e 2 ln cos x = cos 2 x Thus, the solution is y ⋅ cos 2 x = ∫ tan 2 x · cos 2 x d x ⇒ y ⋅ cos 2 x = ∫ 1 - cos 2 x 2 d x = x 2 - sin 2 x 4 + C ⇒ f x = cos 2 x ⇒ f π 3 = 1 2 2 = 1 4