JEE Main201910 Apr 2019Evening ShiftMathematicsDifferential EquationsActual
Let y = y x be the solution of the differential equation, d y d x + y tan x = 2 x + x 2 tan x , x ∈ - π 2 , π 2 , such that y 0 = 1 . Then
Options
- Ay ' π 4 - y ' - π 4 = π -   2
- By ' π 4 + y ' - π 4 = -   2
- Cy π 4 - y - π 4 =   2
- Dy ' π 4 + y ' - π 4 = π 2 2 + 2
Correct answer
A. y ' π 4 - y ' - π 4 = π -   2
Step-by-step solution
Given d y d x + y tan x = 2 x + x 2 tan x This is a linear differential equation of the type d y d x + P y = Q , where P = tan x   &   Q = 2 x + x 2 tan x Now, the integrating factor I . F . = e ∫ P d x = e ∫ tan x   d x = e lnsec x = sec x And, the general solution is y I . F . = ∫ Q I . F . d x + c y ⋅ sec x = ∫ 2 x + x 2 tan x sec x d x + c ⇒ y s e c x = ∫ 2 x s e c   x d x + ∫ x 2 sec x ⋅ tan x d x + c Using integration by parts in th