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JEE Main202117 Mar 2021Evening ShiftMathematicsDifferential EquationsActual

Let y = y ( x ) be the solution of the differential equation cos x ( 3 sin x + cos x + 3 ) d y = ( 1 + y sin x ( 3 sin x + cos x + 3 ) ) d x , 0 ≤ x ≤ π 2 , y 0 = 0 . Then, y π 3 is equal to:

Options

  1. A2 log e 2 3 + 9 6
  2. B2 log e 2 3 + 10 11
  3. C2 log e 3 + 7 2
  4. D2 log e 3 3 - 8 4

Correct answer

B. 2 log e 2 3 + 10 11

Step-by-step solution

Given cos x ( 3 sin x + cos x + 3 ) d y = ( 1 + y sin x ( 3 sin x + cos x + 3 ) ) d x ⇒ d y d x = ( 1 + y sin x ( 3 sin x + cos x + 3 ) ) cos x ( 3 sin x + cos x + 3 ) ⇒ d y d x = 1 cos x ( 3 sin x + cos x + 3 ) + y sin x ( 3 sin x + cos x + 3 ) ) cos x ( 3 sin x + cos x + 3 ) ⇒ d y d x - ( tan x ) y = 1 ( 3 sin x + cos x + 3 ) cos x This is a linear differential equation of the type d y d x + P y = Q , where P = - tan x and Q = 1 cos x 3 sin x + cos x + 3 . Now, the integrating factor I . F . = e

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