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JEE Main202227 Jun 2022Morning ShiftMathematicsDifferential EquationsActual

If d y d x + 2 x - y 2 y - 1 2 x - 1 = 0 , x , y > 0 , y 1 = 1 , then y 2 is equal to

Options

  1. A2 + log 2 3
  2. B2 + log 2 2
  3. C2 - log - 2 3
  4. D2 - log 2 3

Correct answer

D. 2 - log 2 3

Step-by-step solution

Given, d y d x + 2 x - y 2 y - 1 2 x - 1 = 0 x ,   y > 0 ,   y 1 = 1 Now rearranging and integrating both side of d y d x = - 2 x 2 y - 1 2 y 2 x - 1 , we get ⇒ ∫ 2 y 2 y - 1 d y = - ∫ 2 x 2 x - 1 d x ⇒ 1 ln 2 ∫ 2 y ln 2 2 y - 1 d y = - 1 ln 2 ∫ 2 x ln 2 2 x - 1 d x ⇒ 1 ln 2 ln 2 y - 1 = - 1 ln 2 ln 2 x - 1 + C At x = 1 , y = 1 Putting this values in above equation we get C = 0 So, ln 2 y - 1 + ln 2 x - 1 = 0 ⇒ 2 x - 1 2 y - 1 = 1 ⇒ 2 y - 1 = 1 2 x

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