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JEE Main202120 Jul 2021Morning ShiftMathematicsDifferential EquationsActual

Let y = y x be the solution of the differential equation e x 1 - y 2 d x + y x d y = 0 , y 1 = - 1 Then the value of y 3 2 is equal to:

Options

  1. A1 - 4 e 3
  2. B1 - 4 e 6
  3. C1 + 4 e 3
  4. D1 + 4 e 6

Correct answer

B. 1 - 4 e 6

Step-by-step solution

We have, e x 1 - y 2   d x + y x d y = 0 ,   y 1 = - 1 ⇒     e x 1 - y 2   dx = - y x d y ⇒       ∫ - y 1 - y 2   d y = ∫ x e x d x ⇒       1 2 ∫ d 1 - y 2 1 - y 2   d y = ∫ x e x d x ⇒   1 - y 2 = e x x - 1 + c Now, at x = 1 , y = - 1 , then ⇒     0 = 0 + c ⇒ c = 0 ∴     1 - y 2 = e x x - 1 At x = 3 , then 1 - y 3 2 = 2 e 3 ⇒     1 - y 3 2 = 4 e

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