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JEE Main20209 Jan 2020Evening ShiftMathematicsDifferential EquationsActual

If d y d x = x y x 2 + y 2 ; y 1 = 1 ; then a value of x satisfying y x = e is:

Options

  1. A1 2 3 e
  2. Be 2
  3. C2 e
  4. D3 e

Correct answer

D. 3 e

Step-by-step solution

Put y = v x d y d x = v + x d v d x v + x d v d x = v x 2 x 2 + v 2 x 2 ⇒ 1 + v 2 v 3 d v = - 1 x d x ⇒ ∫ 1 v 3 + 1 v d v = ∫ - 1 x d x ⇒ - 1 2 1 v 2 + l n v = - l n x + c ⇒ - x 2 2 y 2 = - l n y + c When x = 1 , y = 1 then - 1 2 = c ⇒ x 2 = y 2 1 + 2 l n y ⇒ x 2 = e 2 3

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