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JEE Main20207 Jan 2020Evening ShiftMathematicsDifferential EquationsActual

Let y = y x be the solution curve of the differential equation, y 2 - x d y d x = 1 , satisfying y 0 = 1 . This curve intersects the X - axis at a point whose abscissa is

Options

  1. A2 - e
  2. B- e
  3. C2
  4. D2 + e

Correct answer

A. 2 - e

Step-by-step solution

d x d y + x = y 2 This equation is a linear differential equation of the type d x d y + P x = Q , where P = 1 and Q = y 2 . Integrating Factor I.F. = e ∫ 1 d y = e y Now solution of the differential equation is x I . F . = ∫ P I . F . d y + C x · e y = ∫ y 2 · e y · d y = y 2 · e y - ∫ 2 y · e y · d y ⇒   x · e y = y 2 · e y - 2 y · e y + 2 e y + c ⇒   x = y 2 - 2 y + 2 + c · e - y Put x = 0 ,   y = 1 ⇒  

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