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JEE Main202227 Jun 2022Evening ShiftMathematicsDifferential EquationsActual

If the solution curve of the differential equation tan - 1 y - x d y = 1 + y 2 d x passes through the point 1 , 0 then the abscissa of the point on the curve whose ordinate is tan 1 is

Options

  1. A2
  2. B2 e
  3. C3 e
  4. D2 e

Correct answer

B. 2 e

Step-by-step solution

tan - 1 y - x d y = 1 + y 2 d x ⇒ d x d y + 1 1 + y 2 x = tan - 1 y 1 + y 2 I.F. = e ∫ 1 1 + y 2 d y = e tan - 1 y General solution will be x e tan - 1 y = ∫ e tan - 1 y × tan - 1 y 1 + y 2 d y ⇒ x e tan - 1 y = tan - 1 y e tan - 1 y - e tan - 1 y + C Since the curve passes through 1 , 0 ⇒    C = 2 So, the curve reduces to x e tan - 1 y = tan - 1 y e tan - 1 y - e tan - 1 y + 2 when y = tan 1 ,   x e = 1 e - e + 2 ⇒     x = 2 e

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