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JEE Main202622 January 2026Morning ShiftMathematicsDifferential EquationsActual

Let the solution curve of the differential equation x d y-y d x= x²+y² d x, x>0 , y(1)=0 , be y=y(x) . Then y(3) is equal to

Options

  1. A1
  2. B4
  3. C2
  4. D6

Correct answer

B. 4

Step-by-step solution

xdy - ydx = x^2+y^2 ,dx . Put y = vx : x^2 dv = x 1+v^2 ,dx . dv 1+v^2 = dx x . Integrating: (v+ 1+v^2 ) = x + C . y(1) = 0 v = 0 at x = 1 : C = 0 . v + 1+v^2 = x y + x^2+y^2 = x^2 . At x = 3 : y + 9+y^2 = 9 9+y^2 = (9-y)^2 18y = 72 y = 4 .

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