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JEE Main202623 January 2026Morning ShiftMathematicsDifferential EquationsActual

Let y=y(x) be the solution of the differential equation x⁴ ~d y+ (4 x³ y+2 x ) d x=0, x>0, y ( 2 )=0 . Then ⁴ y ( 3 ) is equal to :

Options

  1. A72
  2. B92
  3. C64
  4. D81

Correct answer

D. 81

Step-by-step solution

x^4 ,dy + 4x^3 y ,dx + 2 x ,dx = 0 d(x^4 y) + 2 x ,dx = 0 . Integrating: x^4 y - 2 x = C . y( /2) = 0 : ( /2)^4 0 - 2 ( /2) = 0 C = 0 . x^4 y = 2 x . At x = /3 : ( /3)^4 y( /3) = 2 ( /3) = 1 . ^4 y( /3) = 81 ( /3)^4 y( /3) 1 ( /3)^4 ^4 81 ... Directly: y( /3) = 81 ^4 , so ^4 y( /3) = 81 .

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