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Differential Equations question

2022 · 28 Jun · Shift 1 · Q32
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  5. /2022 · 28 Jun · Shift 1 · Q32

Differential Equations question

2022 · 28 Jun · Shift 1 · Q32

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation x(1−x2)dydx+(3x2y−y−4x3)=0x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0x(1−x2)dxdy​+(3x2y−y−4x3)=0, x>1x \gt 1x>1, with y(2)=−2y(2) = - 2y(2)=−2. Then y(3) is equal to :
  1. A
    −-− 18
  2. B
    −-− 12
  3. C
    −-− 6
  4. D
    −-− 3
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