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JEE Main 2019 · 12 Apr · Shift I

JEE Main 2019 Mathematics question on Differential Equations

Consider the differential equation, y2dx+(x−1y)dy=0y^2dx + \left(x - \dfrac{1}{y}\right)dy = 0.

If value of yy is 1 when x=1x = 1, then the value of xx for which y=2y = 2, is
  1. A.
    52+1e\dfrac{5}{2} + \dfrac{1}{\sqrt{e}}
  2. B.
    32−1e\dfrac{3}{2} - \dfrac{1}{\sqrt{e}}
  3. C.
    12+1e\dfrac{1}{2} + \dfrac{1}{\sqrt{e}}
  4. D.
    32−e\dfrac{3}{2} - \sqrt{e}
Show answer

Correct option: B

[JEE Main 2019 · 12 Apr · Shift I]

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