Consider the differential equation, y^ 2 d x+ (x- 1 y ) d y=0 , If value of y is 1 when x…

Mathematics · JEE Main · NTA ExamsDifferential Equations

Consider the differential equation, \(y^{2} d x+\left(x-\frac{1}{y}\right) d y=0\), If value of y is 1 when x = 1, then the value of x for which y = 2, is:
  1. \(\frac{3}{2}-\frac{1}{\sqrt{e}}\)
  2. \(\frac{3}{2}-\sqrt{e}\)
  3. \(\frac{1}{2}+\frac{1}{\sqrt{e}}\)
  4. \(\frac{5}{2}+\frac{1}{\sqrt{e}}\)

Answer

(A) 3 2 - 1 e

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