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 Exams — Differential 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:
- \(\frac{3}{2}-\frac{1}{\sqrt{e}}\)
- \(\frac{3}{2}-\sqrt{e}\)
- \(\frac{1}{2}+\frac{1}{\sqrt{e}}\)
- \(\frac{5}{2}+\frac{1}{\sqrt{e}}\)
Answer
(A) 3 2 - 1 e
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