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Differential Equations question

2019 · 12 Apr · Shift 1 · Q41

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JEE Main
Mathematics
Differential Equations
2019 · 12 Apr · Shift 1 · Q41

Differential Equations question

2019 · 12 Apr · Shift 1 · Q41

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Consider the differential equation,

{y^2}dx + \left( {x - {1 \over y}} \right)dy = 0

, If value of y is 1 when x = 1, then the value of x for which y = 2, is :

A
${3 \over 2} - {1 \over {\sqrt e }}$
B
${1 \over 2} + {1 \over {\sqrt e }}$
C
${5 \over 2} + {1 \over {\sqrt e }}$
D
${3 \over 2} - \sqrt e$

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