JEE Math Practice Question
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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