JEE Main 1-Sep-2021 Shift 2 Solved Paper

Section: Mathematics

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Question : 67 of 90

Marks: +1, -0

If y=y(x) is the solution curve of the differential equation x2dy+(y−

1

x

)dx=0 ; x>0
and y(1)=1, then y(

1

2

) is equal to

[1 Sep 2021 Shift 2]

3
2
−
1
√e
3+
1
√e
3+e
3−e

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