The solution curve of the differential equation,(1+ {e}^{- {x}} ) (1+ {y}^{2} ) {{dy}}/{ {dx}}= {y}^{2}, which passesthrough the point (0,1), is : [3 Sep 2020 Shift 1]

Correct Answer is : y2=1+yloge(1+ex2)

{y}^{2}=1+ {y} \log _{ {e}}({1+ {e}^{ {x}}}/{2} )

{y}^{2}+1= {y} (\log _{ {e}}({1+ {e}^{{x}}}/{2} )+2 )

{y}^{2}=1+ {y} \log _{ {e}}({1+ {e}^{ {-x}}}/{2} )

{y}^{2}+1= {y} (\log _{ {e}}({1+ {e}^{{-x}}}/{2} )+2 )

JEE Mains 03-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 65 of 75

Marks: +1, -0

The solution curve of the differential equation,(1+e−x)(1+y2)

=y2, which passesthrough the point (0,1), is :

[3 Sep 2020 Shift 1]

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