JEE Main 12 Jan 2019 Shift 1 Solved Paper

Section: Mathematics
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Question : 58 of 90
 
Marks: +1, -0
Lety = y(x) be the solution of the differentialequation, x
dy
dx
+ y = x loge x , (x > 1). If 2y (2) = loge4−1 , then y (e) is equal to :
[12 Jan 2019 Shift 1]
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