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Let y = y ( x ) be the solution of the differential equation x log e x dx dy + y = x 2 log e x, (x > 1). If y

1. = 2, then y(e) is equal to
2. 2 e 1 2 
3. 4 e 1 2 
4. 4 e 4 2  (4) 2 e 2 2 

Correct answer: 4 e 1 2 

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