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Let y(x) be the solution of the differential equation (x log x) dy/dx + y = 2x log x, (x ≥ 1). Then y(e) is equal to:

1. 2
2. 2e
3. e
4. (d)

Correct answer: 2

Solution

Dividing by x log x gives a linear ODE with integrating factor ln x: d/dx(y ln x)=2 ln x, so y ln x = 2(x ln x - x)+C. Using y(1) finite gives C=2; at x=e, y*1 = 2(e-e)+2 = 2, so y(e)=2.

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