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

- 0
- 2e
- e
- 2

**Answer**

(dy/dx) + (y/xlogx) = 2 and at x = 1, y = 0

Integration factor = e^{(∫(1/xlogx))} = e^{(log(logx))} = log x

So, y(log x) = 2∫(log x) dx

y(log x) = 2(x log x - x) + c

Put x = 1 to get c = 2

Now, y(e) : y(log e) = 2(e log e - e) + 2 = 2

**The correct option is D.**