Question

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

(A) 0 

(B) 2 

(C) 2e 

(D) e

Answer 1

Answer is (B) 2

It is a linear differential equation of first order of the form

Therefore, solution of given differential equation is given by

Note: Since we need to put x = 1 in order to find the value of the constant and at x = 1, P is not defined, so this whole question is conceptually incorrect. Although if we still solve it we get the general solution as

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

⇒ y(e) = 2e - 2e + 2

⇒ y(e) = 2