Question

Given that dy/dx = ye^x such that x = 0, y = e. The value of y(y > 0) when x = 1 will be

(A) e 

(B) 1/e 

 (C) e^e 

(D) e²

Answer 1

Answer is (C) e^e

dy/y = e^xdx ⇒ ln y = e^x + c

At x = 0, y = e. So, c = 0.

ln y = e^x

Therefore, at x = 1, y = e^e.