Let y = y x be the solution of the differential equation d y = e α x + y d x ; α ∈ N . If y log e 2 = log e 2 and y 0 = log e 1 2 , then the value of α is equal to ___.

Step-by-step solution

We have, ∫ e - y d y = ∫ e α x d x ⇒ - e - y = e α x α + c             . . . 1 Put x , y = ℓ n 2 , ℓ n 2 - 1 2 = 2 α α + c               . . . 2 Put x , y ≡ 0 , - ℓ n 2 - 2 = 1 α + c From equation 1 - 2 , we get 2 α - 1 α = 3 2 ⇒ α = 2 ( as α ∈ N ) .