If the solution curve f ( x , y ) = 0 of the differential equation 1 + log e x d x d y - x log e x = e y , x > 0 , passes through the points ( 1 , 0 ) and ( a , 2 ) , then a a is equal to

Step-by-step solution

The given differential equation is 1 + ln x d x d y - x ln x = e y Put x ln x = t ⇒ 1 + ln x d x = d t ⇒ dt dy - t = e y which is a linear differential equation. I . F = e - ∫ d y = e - y The solution of the differential equation is ⇒ t × e - y = ∫ e y × e - y d y + c ⇒ t × e - y = y + c ⇒ x ln x = y   e y + c   e y Put x = 1 ,   y = 0 ⇒   c = 0 Put x = a ,   y = 2 a ln a = 2 e 2 ∴   a a = e 2 e 2 Hence this is the co