If the solution curve y = y ( x ) of the differential equation 1 + y 2 1 + log e x d x + x d y = 0 , x > 0 passes through the point ( 1 , 1 ) and y e = α - tan 3 2 β + tan 3 2 , then α + 2 β is

Step-by-step solution

Given: 1 + y 2 1 + log e x d x + x d y = 0 ⇒ 1 + log e x x d x = - 1 1 + y 2 d y ⇒ ∫ 1 x + log x x d x + ∫ d y 1 + y 2 = 0 ⇒ log x + ( log x ) 2 2 + tan - 1 y = C This curve passes through the point 1 , 1 . ⇒ log 1 + ( log 1 ) 2 2 + tan - 1 1 = C ⇒ C = π 4 ⇒ log x + ( log x ) 2 2 + tan - 1 y = π 4 Now, putting x = e ⇒ log e + ( log e ) 2 2 + tan - 1 y = π 4 ⇒ tan - 1 y = π 4 - 3 2 ⇒ y = tan π 4 - 3 2 ⇒ y = tan π 4 - tan 3 2 1 + tan π 4 tan 3 2 ⇒ y = 1 - tan 3 2 1 + tan 3 2 Hence, on comparing we get, ⇒ α = β = 1 ⇒