If y = y x is the solution of the differential equation 1 + e 2 x d y d x + 2 1 + y 2 e x = 0 and y 0 = 0 , then 6 y ' 0 + y log c 3 2 is equal to:

Step-by-step solution

Given, 1 + e 2 x d y d x + 2 1 + y 2 e x = 0 ⇒ d y 1 + y 2 + 2 e x 1 + e 2 x d x = 0 .........(i) Now integrating both side we get, ⇒ ∫ d y 1 + y 2 + ∫ 2 e x 1 + e 2 x d x = ∫ 0 ⇒ tan - 1 y + 2 tan - 1 e x = c ∵ y 0 = 0 so, C = π 2 ⇒ tan - 1   y + 2   tan - 1 e x = π 2 ....(ii) Now from equation (i), we get d y d x x = 0 = - 1 From equation (ii) we get, y (In 3 ) = - 1 3 So, 6 y ' 0 + ( y In 3 ) 2 = 6 - 1 + 1 3 = - 4 .