Let α x = exp x β y γ be the solution of the differential equation 2 x 2 y d y - 1 - x y 2 d x = 0 , x > 0 , y 2 = log e 2 . Then α + β - γ equals :

Step-by-step solution

Given the solution of differential equation, α x = e x β · y γ . . . . . . ( 1 ) Now calculating the solution of differential equation, 2 x 2 y d y d x = 1 - x y 2 Let y 2 = t ⇒ 2 d y d x = d t d x ∴   x 2 d t d x = 1 - x t dt dx + t x = 1 x 2   . . . . . ( 2 ) Above differential equation is linear in t We know solutions of differential equation of the form d y dx + P y = Q is given by, y . e ∫ P d x = ∫ Q . e ∫ P d x + C