If a curve y = f x , passing through the point 1 , 2 , is the solution of the differential equation 2 x 2 d y = 2 x y + y 2 d x , then f 1 2 is equal to

Step-by-step solution

d y d x = y x + y 2 2 x 2 ⇒ y - 2 d y d x - 1 y · 1 x = 1 2 x 2 Put - 1 y = t    ⇒    1 y 2 d y d x = d t d x ⇒    d t d x + 1 x t = 1 2 x 2 This is a linear differential equation, with Integrating Factor : e ∫ 1 x d x = e ln x = x So, solution of the linear differential equation is t x = ∫ 1 2 x 2 · x d x + C    ⇒ - x y = 1 2 ln x + C The curve passes through 1 , 2 ⇒ - 1 2 = 1 2 ln 1 + C ⇒ C = - 1 2 Hence, the particu