The solution of the differential equation 1 − x 2 ⋅ d y d x + x y = x − x 3 y 1 2 , ∀ x < 1 is 9 y = − f x + c 1 − x 2 1 4 , where c is an arbitrary constant and f 1 2 = 3 4 . Then, f x is

Step-by-step solution

Given equation is 1 y d y d x + x 1 - x 2 y = x Let 2 y = ν ⇒ 1 y d y d x = d ν d x Thus, we have d ν d x + x 2 1 - x 2 ⋅ ν = x ∴ I . F . = e ∫ x 2 1 - x 2 d x = e - 1 4 ln ⁡ 1 - x 2 = 1 - x 2 - 1 4 Thus, the solution is ν 1 - x 2 - 1 4 = ∫ x 1 - x 2 - 1 4 d x or ν ⋅ 1 - x 2 - 1 4 = - 2 3 1 - x 2 3 4 + c " or 2 y = - 2 3 1 - x 2 + c " 1 - x 2 1 4 ⇒ y = - 1 - x 2 3 + c ' 1 - x 2 1 4 ⇒ 9 y = - 1 - x 2 + c 1 - x 2 1 4 ⇒ f x = 1 - x 2 ∴ f x is an even function