The solution of the differential equation d y d x + x y 1 − x 2 = x y , x 1 is y = - f x 3 + C 1 - x 2 1 4 , where f 1 2 = 3 4 and C is an arbitrary constant. Then, the value of f - 1 2 is

Question

The solution of the differential equation d y d x + x y 1 − x 2 = x y , x 1 is y = - f x 3 + C 1 - x 2 1 4 , where f 1 2 = 3 4 and C is an arbitrary constant. Then, the value of f - 1 2 is, - 3 4, 3 4, 1 4, 3 2

MARKS App · Accepted Answer

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 ∫ - d 1 - x 2 4 1 - x 2 = 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 ⇒ f x = 1 - x 2 ⇒ f - 1 2 = 1 - 1 4 = 3 4

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