Let y = f x be a solution of the differential equation d y d x = y 2 - x 2 2 x y ∀ x , y 0 . If f 1 = 2 , then f ′ 1 is equal to

Question

Let y = f x be a solution of the differential equation d y d x = y 2 - x 2 2 x y ∀ x , y > 0 . If f 1 = 2 , then f ′ 1 is equal to, 2, 5 2, 5 4, 3 4

MARKS App · Accepted Answer

Given, d y d x = y 2 - x 2 2 x y Let y = v x Then, v + x d v d x = v 2 - 1 2 v ⇒ x d v d x = v 2 - 1 - 2 v 2 2 v ⇒ x d v d x = - 1 2 v 2 + 1 v So, 2 v v 2 + 1 d v + d x x = 0 ⇒ x v 2 + 1 = C ⇒ x 2 + y 2 = C x Since, f 1 = 2 ⇒ C = 5 i.e. f x = 5 x − x 2 ⇒ f ' x = 5 − 2 x 2 5 x − x 2 ⇒ f ' 1 = 3 4

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