NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
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
Options
- A2
- B5 2
- C5 4
- D3 4
Correct answer
D. 3 4
Step-by-step solution
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