NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The differential equation of the family of curves p y 2 = 3 x - p is (where p is an arbitrary constant) is
Options
- Ay d y d x = y + x
- By d y d x = 1
- Cy 2 = d y d x
- Dy 2 = 2 x y d y d x - 1
Correct answer
D. y 2 = 2 x y d y d x - 1
Step-by-step solution
Given family of curves is p 3 = x y 2 + 1 On differentiating with respect to x , we get, 0 = y 2 + 1 ⋅ 1 − x 2 y ⋅ y ′ y 2 + 1 2 ⇒ 2 x y y ′ = y 2 + 1 ⇒ y 2 = 2 x y d y d x - 1