NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The differential equation of the family of curves y = k 1 x 2 + k 2 is given by (where, k 1 and k 2 are arbitrary constants and y 1 = d y d x , y 2 = d 2 y d x 2 )
Options
- Ay 1 = x 2 y 2
- By 1 2 = x y 2
- Cx y 2 = y 1
- Dy 1 y 2 = x
Correct answer
C. x y 2 = y 1
Step-by-step solution
Differentiating with respect to x , we get, d y d x = 2 k 1 x ⇒ y 1 x = 2 k 1 Differentiating again, we get, x ⋅ y 2 - y 1 x 2 = 0 or x y 2 = y 1