NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
If the differential equation 3 x 1 3 d y + x - 2 3 y d x = 3 x d x is satisfied by k x 1 3 y = x 2 + c (where c is an arbitrary constant), then the value of k is
Options
- A1 3
- B2 3
- C2
- D1
Correct answer
C. 2
Step-by-step solution
The given equation is x 1 3 . d y + 1 3 x - 2 3 d x . y = x d x or d x 1 3 . y = x d x Integrating, we get, x 1 3 . y = x 2 2 + λ Or 2 x 1 3 y = x 2 + C ⇒ k = 2