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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

If the curve satisfying differential equation x d y = y + x 3 d x passes through 1 , 1 , then the equation of the curve is

Options

  1. Ay 2 = x 3 - x
  2. By = x 2 - x
  3. C2 y = x 3
  4. D2 y = x 3 + x

Correct answer

D. 2 y = x 3 + x

Step-by-step solution

The given equation is x d y - y d x x 2 = x d x or d y x = x d x On integrating, we get, y x = x 2 2 + C ⇒ y = x 3 2 + C ⋅ x As it passes through 1 , 1 , 1 = 1 2 + C ⇒ C = 1 2 ∴ The equation of curve is 2 y = x 3 + x

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