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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

The solution of the differential equation x c o s y d y d x + sin ⁡ y = 1 is (Here, x > 0 and λ is an arbitrary constant)

Options

  1. Ax - x c o s x = λ
  2. Bx + x c o s x = λ
  3. Cx - x s i n y = λ
  4. Dx + x c o s y = λ

Correct answer

C. x - x s i n y = λ

Step-by-step solution

Let, sin ⁡ y = t ⇒ cos ⁡ y d y d x = d t d x ∴ the equation becomes x d t d x + t = 1 or x d t d x = 1 - t ⇒ d t 1 - t = d x x On integrating, we get, - ln ⁡ 1 - t = ln ⁡ x + ln ⁡ C or 1 1 - t = C x i.e. 1 - sin ⁡ y x = 1 C = λ s a y

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