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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

The general solution of the differential equation d y d x + sin ⁡ x + y 2 = sin ⁡ x - y 2 is (where c is an arbitrary constant)

Options

  1. Aln tan ⁡ y 2 = c - 2 sin ⁡ x
  2. Bln ⁡ tan ⁡ y 4 = c - 2 sin ⁡ x 2
  3. Cln ⁡ tan ⁡ y 2 + π 4 = c - 2 sin ⁡ x
  4. Dln ⁡ tan ⁡ y 4 + π 4 = c - 2 sin ⁡ x 2

Correct answer

B. ln ⁡ tan ⁡ y 4 = c - 2 sin ⁡ x 2

Step-by-step solution

Given equation d y d x + sin ⁡ x + y 2 = sin ⁡ x - y 2 ⇒ d y d x = sin ⁡ x - y 2 - sin ⁡ x + y 2 ⇒ d y d x = - 2 sin ⁡ y 2 cos ⁡ x 2 ⇒ c o s e c y 2 d y = - 2 cos ⁡ x 2 d x On integrating both sides, we get ∫ c o s e c y 2 d y = - ∫ 2 cos ⁡ x 2 d x ⇒ ln ⁡ ( tan ⁡ y 4 ) 1 2 = - 2 sin ⁡ x 2 1 2 + c ⇒ ln ( tan ⁡ y 4 ) = c - 2 sin ⁡ x 2

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