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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

The solution of the differential equation d y d x = y c o s x - y 2 sin ⁡ x is equal to (where c is an arbitrary constant)

Options

  1. Asin ⁡ x = x - y + c
  2. Bsin ⁡ x = x + y + c
  3. Csin ⁡ x = x y + c y
  4. Dsin ⁡ x x = y + c

Correct answer

C. sin ⁡ x = x y + c y

Step-by-step solution

d y d x = y c o s x - y 2 sin ⁡ x y cos ⁡ xdx - sin ⁡ x d y = y 2 d x y c o s  x . d x - sin ⁡ x . d y y 2 = d x d sin ⁡ x y = d x On integrating, we get, sin ⁡ x y = x + c sin ⁡ x = x y + c y

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