NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation 1 - x 2 d y d x - x y = 1 is (where, x < 1 ,   x ∈ R and C is an arbitrary constant)
Options
- Ay 1 - x 2 = t a n - 1 x + C
- By 1 - x 2 = t a n - 1 x + C
- Cy 1 - x 2 = s i n - 1 x + C
- Dy ⋅ 1 - x 2 = s i n - 1 x + C
Correct answer
C. y 1 - x 2 = s i n - 1 x + C
Step-by-step solution
d y d x - x 1 - x 2 y = 1 1 - x 2 Ι . F . = e - ∫ x 1 - x 2 d x = e 1 2 ∫ - 2 x 1 - x 2 d x = e 1 2 ln 1 - x 2 = 1 - x 2 Hence, the solution of the differential equation is y 1 - x 2 = ∫ 1 - x 2 1 - x 2 d x y 1 - x 2 = ∫ 1 1 - x 2 d x y 1 - x 2 = s i n - 1 x + C