NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation 3 x 2 s i n 1 x + y d x = x c o s 1 x d x - x d y is (where, c is an arbitrary constant)
Options
- As i n 1 x = x y + c
- Bx 3 s i n 1 x + x y = c
- Cx 3 s i n 1 x = x y + c
- Ds i n x = x 3 y + c
Correct answer
B. x 3 s i n 1 x + x y = c
Step-by-step solution
Given equation can be written as d x 3 s i n 1 x + d x y = 0 ∴ solution is x 3 s i n 1 x + x y = c