NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice
The solution of the differential equation x d x + y d y = x d y - y d x x 2 + y 2 is tan ⁡ f x , y - C = y x (where, C is an arbitrary constant). If f 1,1 = 1 , then f π , π is equal to
Options
- A2
- Bπ 2
- C- 1
- Dπ
Correct answer
B. π 2
Step-by-step solution
Given equation is x d x + y d y = d tan - 1 y x On integrating, we get, x 2 + y 2 2 = tan - 1 y x + C or tan x 2 + y 2 2 - C = y x ⇒ f x , y = x 2 + y 2 2 ∴ f π , π = π 2