Manipal MET2018MathematicsDifferential Equations
The general solution of x (1+y^2 )^ 1 / 2 d x+y (1+x^2 )^ 1 / 2 d y=0 is
Options
- A⁻¹ x+ ⁻¹ y=c
- Bx^2+y^2= (1+x^2 )^ 1 / 2 + (1+y^2 )^ 1 / 2 +c
- C(1+x^2 )^ 1 / 2 + (1+y^2 )^ 1 / 2 =c
- D⁻¹ x- ⁻¹ y=c
Correct answer
C. (1+x^2 )^ 1 / 2 + (1+y^2 )^ 1 / 2 =c
Step-by-step solution
Given equation can be rewritten as x (1+x^2 )^ 1 / 2 d x+ y (1+y^2 )^ 1 / 2 d y=0 On integrating, we get 1+x^2 + 1+y^2 =c