JEE Main2003MathematicsDifferential EquationsActual
The solution of the differential equation (1+y^2 )+ (x-e^ ⁻¹ y ) d y d x =0 , is
Options
- Ax e^ 2 ⁻¹ y =e^ ⁻¹ y +k
- B(x-2)=k e^ 2 ⁻¹ y
- C2 x e^ ⁻¹ y =e^ 2 ⁻¹ y +k
- Dx e^ ⁻¹ y = ⁻¹ y+k
Correct answer
C. 2 x e^ ⁻¹ y =e^ 2 ⁻¹ y +k
Step-by-step solution
(1+y^2 )+ (x-e^ ⁻¹ y ) d y d x =0 (1+y^2 ) d x d y +x=e^ ⁻¹ y d x d y + x (1+y^2 ) = e^ ⁻¹ y (1+y^2 ) I.F. =e^ 1 (1+y^2 ) d y =e ⁻¹ yx (e^ ⁻¹ y )= e^ ⁻¹ y 1+y e^ ⁻¹ y d y x (e^ ⁻¹ y )= e^ 2 ⁻¹ y 2 +C 2 x e^ ⁻¹ y =e^ 2 ⁻¹ y +k