COMEDK2021MathematicsDifferential Equations
The solution of the differential equation (1+y² )+ (x-e^ ⁻¹ y ) d y d x =0 is
Options
- A2 x e^ ⁻¹ y =e^ 2 ⁻¹ y +C
- Bx e^ ⁻¹ y = ⁻¹ y+C
- Cx e^ 2 ⁻¹ y =e^ ⁻¹ y +C
- D(x-2)=C e^ - ⁻¹ y
Correct answer
A. 2 x e^ ⁻¹ y =e^ 2 ⁻¹ y +C
Step-by-step solution
The given differential equation is (1+y²) + (x - e^ ⁻¹ y ) dy dx = 0 . Rearranging the terms, we get (x - e^ ⁻¹ y ) dy dx = -(1+y²) . Taking the reciprocal, dx dy = - x - e^ ⁻¹ y 1+y² = - x 1+y² + e^ ⁻¹ y 1+y² . This is a linear differential equation of the form dx dy + P(y)x = Q(y) , where P(y) = 1 1+y² and Q(y) = e^ ⁻¹ y 1+y² . The integrating factor (IF) is e^ P(y) dy = e^ 1 1+y² dy = e^ ⁻¹ y . The solution is given by x IF = Q(y) IF dy + C . x e^ ⁻¹ y = e^ ⁻¹ y 1+y² e^ ⁻¹ y dy + C . Let u = ⁻¹ y , then du = 1 1