COMEDK2023Evening ShiftMathematicsDifferential EquationsActual
The general solution of the differential equation (1+y^2 ) d x= ( ⁻¹ y-x ) d y
Options
- Ax=c ⁻¹ y+e^ - ⁻¹ y
- Bx= ⁻¹ y+c e^ ⁻¹ y
- Cx= ⁻¹ y-1+c e^ - ⁻¹ y
- Dx= ⁻¹ y-1+c e^ ⁻¹ y
Correct answer
C. x= ⁻¹ y-1+c e^ - ⁻¹ y
Step-by-step solution
The given differential equation is (1+y^2) dx = ( ⁻¹ y - x) dy . Rearranging the terms, we get dx dy = ⁻¹ y - x 1+y^2 . This can be written as dx dy + x 1+y^2 = ⁻¹ y 1+y^2 . This is a linear differential equation of the form dx dy + P(y)x = Q(y) , where P(y) = 1 1+y^2 and Q(y) = ⁻¹ y 1+y^2 . The integrating factor (IF) is e^ P(y) dy = e^ 1 1+y^2 dy = e^ ⁻¹ y . The solution is given by x IF = Q(y) IF dy + c . x e^ ⁻¹ y = ⁻¹ y 1+y^2 e^ ⁻¹ y dy + c . Let u = ⁻¹ y , then du = 1 1+y^2 dy . The integral becomes u e^u du