BITSAT2021MathematicsDifferential EquationsActual
The general solution of the differential equation ( ⁻¹ y-x ) d y= (1+y² ) d x is
Options
- Ax= ( ⁻¹ y+1 )+ C e^ - ⁻¹ y
- Bx= ( ⁻¹ y-1 )+ C e^ - ⁻¹ y
- Cx= ( ⁻¹ x-1 )+ C e^ - ⁻¹ x
- Dx= ( ⁻¹ x+1 )+ Ce ^ - ⁻¹ x
Correct answer
B. x= ( ⁻¹ y-1 )+ C e^ - ⁻¹ y
Step-by-step solution
The given differential equation can be written as d x d y + x 1+y² = ⁻¹ y 1+y² (i) Now, eq. (i) is a linear differential equation of the form d x d y + P ₁ x= Q ₁ where P ₁= 1 1+y² and Q ₁= ⁻¹ y 1+y² Therefore, I.F =e^ 1 1+y² d y =e^ ⁻¹ y Thus, the solution of the given differential equation is given by x e^ ⁻¹ y = ( ⁻¹ y 1+y² ) e^ ⁻¹ y dy + C (ii) Let I = ( ⁻¹ y 1+y² ) e^ ⁻¹ y dy On substituting ⁻¹ y=t , so that ( 1 1+y² ) dy = dt , we get I= t e^ t d t=t e^ t - 1 e^ t d t=t e^ t -e^ t =e^ t (t-1) or I =e^ ⁻¹ y (