MHT CET202520 Apr 2025Evening ShiftMathematicsDifferential EquationsActual
The solution of (1+y^2 )+ (x- e ^ ⁻¹ y ) dy d x =0 is
Options
- A2 x e ^ ⁻¹ y = e ^ 2 ⁻¹ y + k , where k is the constant of integration
- Bx e ^ ⁻¹ y = e ^ ⁻¹ y + k , where k is the constant of integration
- Cx e ^ 2 ⁻¹ y = e ^ ⁻¹ y + k , where k is the constant of integration
- Dx=2+ k e ^ - ⁻¹ y , where k is the constant of integration
Correct answer
A. 2 x e ^ ⁻¹ y = e ^ 2 ⁻¹ y + k , where k is the constant of integration
Step-by-step solution
The differential equation is given as (1 + y^2) + (x - e^ ⁻¹ y ) dy dx = 0 . Rearranging to express in terms of dx dy yields: dx dy + 1 1 + y^2 x = e^ ⁻¹ y 1 + y^2 . This is linear in x with P(y) = 1 1 + y^2 and Q(y) = e^ ⁻¹ y 1 + y^2 . The integrating factor is e^ P(y) , dy = e^ ⁻¹ y , leading to the solution: x e^ ⁻¹ y = e^ 2 ⁻¹ y dy 1 + y^2 + C . Let u = ⁻¹ y , so du = dy 1 + y^2 , simplifying the integral to e^ 2u , du = 1 2 e^ 2u = 1 2 e^ 2 ⁻¹ y . Thus: x e^ ⁻¹ y = 1 2 e^ 2 ⁻¹ y + C . Multiplying by 2 and lett