MHT CET202615 April 2026Evening ShiftMathematicsDifferential EquationsActual
The solution of the differential equation e^ -x (y+1) d y + ( ^2 x - 2x)y , d x = 0 , given that y = 1 when x = 0 is
Options
- Ay + 1 y + e^x ^2 x = 1
- By + y + e^x ^2 x = 2
- C(y+1) + e^x ^2 x = 2
- D(y + 1 y ) + e^x ^2 x = 1
Correct answer
B. y + y + e^x ^2 x = 2
Step-by-step solution
The given differential equation is e^ -x (y+1) d y + ( ^2 x - 2x)y , d x = 0 Separating the variables, we get: y+1 y d y + e^x( ^2 x - 2x) d x = 0 (1 + 1 y ) d y + e^x( ^2 x - 2x) d x = 0 Integrating both sides: (1 + 1 y ) d y + e^x( ^2 x - 2x) d x = C Using the standard integral e^x(f(x) + f'(x)) d x = e^x f(x) , where f(x) = ^2 x and f'(x) = - 2x , we get: y + y + e^x ^2 x = C Given that y = 1 when x = 0 , substituting these values: 1 + 1 + e^0 ^2 0 = C 1 + 0 + 1 = C C = 2 Therefore, the solution is: y + y + e^x