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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

  1. Ay + 1 y + e^x ^2 x = 1
  2. By + y + e^x ^2 x = 2
  3. C(y+1) + e^x ^2 x = 2
  4. 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

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