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COMEDK2023Evening ShiftMathematicsDifferential EquationsActual

The particular solution of e^ d y d x =2 x+1 given that y=1 when x=0 is

Options

  1. Ay=(x+1) |2 x+1|-x+1
  2. By= (x+ 1 2 ) |2 x+1|- 1 2 x+1
  3. Cy= (x- 1 2 ) |2 x+1|-x-1
  4. Dy= (x+ 1 2 ) |2 x+1|-x+1

Correct answer

D. y= (x+ 1 2 ) |2 x+1|-x+1

Step-by-step solution

Given the differential equation e^ dy dx = 2x + 1 . Taking the natural logarithm on both sides, we get dy dx = (2x + 1) . Integrating both sides with respect to x , we have y = (2x + 1) dx . Let u = 2x + 1 , then du = 2 dx , which implies dx = 1 2 du . Substituting these into the integral, y = 1 2 (u) du . Using integration by parts, (u) du = u (u) - u + C . Thus, y = 1 2 (u (u) - u) + C = 1 2 ((2x + 1) (2x + 1) - (2x + 1)) + C . y = (x + 1 2 ) (2x + 1) - x - 1 2 + C . Given y = 1 when x = 0 , substituting these va

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