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JEE Main202118 Mar 2021Evening ShiftMathematicsDifferential EquationsActual

Let y = y ( x ) be the solution of the differential equation d y d x = y + 1 y + 1 e x 2 / 2 - x , 0 < x < 2 . 1 , with y ( 2 ) = 0 . Then the value of d y d x at x = 1 is equal to

Options

  1. A- e 3 / 2 e 2 + 1 2
  2. B- 2 e 2 1 + e 2 2
  3. Ce 5 / 2 1 + e 2 2
  4. D5 e 1 / 2 e 2 + 1 2

Correct answer

A. - e 3 / 2 e 2 + 1 2

Step-by-step solution

Let y + 1 = Y ∴ d Y d x = Y 2 e x 2 2 - x Y Put - 1 Y = k ⇒ d k d x + k - x = e x 2 2 I . F . = e - x 2 2 ∴ k = x + c e x 2 / 2 Put k = - 1 y + 1 ∴ y + 1 = - 1 ( x + c ) e x 2 / 2       . . . i when x = 2 ,   y = 0 , then c = - 2 - 1 e 2 Differentiate equation ( i )   & put x = 1 we get, d y d x x = 1 = - e 3 / 2 1 + e 2 2

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