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JEE Main202311 Apr 2023Evening ShiftMathematicsDifferential EquationsActual

Let y = y x be the solution of the differential equation d y d x + 5 x x 5 + 1 y = x 5 + 1 2 x 7 , x > 0 . If y 1 = 2 , then y 2 is equal to

Options

  1. A637 128
  2. B679 128
  3. C693 128
  4. D697 128

Correct answer

C. 693 128

Step-by-step solution

Given: d y d x + 5 x 1 + x 5 y = 1 + x 5 2 x 7 This is linear differential equation. I . F . = e ∫ 5 x 1 + x 5 d x ⇒ I . F . = e ∫ 5 x 5 x 6 1 + x 5 d x ⇒ I . F . = e ∫ 5 x 6 x - 5 + 1 d x ⇒ I . F . = e - ∫ - 5 x 6 x - 5 + 1 d x ⇒ I . F . = e - ∫ d x - 5 + 1 x - 5 + 1 ⇒ I . F . = e - log e x - 5 + 1 ⇒ I . F . = x - 5 + 1 - 1 = x 5 x 5 + 1 So, solution is given by y x 5 x 5 + 1 = ∫ x 5 x 5 + 1 × x 5 + 1 2 x 7 d x ⇒ y x 5 x 5 + 1 = &#8

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