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MHT CET2019Evening ShiftMathematicsDifferential EquationsActual

The particular solution of the differential equation log d y d x = x , when x =0, y = 1 is …..

Options

  1. Ay = e x + 2
  2. By = - e x
  3. Cy = - e x + 2
  4. Dy = e x

Correct answer

D. y = e x

Step-by-step solution

We have, differential equations, log ⁡ d y d x = x ⇒ d y d x = e x ⇒ d y = e x d x Integrating on both sides, we get ∫ d y = ∫ e x d x ⇒ y = e x + C … (i) On putting x = 0, y = 1 is Eq. (i), we get 1 = e 0 + C ⇒ C = 0 Now, particular solution of the given differential is y = e x

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