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JEE Main20204 Sep 2020Evening ShiftMathematicsDifferential EquationsActual

The solution of the differential equation d y d x - y + 3 x log e ( y + 3 x ) + 3 = 0 is (where C is a constant of integration)

Options

  1. Ax - 1 2 log e ( y + 3 x ) 2 = C
  2. Bx - log e ( y + 3 x ) = C
  3. Cy + 3 x - 1 2 log e x 2 = C
  4. Dx - 2 log e ( y + 3 x ) = C

Correct answer

A. x - 1 2 log e ( y + 3 x ) 2 = C

Step-by-step solution

   d y d x - y + 3 x ln ( y + 3 x ) + 3 = 0 d y d x + 3 = y + 3 x ln ( y + 3 x ) d d x ( y + 3 x ) = y + 3 x ln ( y + 3 x ) ∫ ln ( y + 3 x ) ( y + 3 x ) d ( y + 3 x ) = ∫ d x Let ln ( y + 3 x ) = t 1 ( y + 3 x ) d ( y + 3 x ) = d t ∫ tdt = ∫ dx t 2 2 = x + c ( ln ( y + 3 x ) ) 2 2 = x + c

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