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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

The equation of the curve passing through the point 1,1 and satisfying the differential equation d y d x = x + 2 y - 3 y - 2 x + 1 is

Options

  1. Ax 2 - 4 x y - y 2 + 6 x + 2 y - 4 = 0
  2. Bx 2 + 4 x y - y 2 - 6 x + 2 y + 4 = 0
  3. Cx 2 + 4 x y - y 2 - 6 x - 2 y + 4 = 0
  4. Dx 2 + 4 x y + y 2 - 6 x - 2 y - 4 = 0

Correct answer

C. x 2 + 4 x y - y 2 - 6 x - 2 y + 4 = 0

Step-by-step solution

d y d x = x + 2 y - 3 y - 2 x + 1 y d y - 2 x d y + d y = x d x + 2 y d x - 3 d x y d y - 2 x d y + y d x + d y - x d x + 3 d x = 0 y d y - 2 d x y + d y - x d x + 3 d x = 0 On integrating, we get, y 2 2 - 2 x y + y - x 2 2 + 3 x + C = 0 Since, the curve passes through 1,1 ⇒ 1 2 - 2 + 1 - 1 2 + 3 + C = 0 C = - 2 y 2 - 4 x y + 2 y - x 2 + 6 x - 4 = 0 or x 2 + 4 x y + y 2 - 6 x - 2 y + 4 = 0

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