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NTA Abhyas JEE Main2020MathematicsDifferential EquationsPractice

The equation of the curve lying in the first quadrant, such that the portion of the x -axis cut-off between the origin and the tangent at any point P is equal to the ordinate of P , is (where, c is an arbitrary constant)

Options

  1. Ay = c e x y
  2. By e x y = c
  3. Cy e y x = c
  4. Dy = c e y x

Correct answer

B. y e x y = c

Step-by-step solution

Equation of the tangent at P x , y is Y - y = d y d x X - x Putting Y = 0 , we get, X = x - y y ' = y ∴ d x d y - x y = - 1 is a linear differential equation. Ι . F . = e - ln ⁡ y = 1 y Thus, the solution is x 1 y = ∫ - 1 1 y d y or x y = - ln ⁡ y + ln ⁡ c ⇒ c y = e x y Hence, y e x y = c is the equation of the curve

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