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JEE Advanced2015MathematicsDifferential EquationsActual

Consider the family of all circles whose centers lie on the straight line y = x . If this family of circles is represented by the differential equation P y " + Q y ′ + 1 = 0 , where P , Q are functions of x , y a n d y ′ h e r e y ′ = d y d x , y " = d 2 y d x 2 , then which of the following statements is (are) true ?

Options

  1. AP = y + x
  2. BP = y - x
  3. CP + Q = 1 - x + y + y ′ + y ′ 2
  4. DP - Q = x + y - y ′ - y ′ 2

Correct answer

B. P = y - x

Step-by-step solution

Let required circle be x 2 + y 2 + 2 g x + y + c = 0 Differentiation 2 x + 2 y y ′ + 2 g 1 + y ′ = 0 ⇒ x + y y ′ + g 1 + y ′ = 0 ...(i) Again differentiation 1 + y y ′ + y 2 + g y ′ = 0 ...(ii) Putting value of g in equation (ii), 1 + y y ′ + y ′ 2 - x + y y ′ 1 + y ′ y ′ ′ = 0 So, P = y - x , Q = 1 + y ′ + y ′ 2 i.e., P + Q = 1 - x + y + y ′ y ′ 2

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