Let C 1 be the curve obtained by the solution of differential equation 2 x y d y d x = y 2 - x 2 , x > 0 . Let the curve C 2 be the solution of 2 x y x 2 - y 2 = d y d x . If both the curves pass through 1 , 1 , then the area (in sq. units) enclosed by the curves C 1 and C 2 is equal to :

Step-by-step solution

d y d x = y 2 - x 2 2 x y ,    x ∈ 0 , ∞ put y = v x x d v d x + v = v 2 - 1 2 v 2 v v 2 + 1 d v = - d x x Integrate, l n v 2 + 1 = - l n x + C l n y 2 x 2 + 1 = - l n x + C put x = 1 , y = 1 , C = l n 2 l n y 2 x 2 + 1 = - l n x + l n 2 ⇒ x 2 + y 2 - 2 x = 0 (Curve C 1 ) Similarly, d y d x = 2 x y x 2 - y 2 Put y = v   x x 2 + y 2 - 2 y = 0 Required area = 2 ∫ 0 1 2 x - x 2 - x d x = π 2 - 1 sq. units