The differential equation d y d x = 1 - y 2 y represents the arc of a circle in the second and the third quadrant and passing through - 1 2 , 1 2 . Then, the radius (in units) of the circle is

Step-by-step solution

∫ y 1 - y 2 d y = ∫ d x ⇒ 1 2 ∫ 2 y d y 1 - y 2 = ∫ d x ⇒ - 1 2 ∫ d 1 - y 2 1 - y 2 = ∫ d x ⇒ - 1 - y 2 = x + C As the curves passes through - 1 2 , 1 2 ; C = 0 Thus, we have - 1 - y 2 = x ⇒ 1 - y 2 = x 2 ⇒ x 2 + y 2 = 1 , which is a circle with radius ‘ 1 ’ unit