The correct solution of the differential equation (dy/dx) + y =1 where y≠1 is-

y= 1 + (1/C) e-x y= 1 -(1/C) e-x y= -1 -(1/C) e-x y= 1 -(1/C) ex

y= 1 -(1/C) e-x

Given that differential equation (dy/dx) + y =1 ⇒ dy = (1-y) dx separating the variables, we get: dy/(1-y) = dx now, integrating both sides we get: ∫dy/(1-y) =∫ dx ⇒ log(1-y) = x + log C ⇒ -log C - log(1-y) = x ⇒ log C(1-y) = -x ⇒ C(1-y)= e-x y= 1 -(1/C) e-x