(4)xy'-y-sqrt(y^2)-x^(2)=0;

将原方程改写为： \[ y' = \frac{y}{x} + \sqrt{\left(\frac{y}{x}\right)^2 - 1}. \] 令 $ t = \frac{y}{x} $，则 $ y = xt $，$ y' = t + x \frac{dt}{dx} $。代入得： \[ t + x \frac{dt}{dx} = t + \sqrt{t^2 - 1}, \] 化简得： \[ x \frac{dt}{dx} = \sqrt{t^2 - 1}. \] 分离变量并积分： \[ \int \frac{dt}{\sqrt{t^2 - 1}} = \int \frac{dx}{x}, \] \[ \ln \left| t + \sqrt{t^2 - 1} \right| = \ln |x| + C, \] \[ t + \sqrt{t^2 - 1} = Cx. \] 代回 $ t = \frac{y}{x} $： \[ \frac{y}{x} + \sqrt{\left( \frac{y}{x} \right)^2 - 1} = Cx, \] \[ y + \sqrt{y^2 - x^2} = Cx^2. \] **答案：** \[ \boxed{y + \sqrt{y^2 - x^2} = Cx^2} \]