Problem in Differential Equation (How to proceed?) $$y\frac{dy}{dx}=\sqrt{1-y^2}$$, y=0 when x=0
$$\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}=1$$
$$\int\frac{\frac{dy}{dx}y}{\sqrt{1-y^2}}dx=\int 1dx$$
$$-\sqrt{1-y^2}=x+c_1$$
$$1-y^2=(x+c_1)^2$$
$$y^2=1-(x+c_1)^2$$
$$y=\pm \sqrt{1-(x+c_1)^2}$$
When x=0, y=0
$$0=\pm \sqrt{1-c^2_1}$$

$$\int_0^y \frac{y\, dy}{\sqrt{1-y^2}} = \int_0^x dx \to \sqrt{1-y^2} - 1=x$$ solving it gives you $$y = \pm \sqrt{2x-x^2}$$ is the circle or radius $1$ centered at $(1,0).$
$\bf note:$ use the definite integral whenever possible.
