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
$$0=\pm \sqrt{1-c^2_1}$$

Square both sides:
$$1-y^2=(x+c_1)^2$$
rearrange for $y^2$:
$$y^2=1-(x+c_1)^2$$
take $\pm$ square root:
$$y=\pm\sqrt{1-(x+c_1)^2}$$
