[tex]y' = \frac{dy}{dx} [/tex]
seperable differential equations will have the form
[tex]\frac{dy}{dx} = F(x) G(y)[/tex]
what you do from here is isolate all the y terms on one side and all the X terms on the other
[tex]\frac{dy}{G(y)} = F(x) dx[/tex]
just divided G(y) to both sides and multiply dx to both sides
then integrate both sides
[tex]\int \frac{1}{G(y)} dy = \int F(x) dx
[/tex]
once you integrate, you will have a constant. use the initial value condition to solve for the constant, then try to isolate x or y if the question asks for it
In your problem,
[tex]G(y) = \sqrt{1-y^2}
F(x) = 2x[/tex]
so all you need to integrate is
[tex]\int \frac{1}{\sqrt{1-y^2}} dy = \int 2x dx[/tex]
