xy'=(1-y^2)^(1/2)
write y' into dy/dx then we get
x * (dy/dx) = (1-y^2)^(1/2)
Then we rearrange the equation, let x,dx and y,dy on the same side.
we get 1/x * dx = (1-y^2)^(-1/2) * dy
by taking integral on both side,
ln|x|+C = arcsin(y)
therefore y = sin(ln|x|+C)
