Given the following differential equation:
$$y^2+(2xy+1)\frac{dy}{dx}=0$$
I have found that the solution is given implictly by:
$$ (\ast) \quad xy^2(x)+y(x)=k \: ; k \in \mathbb{R}$$
If I was interested in a domain that doesn't contain $0$, then I'd write that: $$y(x)=\frac{-1 \pm \sqrt{1+4kx}}{2x}$$ and $D_y=]-\frac{1}{4k},0[; k>0 \quad \quad D_y=]0,-\frac{1}{4k}[; k<0$
For $k=0$; $y(x)=0 \lor y(x)=\frac{1}{x}$ would be two solutions for $(\ast)$. $$\\$$ But now if for all $k \in \mathbb{R}$; I was interested in a solution in an interval that contains $0$, what would be the general procedure?
The solution $$ y = \frac{-1 + \sqrt{1 + 4 k x}}{2x} $$ actually has a removable singularity at $x=0$. It can be written in the form $$ y = \frac{2k}{1 + \sqrt{1+4k x}}$$ to make this clear. When $k > 0$ this is defined for $-1/(4k) < x < \infty$, and for $k < 0$ it is defined for $-\infty < x < -1/(4k)$.