Singular solution to $(x+2y)y'=1$

Singular solution to $(x+2y)y'=1$

I have a problem and I got most of the solution, but don't understand how
to proceed. The problem is to solve:
$$(x+2y)y'=1, \qquad y(0)=-1.$$
Here is my reasoning:
Substitute $z = x+2y$. Then $z'=1+2y'$. So $y'=\frac{z'-1}{2}$. So
$\frac{z'-1}{2} = \frac{1}{z}$. So $z' = \frac{2+z}{z}$. So
$\frac{dz}{dx}$ ($x$ taken without loss of generality) = $\frac{2+z}{z}$.
So $\frac{zdz}{2+z} = dx$. So $z - 2\ln|z+2| = x+C$. Therefore, $$(x+2y) -
2\ln|x+2y+2| = x+C.$$
Am I correct that there is no member of this one-parameter family of
solutions that satisfies the initial condition $y(0)=-1$, because plugging
in does not work?
Also, is there a singular solution, and how to find it??
Thanks.