The differential equation $y'=y+1$ has no singular solution?

Question

$\frac{dy}{dx}=y+1$

Solving the above given differential equation, yields the following general solution. $y+1=e^{x+C}$

$y=Ce^{x}-1$

$\implies$ Solution $y=-1$ at $C=0$

Can I say that $y= -1$ is a singular solution?

Answer 1

$y=-1$ is not a singular solution because it is included in the general solution $y=Ce^x-1$ when $C=0$. Solutions are only called singular when they are not attainable from the general solution form. In fact, I believe all linear first order homogenous equations have no singular solutions.

Answer 2

$y=-1$ is not a singular solution, because it does not pass by any point $(x_0,y_0)$ such that the initial value problem $\begin{cases} y'=y+1\\ y(x_0)=y_0\end{cases}$ has more than one solution: in point of fact, the Cauchy problem $$\begin{cases}y'=y+1\\ y(x_0)=-1\end{cases}$$ has only the one maximal solution $y=-1$, whatever $x_0$ is.