# \frac{dy}{dx}=y+1 Solving the above given differential equation, yields the following

$\frac{dy}{dx}=y+1$ Solving the above given differential equation, yields the following general solution. $y+1={e}^{x+C}$
$y=C{e}^{x}-1$
$⇒$ Solution $y=-1$ at $C=0$
Can I say that $y=-1$ is a singular solution?
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timbreoizy
Step 1
$y=-1$ is not a singular solution because it is included in the general solution $y=C{e}^{x}-1$ 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.
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Giancarlo Brooks
Step 1
$y=-1$ is not a singular solution, because it does not pass by any point such that the initial value problem
$\left\{\begin{array}{l}{y}^{\prime }=y+1\\ y\left({x}_{0}\right)={y}_{0}\end{array}$
has more than one solution: in point of fact, the Cauchy problem
$\left\{\begin{array}{l}{y}^{\prime }=y+1\\ y\left({x}_{0}\right)=-1\end{array}$
has only the one maximal solution $y=-1$, whatever ${x}_{0}$ is.