solution of DE :- $\frac{dy}{dx}=y$ Solve the following DE -$$\frac{dy}{dx}=y$$
Quite simple right ? $y=ce^x$ is the solution where c is an arbitrary constant .
But if we observe y=0 is also a solution of this DE which is not obtained while solving it (just a mere observation) . Why does this happen that the variable seperable method misses this solution?
Will we call this as general solution , particular solution or singular solution ??
 A: It is an interesting question.
If you solve through the integrating factor method, you are going to observe the particular solution $y\equiv 0$ is included. Indeed, if we multiply both sides by $\exp(-x)$, we arrive at
\begin{align*}
y' = y & \Longleftrightarrow y' - y = 0\\\\
& \Longleftrightarrow \exp(-x)y' - \exp(-x)y = 0\\\\
& \Longleftrightarrow [\exp(-x)y]' = 0\\\\
& \Longleftrightarrow \exp(-x)y = k\\\\
& \Longleftrightarrow y = k\exp(x)
\end{align*}
and we are done.
On the other hand, if we solve it through the separation of variables method, we have to suppose that $y\not\equiv 0 $. That being the case, we are allowed to divide both sides by $y$ in order to get:
\begin{align*}
y' = y & \Longleftrightarrow \frac{y'}{y} = 1\\\\
& \Longleftrightarrow \ln|y| = x + c\\\\
& \Longleftrightarrow |y| = \exp(x+c)\\\\
& \Longleftrightarrow |y| = \exp(c)\exp(x)\\\\
& \Longleftrightarrow y = \pm\exp(c)\exp(x)\\\\
& \Longleftrightarrow y = k\exp(x) 
\end{align*}
where $k\in\mathbb{R}\backslash\{0\}$.
However, if we notice (as you did) that $y\equiv 0$ is also a solution, we can assume that $k\in\mathbb{R}$.
Then both methods hold the same solution to the proposed DE.
Hopefully this helps!
