Why does the general solution of $y'=y$ not covering $y=0?$ It is regarding the ODE $y'=y.$ 
Usually we try to find its general solution using separation of variables as 
$\frac{dy}{y}=dx\implies\log y=x+c\implies y=e^{x+c}$ ($c$ being arbitrary constants).
Please tell me why does the general solution does not cover the case $y=0.$
 A: Reason: your method "separation of variables".
If you write $\frac{dy}{y}$, then you should have $y \ne 0$. Furthermore $ \log$ is only defined for positive values.
A: So the way to get all the solutions is to write this as $y'-y=0$ and treat it as a linear equation with trial solution $y=e^{\alpha x}$ from which we obtain $\alpha=1$.
As a linear equation we know that the solutions are linear combinations of the relevant trial solutions and therefore $y=Ae^x$ is the general form.
Note that the form $y=e^{x+c}$ does not cover solutions with $y\lt 0$ either.

Another route which finds a general solution is to take $$z=ye^{-x}$$ then $$z'=y'e^{-x}-ye^{-x}=\text{ [y'=y] }0$$ So $z=A$ is constant and $y=Ae^x$. This can be used to justify the trial solution method, but by coming down to $z'=0$ it reverts to basic theorems in calculus to establish existence and uniqueness.
Separation of variables is a computational convenience requiting care to avoid implicit assumptions.
A: The general solution of $y'=y$ is : 
$$y(x)=C\:e^x$$ 
any value of $C$. 
Especially in case $C=0$ we get $y(x)=0$. So this particular solution isn't forgotten in the above general solution.
Your solution $y=e^{x+c}$ is uncomplete because $e^{x+c}>0$ which forgets the solutions negative or nul. 
With you method of solving, at first you should write "The solutions other than $y(x)=0$ ". Then you are allowed to divide by $y$. And at end include the solution $y=0$ into the set of functions found. Also you should write $\ln|y|$ instead of $\ln(y)$ in order to not forget the negative functions. 
Your result will be :
$$y(x)=\begin{cases}
e^ce^x \\
0 \\
-e^ce^x
\end{cases}$$
which can be gathered into $\quad y(x)=C\:e^x$.
A: In the first step, you divided by $y.$
A: Generally, when solving ODEs, you are looking for a non-trivial solution; i.e. $y$ that is not identically zero. Note that the trivial solution ($ y = 0$) is quite often a valid solution, but it just isn't interesting from either a mathematical or a practical point of view. Moreover, as others have mentioned, the method breaks down if we do not assume that $y$ is not zero.
A: $y'=y$ and hence $dy/dx=y$
We need to look at two cases:
When $y=0$ and y$\neq0$
When y=0, the solution is trivially 
$y'=0$. Integration yields $y=A=0$.
Therefore $y=0$ is a solution 
