Solution to $\frac{dy}{dx} = ky $ Solving the differential equation $\frac{dy}{dx} = ky$ gives you $ y = Ce^{kx}$ through separation of variables. I can do that problem without any issue, but is that the only solution? Why doesn't  $ y = C_1e^{kx} + C_2$ also satisfy the equation, as $C_2$ should immediately drop out when differentiating?
 A: Just plug it in:
$$\frac{dy}{dx} = kC_1e^{kx}$$
$$ky = kC_1e^{kx} + kC_2$$
You can see that the two are not equal: the constant factor (which, as you noted, drops out in the derivative term) does not disappear from the $ky$ term.  While I sometimes find that 'seeing' that a constant term will drop out after integration can be a useful way to solve simple differential equations, you usually can't get the general solution just by tacking on a constant factor to a particular solution.
A: Because 
$$y' = ky \implies \frac{y'}{y} = k \implies \ln |y|= kx + \color{#05f}{C_1}\implies e^{\ln |y|} = e^{kx + C_1} \implies |y| = \color{red} C e^{kx} $$
where $C = e^{C_1}$.
Note: The solution on this form causes problem on seeing certain solutions, as $y=0$, $(C = -\infty)$. 
A: The solution is better written as:
$$\log y = kx + C$$
Essentially, since $$\frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}(\log |y|)$$ you are really performing a simple integral for $\log |y|$ and the $C$ comes in that solution. Then solving by $y$ gives you $y=\pm(e^C)e^{kx}=C'e^{kx}$.
This has problems - for example, you have to allow for $C=-\infty$ to allow for $y=0$ as an answer. 
But the point is that there isn't a second constant involved - the transition from $\log|y|$ to $y=\dots$ is not an integral, but simply an inverse function. 
A: An o.d.e. has (locally) unique solution as long as you give an initial condition. Given the initial condition, you will be able to evaluate the constant $C_2$. If the initial condition is not given, you can give a "general solution" $y(x)=C_1e^{kx} + C_2$.
A: If $\varphi : I\to\mathbb R$ is a solution, where $I$ is an interval, then
$$
\varphi'(x)=k\,\varphi(x)\quad\Longleftrightarrow\quad \mathrm{e}^{-kx}
\big(\varphi'(x)-k\,\varphi(x)\big)=0 \quad\Longleftrightarrow\quad 
\big(\mathrm{e}^{-kx}\varphi(x)\big)'=0,
$$
and equivalently there exists a constant $c$, such that $\mathrm{e}^{-kx}\varphi(x)=c$ or equivalently
$\varphi(x)=c\mathrm{e}^{kx}$.
Note, that the above expression is obtain via equivalences.
A: Noting that 
$$
\frac{d}{dx}(e^{-kx}y(x))=e^{-kx}(y'(x)-k·y(x))=0
$$
the last per the differential equation, one sees that the modified differential equation only has constant solutions so that
$$
e^{-kx}y(x)=C\implies y(x)=C·e^{kx}
$$
are the only solutions.
A: suppose you have another solution $g \ne Ce^{kx}$ for any $k$ satisfying $$g'(x) = kg(x) $$ in some open interval $(a,b).$ 
consider 
$$y = g(x)e^{-kx}.$$ if you take the derivative, you find that $$y' = g'(x)e^{-kx}-ke^{-kx} g(x) = (g'-kg)e^{-kx} = 0 \text{ for  } a < x < b.$$
then the mean value theorem tells you that $$y = constant \implies g(x) = constant  \times e^{kx}$$ leading to contradiction. and that proves that the unique solution of $$y' = ky  $$ is $$y = Ce^{kx}. $$
