Motivating $y'=y \implies y=Ce^x$ Is there some intuitive reason why one should think that a function which is its own derivative should be of the form $Ca^x$ for some number $a$?  Of course I can prove that the unique solution set to $f'(x) = f(x)$ is the set $\{f\mid f(x) = Ce^x\}$, but pretending I don't know this, what would be a good rationale for checking a function of this type?
 A: One way to intuitively see this is to write formally the approximation of the derivative as
$$\frac{dy}{dx}\approx\frac{y(x+\Delta x)-y(x)}{\Delta x} \tag 1$$
Then, using $(1)$ reveals that 
$$\frac{y(x+\Delta x)-y(x)}{\Delta x}\approx y(x)\implies y(x+\Delta x)=(1+\Delta x)y(x) \tag 2$$
Setting $x=0$ in $(2)$ and iterating yields
$$\begin{align}
y(\Delta x)&\approx (1+\Delta x)y(0) \\\\
y(2\Delta x)&\approx (1+\Delta)^2y(0)\\\\
y(n\Delta x)&\approx (1+\Delta x)^ny(0) \tag 3
\end{align}$$
Then, setting $n\Delta x=x_0$ in $(3)$, we find that 
$$y(x_0)\approx \left(1+\frac{x_0}{n}\right)^ny(0) \tag 4$$
Recalling that the limit definition of $e^x$ is $e^x=\lim_{n\to \infty}\left(1+\frac{x}{n}\right)$, taking the limit as $n\to \infty$ of $(4)$, and noting that the approximation in $(4)$ becomes exact as $\Delta x\to 0$ reveals
$$y(x_0)=y(0)e^{x_0}$$
A: Here is one possibility: start with a difference equation instead of a differential equation, and take this simple example:  
$$y(x+1) - y(x) = y(x)$$
or
$$y(x+1) = 2 y(x)$$
Now we have the sequence  
$y(1) = 2 y(0)$
$y(2) = 2 y(1) = 2^2 y(0)$
$y(3) = 2 y(2) = 2^2 y(1) = 2^3 y(0)$
$\dots$
$y(k) = 2^k y(0)$  
This is exactly the form you were looking for with $a = 2$ and $C = y(0)$.
More generally write
$$y(x+h) - y(x) = h y(x)$$
$$y(x+h) = y(x) (1+h)$$
Then we have as before ($k = 1, 2, 3, ...$)
$$y(k h) = (1+h)^k y(0)$$
New let
$x = k*h$ and inserting $k = x/h$   
we have
$$y(x) = y(0) \left((h+1)^{1/h}\right)^x$$
For $h\to 0$ we have
$(h+1)^{1/h} \to e$
so that
$$y(x) = y(0) e^x$$
