Need explanation for simple differential equation I can't figure out this really simple linear equation:
$$x'=x$$
I know that the result should be an exponential function with $t$ in the exponent, but I can't really say why. I tried integrating both sides but it doesn't seem to work. I know this is shameful noob question, but I would be grateful for any hints.

Proof using only differentiation: Try the family of functions
$$
x(t) = x(0) e^t
$$ 
as solution (just differentiate)
$$
x'(t) = x(0) \frac{d}{dt} e^t = x(0) e^t = x(t)
$$
and proof the uniqueness. 
This is usually done by assuming a second solution $y(t)$ with
$$
y'(t) = y(t)
$$
then
$$
\frac{d}{dt} \frac{y(t)}{x(t)} = 
\frac{y'(t) x(t) - y(t) x'(t)}{x^2(t)} =
\frac{y(t) x(t) - y(t) x(t)}{x^2(t)} =
0
$$
for $x(t) \ne 0$, which implies
$$
y(t) = C \, x(t) \\
y(0) = C \, x(0)
$$
for some constant $C$, thus if a second solution exists, it is unique
if it has the same initial condition 
otherwise it is of the same form as the first one, 
only differing by the multiplicative constant. 
This proof is derived from Grönwald's lemma.
Proof using Integration: 
$$
x'(t) = x(t) \iff \\
0 = 
x'(t) - x(t) = 
\left( x'(t) - x(t) \right) \, e^{-t} =
\frac{d}{dt} \left( x(t) \, e^{-t} \right) \iff \\
x(t) \, e^{-t} = C \iff \\
x(t) = C \, e^t
$$
This one used the integrating factor $e^{-t}$.
