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.
 A: Hint:
$$\frac{dx}{dt}=x \Leftrightarrow  dt=\frac{1}{x}dx$$
A: $\dfrac{dx(t)}{dt}=x(t)$, so: $\dfrac{dx(t)}{x(t)}=dt$, integrating the two sides you get:$\ln(x(t))-\ln(x(0))=t$ and finally:
$$x(t)=x(0)\exp(t)$$
A: $$x'=x \Rightarrow \frac{dx}{dt} \Rightarrow \frac{1}{x} dx = 1 \ dt \Rightarrow \int \frac{1}{x} dx = \int 1 \ dt \Rightarrow \ln(x)= t + C \Rightarrow x(t)=e^te^c.$$
$\cdot \ \text{Let A}=e^c \ \text{then} \ x(t) = Ae^t$
A: Note that it's unnecessary to treat $dx$ and $dt$ as independent entities ("infinitesimals") when solving a differential equation like this.
One way to solve it (other than just guessing the solution) is this:
\begin{align}
& x'(t) = x(t) \\
\implies & x'(t) - x(t) = 0 \\
\implies & x'(t) e^{-t} - x(t) e^{-t} = 0 \qquad (\heartsuit)\\
\implies & x(t) e^{-t} = C \quad \text{for some } C \in \mathbb R \quad (\spadesuit)\\
\implies & x(t) = Ce^t.
\end{align}
To go from ($\heartsuit$) to ($\spadesuit$), we took antiderivatives of both sides, using the product rule in reverse.
This is a standard trick which can be used to solve any first order linear ODE.
A: 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 $e^{-t}$.
A: I like to remember that $\frac{d}{dt}ln x=\frac{x'}{x}$.  Thus, very quickly we have $$\frac{d}{dt}lnx=1 \rightarrow ln x=t+C \rightarrow x=Ce^t$$
