Differential Equation : $f '' = f '$ I came across this question from my textbook, but in the text it only talked about $f ''+f =0$ when discussing trig functions. Now this questions at the end of the log and exp chapter with no discussion of it in this chapter. If anybody could please help.
It says: Find all functions f which are twice differentiable and satisfy the equation $f '' = f '$.
 A: The ODE is a homogeneous linear OE with constant coefficients. First set the equation $m^2-m=0$ and then the general solution will be $$y_c=C_1x^0+C_2\exp(x)$$
A: Suppose that $f'' \equiv  0$. Then, we have that
$$f' = 0  \implies f(x) = c$$
for some constant $c$. If $f \not \equiv 0$ then rewrite your relation as:
$$\displaystyle \frac{f'}{f''} = 1$$
Integrating both sides, you obtain:
$$\displaystyle \int \frac{f'}{f''} = \int 1 \implies \log(|f'|) = x + c \implies |f'| = e^{x+c} \implies f' = e^{x+c}$$ since $e^{x+c} > 0 \ \forall x \in \mathbb{R}$. Note that $c \in \mathbb{R}$ is some arbitrary constant. You can in fact rewrite the above as 
$$f' = Ce^x \ \ \text{where} \ C = e^c$$
Re-integrating:
$$\int f' = C\int e^x \implies f = Ce^x + a$$
So any function satisfying the given differential equation must be of the form $f(x) = Ce^x + a$
A: From this it implies,
$$\frac{df'(x)}{dx}=f'(x)$$
$$\Rightarrow \frac{df'(x)}{f'(x)}=dx$$
Integrating we have,
$$\ln(f'(x))=x+c$$
$$\Rightarrow f'(x)=e^x.e^c$$
$$\Rightarrow c_1(\frac{df(x)}{dx})=e^x$$
$c_1=1/e^c$
$$c_1d(f(x))=e^xdx$$
Integrating we have,
$$c_1f(x)=e^x+c_2$$
$$\Rightarrow f(x)=c_3e^x+c_2$$
$c,c_1,c_2,c_3$ are integration constant.
A: @Chris Taylor and @Orest Xherija I think that the proof is not rigorous since we can't write $$\frac{f'}{f''}=1$$ if we do not suppose  that $f''$ does not vanish. To avoid this problem and without using the characteristic polynomial of differential equation I suggest:
We pose $u=f'$ and $v=e^{-x}u$, so we can easily verify that $v'=0$ and that $v$ is a constant say $c_1$, and we conclude  that $u=c_1 e^x$ and with integration $f=c_1 e^x+ c_2$.
