Second Integral of a function We have a function $f(x)$. We know that $f(0)=0$, that $f(x)=kf^{'}(x)$. How could we approach finding $f(x)$?
 A: This is what is called an 'ordinary differential equation', or ODE.
An algebraic equation consists of a variable which is a number $x$, and the equation gives you a relation the number satisfies, such as
$$ x^2 - 4x = 0$$
Meaning, "which number when you multiply it by itself and 4, and then subtract,  we'll get 0?"
The answer is $x=0 , x=4$ , there are 2 numbers which satisfy the equation above.
But in an ODE, the variable is a function and the equation gives us a relation between the function and its derivatives. For example, "Which function is equal to its derivative?", which can be written as
$$ f'(x) = f(x)$$
One can see that the solution is $f(x)=e^x$, which stays the same no matter how many times you derive it. More precisely, when you multiply $e^x$ by a constant number $A$ the equation still holds, so the solution is 
$$ f(x) = A e^x $$
But how can you solve such equations when you don't 'just know' the answer?
Like regular equations, we know how to solve a very little amount of them, only specific types of ODE's are solvable in analytical methods. 
Fortunately, your equation is one of them.
For any ODE of the following form:
$$ y' = g(x) \cdot y $$
you can solve like this:
$$ \frac{y'}{y} = g(x) $$
$$ (\ln y)' = g(x) $$
$$ \ln y = G(x) + c $$
$$ y = e^{G(x)+c} =e^c \cdot  e^{G(x)} $$
Where $c$ is a constant and $G(x)$ is the anti-derivative of $g(x)$. Because $c$ is an arbitrary number, so is $e^c$, and we can call it $A$, giving the result
$$ y= A \cdot e^{G(x)} $$
The world of ODE's is much wider than you think and is very useful in physics, where you see that the whole world is just a bunch of ODEs and PDEs (Partial Differnetial Equations, which I will not talk about).
I hope this helped!
A: Hint. Assume $k\ne 0$. One may consider
$$
g(x)=e^{-x/k}f(x)
$$ we have, for all $x \in \mathbb{R}$,
$$
g'(x)=-\frac1k[f(x)-k\cdot f'(x)]\cdot e^{-x/k}=0
$$What can be said about $g$?
Can you finish it?
A: From $f(x)= kf'(x)= k\frac{df}{dx}$, we have, $\frac{df}{f}= \frac{1}{k}dx$.  Integrating both sides, $ln(f)= \frac{x}{k}+ C$, where C is the constant of integration.  Take the exponential of both sides to get $f(x)= e^{\frac{x}{k}+ C}= C'e^{\frac{x}{k}}$ where $C'= e^C$.  Since $f(0)= C'e^0= C'= 0$, The solution to this problem is f(x)= 0 for all x.
A: An idea assuming $\;f(x)>0\;$ in some open interval (it is possible to fix this assumption and still work, yet I am trying a rather easy way and avoid absolute values and stuff):
$$f(x)=kf'(x)\implies\frac{f'(x)}{f(x)}=\frac1k\stackrel{\text{integ. in both sides}}\implies\log f(x)=\frac xk+c\implies f(x)=Ce^{x/k} $$
with $\;C\;$ a constant. You can now input the condition $\;f(0)=0\;$ to find the constant...
