TITLE: General solution to $\frac{df}{dx}=3f$
QUESTION [5 upvotes]: I am learning how to do calculus and was presented with an example I am struggling a bit to understand. Why does $\frac{df}{dx}=3f$ have the general solution of $f(x)=Ce^{3x}$?

REPLY [5 votes]: If we are worried about the hazards of division by $0$, let
$$g(x)=\frac{f(x)}{e^{3x}}.$$
Differentiate, using the Quotient Rule. We get
$$g'(x)=\frac{e^{3x}f'(x)-f(x)(3e^{3x})}{(e^{3x})^2}.\qquad\qquad(\ast)$$
Using $f'(x)=3f(x)$ we can see that the numerator in $(\ast)$ is $0$.
So $g'(x)$ is identically $0$, and therefore $g(x)$ is a constant function $C$. Thus $$C=\frac{f(x)}{e^{3x}},$$
and $f(x)=Ce^{3x}$.