Solve $f(x)f'''(x)=f'(x)f''(x)$ How can I solve $f(x)f'''(x)=f'(x)f''(x)$ when f(x) is a function from reals to reals? I figured by trial and error that exp(ax) and sin(ax) are solutions, but I don't know a method to find these solutions without guessing.
 A: If you divide both sides by $f$ and $f''$, you get the differential equation 
$$\frac{f'''}{f''} = \frac{f'}{f} $$
The left hand side can be recognized as $(\log f'')'$ and the right hand side can be recognized as $(\log f)'$. Can you take it from here? 
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
Lets $\ds{y \equiv \fermi\pars{x}}$:

\begin{align}
{y''' \over y''} = {y' \over y}\quad\imp\quad \ln\pars{y''} = \ln\pars{y} + \mbox{a constant}
\end{align}

$$
y" - Ay = 0\quad\mbox{where}\quad A\ \mbox{is a constant}
$$

$$
y=\fermi\pars{x} = B\sinh\pars{\root{A}x} + C\cosh\pars{\root{A}x}
$$
  $B$ and $C$ are constants.

