Weird ordinary differential equation I am searching for the functions $f,g:[0,\infty) \to [0,\infty)$ which are both increasing, $f'$ is strictly increasing, $g$ is the inverse of $f$ and
$$ g(x)^2=f(x)^2\cdot g'(x)$$
My approach was to get rid of $f$ by plugging $x=g(x)$ and obtain something which contains only $g$:
$$ (g\circ g)^2(x) = x^2 \cdot g' \circ g(x)$$
but this seems to lead nowhere. Do you have some ideas?
I can't even find an example, since $f(x)=x$ doesn't work because $f'$ should be strictly increasing...
 A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$$
\totald{}{x}\bracks{1 \over {\rm g}\pars{x}} = -\,{1 \over \fermi^{2}\pars{x}}
\quad\imp\quad
{1 \over {\rm g}\pars{x}} = -\int{\dd x \over \fermi^{2}\pars{x}} + \mbox{a constant}.
$$
