Functional equation for scale invariant utility functions

Two utility functions $u,v:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ (giving the utility of, say, an amount of money) are considered equivalent if $u(x)$ is given by $m\,v(x)+c$, for some constants $c$ and $m$ with $m>0$. They then give the same preference ordering on any set of gambles or investments. So we could call a utility function scale invariant if there exist functions $c:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ and $m: \mathbb{R}_{>0}\rightarrow\mathbb{R}_{>0}$ such that $$u(ax)=m(a)\,u(x)+c(a).$$ Which utility functions are scale invariant?

I thought of this question while reading a justification of the Kelly Criterion. It was assumed that a gambler would bet a constant fraction of their wealth, regardless of how much money they currently had.

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What I've attempted so far:

There are solutions $u(x)=x^\beta$ along with their equivalents (i.e. of the form $\alpha x^\beta+\gamma$). We also get $u(x)=\log(x)$ which we can think of as a limit of $x^\beta$ as $\beta\rightarrow 0$ (i.e. it's the pointwise limit of $\frac{x^\beta-1}{\beta}$), and its equivalents. So far these are the only solutions I've found.

As far as solving the equation goes, my first step was to sub in $x=1$ and then $x=1/a$ and use the two resulting equations to eliminate $m$ and $c$. This got too messy to write down. Instead I tried first replacing $u$ by $f(x)=u(x)-u(1)$ so that $f(1)=0$. This eventually gives $$f(ax)f(1/a)-f(a)f(1/a)+f(a)f(x)=0$$ or equivalently $$f(ax)=f(a)+\frac{-f(a)}{f(1/a)}f(x)$$ [which is beginning to look like the Cauchy functional equation $$f(ax)=f(a)+f(x)$$ which has as its solutions $\log(x)$ to different bases (and some pathologies which we can safely ignore since no-one would ever pick them as utility functions).]

Can anyone find the complete set of solutions? Feel free to assume continuity, differentiability etc.

Answer

Answering my own question:

Take the equation $$f(ax)f(1/a)-f(a)f(1/a)+f(a)f(x)=0$$ and differentiate wrt $x$ to give $$af'(ax)f(1/a)+f(a)f'(x)=0$$ and hence $$af'(ax)f(1/a)=-f(a)f'(x).$$ Differentiating a second time gives $$a^2f''(ax)f(1/a)=-f(a)f''(x).$$ Taking the ratio of the last two equations and setting $x=1$ gives $$\frac{f'(a)}{f''(a)}=a\frac{f'(1)}{f''(1)}.$$ So we have reducced the problem to a differential equation. Solving this gives precisely the functions I mentioned in the question. These are precisely the CRRA utility functions.