So recently I have been scrolling through Youtube (mainly to find math videos for entertainment, I'll attempt a question on my own every now and then) when I came across this video by Michael Penn solving the differential and functional equation$$f'(x)=f(1/x)$$which I thought might be a nice challenge for me. Here is my attempt:
Right away we can see that $f'(1)=f(1)$ from plugging in $x=1$. But, this didn't really tell me anything, however I decided to keep this in mind for later.
To make some actual progress we can see that$$\begin{align}f''(x)=\dfrac d{dx}f'(x)=\dfrac d{dx}f(1/x)\\=f'(1/x)\cdot\dfrac d{dx}\dfrac1x=-\dfrac{f(x)}{x^2}\end{align}$$Therefore$$f''(x)=-\dfrac{f(x)}{x^2}$$Now, we can set up 2 differential equations that our function must satisfy:$$x^2y''+y=0$$and$$y'(1)=y(1)$$I'll try solving our first one first: Plugging in $y=x^r$,we have$$x^2x^{r-2}(r^2-r)+x^r=0\\\implies x^r(r^2-r)+x^r=0\\x^r(r^2-r+1)=0\\\implies r^2-r+1=0$$So solving our quadratic equation gets us$$r=\dfrac{1\pm i\sqrt3}2$$We now let$$r_1=\dfrac{1+i\sqrt3}2,r_2=\dfrac{1-i\sqrt3}2\implies y=c_1x^{r_1}+c_2x^{r_2}$$which we can write as[1]$$x^{r_1}=\sqrt x\left(\cos\left(\dfrac{\sqrt3}2\ln x\right)+i\sin\left(\dfrac{\sqrt3}2\ln x\right)\right)=z\sqrt x\\x^{r_2}=\sqrt x\left(\cos\left(\dfrac{\sqrt3}2\ln x\right)-i\sin\left(\dfrac{\sqrt3}2\ln x\right)\right)=\overline z\sqrt x$$which is where I am currently stuck because I am unsure about what to do from here. So, my question is:
Is $f'(x)=f(1/x)$ solvable, or am I doing math manipulations with no real meaning?
[1]I decided to write everything that was being multiplied by $\sqrt x$ as $z$ and $\overline z$ for $r_1$ and $r_2$ respectively just so I didn't have to deal with the huge mess that the expression was. Also note that I am leaving how I expanded $x^{r_1}$ and $x^{r_2}$ as an exercise for the reader.
Everything you're doing is fine, although I would apply the known initial(ish) condition $f'(1) = f(1)$ before expanding everything out.
Doing this, we get
$$\begin{eqnarray} f(x) & = & c_1 x^{r_1} + c_2 x^{r_2} \\ f\left(\frac{1}{x}\right) & = & c_1 x^{-r_1} + c_2 x^{-r_2} \\ f'(x) & = & c_1 r_1 x^{r_1 - 1} + c_2 r_2 x^{r_2 - 1} \\ c_1 r_1 x^{r_1 - 1} + c_2 r_2 x^{r_2 - 1} & = & c_1 x^{-r_1} + c_2 x^{-r_2} \end{eqnarray}$$
However, noticing that $-r_1 = r_2 - 1$ and vice versa, we can match coefficients to discover that $c_1 = r_2 c_2$ and $c_2 = r_1 c_1$ (these are actually redundant, since $r_1 r_2 = 1$. If we set $c = c_1$ and $r = r_1$ (choosing one of the roots arbitrarily), then that means we can write our solution as
$$f(x) = c_1 x^{r_1} + c_1 r_1 x^{r_2} = c \left(x^r + r x^{1-r}\right)$$
which you can then rearrange to get a final solution similarly to your approach. With a bit of extra work you can write it in a form that doesn't include any imaginary terms.