Let $c \in \mathbb{C}$.
How to find a general form of $f=f(x),g=g(x) \in \mathbb{C}[x]$ that satisfy $f'(x)g(x)-f(x)g'(x)=c$?
I think I can solve this algebraically by writing $f=a_nx^n+\cdots+a_1x+a_0$ and $g=b_mx^m+\cdots+b_1x+b_0$, where $n,m \geq 0$, $a_i,b_j \in \mathbb{C}$, with $a_nb_m \neq 0$.
Then $c=f'g-fg'=(a_nx^n+\cdots+a_1x+a_0)'(b_mx^m+\cdots+b_1x+b_0)-(a_nx^n+\cdots+a_1x+a_0)(b_mx^m+\cdots+b_1x+b_0)'=
(na_nx^{n-1}+\cdots+a_1)(b_mx^m+\cdots+b_1x+b_0)-(a_nx^n+\cdots+a_1x+a_0)(mb_mx^{m-1}+\cdots+b_1)=a_nb_m(n-m)x^{n+m-1}+\cdots$.
There are two cases:
(1) $n+m-1=0$ and $n-m \neq 0$: Then $n+m=1$ and $n-m \neq 0$.
This implies $n=1,m=0$ or $n=0,m=1$.
Therefore, $f=a_1x+a_0, g=b_0$ with $a_1b_0=c$
or $f=a_0, g=b_1x+b_0$ with $-a_0b_1=c$.
(2) $n+m-1 > 0$, hence $n-m=0$, so $n=m$,
and also we should check all the other conditions until $a_1b_0-a_0b_1=c$.
This yields, for example:
$f=x+a, g=x+b$ with $b-a=c$.
However, I wonder if there is a solution involving differential equations.
I wish to integrate both sides, but do not know how:
Divide both sides of $f'(x)g(x)-f(x)g'(x)=c$ by $g^2(x)$ ($c \neq 0$ implies that $g(x) \neq 0$), so $(\frac{f}{g})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}=\frac{c}{g^2(x)}$. But now I am not sure if it will help multiplying both sides by $-g'(x)$ since this will 'ruin' the LHS.
Perhaps this will help somehow.
Any help in solving this equation not algebraically is welcome; thank you!
Edit: In case (2) I have not brought all the details; here is an example where $n=m=2$: $f=x^2+ax+b$ and $g=x^2+cx+d$, there are no such $f,g$ of degree $2$ (the original $c$ is now changed to $\tilde{c}$, since we already use $c$ as one of the coefficients of $g$):
$\tilde{c}=f'g-fg'=(c-a)x^2+2(d-b)x+(ad-bc)$,
hence: $c=a$, $d=b$ and $ad-bc=\tilde{c} \neq 0$,
but applying the first and second conditions in the third yield:
$0 \neq \tilde{c}=ad-bc=ab-ba=0$, a contradiction.
Similarly, I have checked $n=m=3$, writing $f=x^3+ax^2+bx+c$,
$g=x^3+dx^2+ex+f$, and again the last condition, based on former conditions, yields a contradiction.
So it seems that a general form is necessarily that of case (1):
$f=a_1x+a_0, g=b_0$ with $a_1b_0=c$
or $f=a_0, g=b_1x+b_0$ with $-a_0b_1=c$.
(Since case (2) is impossible).
Best Answer
Let $f=f(x), g=g(x)$, then as you've mentioned in the post,
$$\frac{f'g-fg'}{g^2} = \left(\frac{f}{g}\right)'$$
So
$$f'g-fg'=\left(\frac{f}{g}\right)' g^2 = c \in \mathbb{C} \setminus \{0\}$$
We can further manipulate this equation
$$\begin{align*} \left(\frac{f}{g}\right)' g^2 &= c \\ \left(\frac{f}{g}\right)' &= \frac{c}{g^2} \\ \frac{f}{g} &= \int \frac{c}{g^2} dx + k \\ f(x) &= g(x) \left(\int \frac{c}{g(x)^2} dx + k\right) \end{align*}$$
Where $k \in \mathbb{R}$ is any constant.
These were equivalent transformations, which means we have obtained a formula for $f$ using only $g$.
Therefore, for any function $g$, for which $\frac{1}{g^2}$ is integrable, we have obtained the all pairs of solutions
$$\boxed{\left(\underbrace{g(x) \left(\int \frac{c}{g(x)^2} dx + k\right)}_{f(x)}, \ g(x)\right), \ \ k\in\mathbb{R}}$$
Meaning that the above formula can generate all solutions for any input function $g$ (for which $\frac{1}{g^2}$ is integrable).
Examples (with $k=0$):
This means that the "general form of $f$ and $g$ that solve $f'g-fg'=c$", is "anything", as at least one of $f$ or $g$ can be any function (as long as the reciprocal of its square is integrable - but for our most common functions, that holds). However once one of $f$ or $g$ is decided, the other function only has one degree of freedom (represented by $k\in\mathbb{R})$.