Let us rename your parameters in order to write the equation in a more usual form: $$m\ddot{x}+c\dot{x}+F(x)=0 \qquad m>0,c\geq 0$$ For a suitable restoring force $F(x)$ to force the system to exert a harmonic motion.
For the sake of simpicity let the system exert small oscillations and therefore the function $F(x)$ can be expanded close to its minimum $F(x)\approx kx$ with $k> 0$. Let us also divide your equation by the mass $m$, defining two new quantities: $$\gamma=\frac{c}{m}\qquad \omega^2 = \frac{k}{m}$$ Hence $$\ddot{x}+\gamma\dot{x}+\omega^2f(x)=0\tag1$$ where $\omega^2$ is written here for convenience.$$\ddot{x}+\gamma\dot{x}+\omega^2x=0\tag1$$
Multiplying $(1)$ by $\dot{x}$ we have $$\frac{1}{2}\frac{d}{dt}\left(\dot{x}^2+\omega^2x^2\right)=-\gamma\dot{x}\dot{x}\tag2$$ We can say that if $\gamma$ is zero the energy of the system $$E=\frac{1}{2}\left(\dot{x}^2+\omega^2x^2\right)$$ is conserved. Note that the RHS of $(2)$ is strictly negative throughout the motion.
It is clear that $(2)$ is the generalisation of the equation $(1)$ you proposed. Answering your question, any function that is strictly positive once multiplied by $\dot{x}$, throughout the motion may serve for this purpose.
Therefore if the energy $E$ must decrease throughout the motion it is mandatory that the friction coefficient be an even function in $\dot{x}$ $$\gamma = \gamma_0+\gamma_2\dot{x}^2+\gamma_4\dot{x}^4+...$$
Hope this helps
P.S. The solution you gave for constant $F$ is wrong, if $F=constant$, changing variables to $x=y-F/kt$ results that y must verify $$\frac{d}{dt}\left(m\dot{y}+ky\right) = 0$$ Solving for $y$ we have $$m\dot{y}+ky=constant \rightarrow y=c_0 + c_1\exp{(-k/mt)}$$ Being $$x = c_0 + c_1\exp{(-k/mt)}-F/kt$$