What can we say about p(t) and q(t) if we suppose the linear differential equation $\dot{x}+p(t)x=q(t)$ is autonomous? This is a problem from MIT OCW 18.03SC, Differential Equations. The answer is simply that "p(t) and q(t) must be constants". That was indeed a possibility that came to my mind, but it is not clear to me that that is the only possibility.
$$\dot{x}=q(t)-p(t)x=f(x,t)$$
Is it true that if the equation is autonomous then $f$ is a function that only changes with x, but not t?
My first attempt was
$$f_t(x,t)=\dot{q}-\dot{p}x-p\dot{x}=0$$
Though I think that in fact this does  not have to be zero because of the $\dot{x}$ term.
If I were to continue with this reasoning I'd get
$$\dot{x}=\frac{\dot{q}}{p}-\frac{\dot{p}}{p}x$$
We already have an equation for $\dot{x}$, so can I equate the two:
$$\frac{\dot{q}}{p}-\frac{\dot{p}}{p}x=q-px$$
$$\implies p=\frac{\dot{p}}{p},\ q=\frac{\dot{q}}{p}$$
At this point, given that I am studying linear equations I don't think I know how to proceed. I see a topic called system of differential equations. Is this one of them?
Is there a mistake in this entire line of reasoning? Is the answer to the original question in fact as simple as $p(t)$ and $q(t)$ must be constant, and if so how is this actually proved?
 A: 
$$f(x,t)=q(t)-p(t)x$$
$$f_t(x,t)=\dot{q}-\dot{p}x-p\dot{x}$$

In this differentiation, you think of $x$ as a function of $t$. However, $x$ and $t$ are two independent variables here. $f$ takes in any pair of numbers, $x$ and $t$, and takes them to another value. To show that $f$ doesn't depend on $t$, we fix $\bf{x}$, and show that $f$ doesn't change with $t$.
To build more intuition, think of the $t$-$x$ plane. $f$ is defined everywhere on its domain on this plane. The solutions to the ODE, $x(t)$, on the other hand give just a curve on this plane, such that at each point the rate of change in $x$ with respect to $t$ on this curve is equal to the value of $f$ at that point of the plane. But you can always keep $t$ constant and change $x$ to obtain the value of $f$ at other points on the plane.
So basically, $\frac{\partial x}{\partial t}$ should be $0$ in your calculation:
$$f_t(x,t)=\dot{q}-\dot{p}x=0$$
And this must hold for all constant $x$. You can put in any two values for $x$ to show $\dot{q}=\dot{p}=0$
