Hi, I'm not sure how Luke Titus would answer, but since "autonomous" is a fairly standardized concept across mathematics, I'll share my take. Let's say the independent variable is
$t$, as in your first example
$y'(t) = 5 + y(t)$. Then for a first-order ODE, if the equation can be expressed in terms of
$y(t)$ and
$y'(t)$ with no other appearance of
$t$, then the ODE is autonomous. For higher-order equations, the ODE may contain
$y''(t)$,
$y^{(3)}(t)$, etc., by no other instances of t.
If we change the independent variable to
$x$, then
$y'=f(y)$ is autonomous, but
$y'=f(x)$ is not (unless
$f(x)$ is a constant).
Another characterization involves rearranging the ODE in the form
$F(t,y,y')=0$, or
$y'(t)-5-y(t)=0$ for the original example. We then consider
$F(t,y_0,y_1) $,
or
$y_1-5-y_0$ for the example. If no
$t$ appears in this formula, then the ODE is autonomous.
Finally, if one wants to push abstraction a little further, we can characterize "autonomous" as follows:
If there exists a function
$G(y_0,y_1)$ such that
$F(t,y,y')=G(y,y')$ for all
$t$, then
$F$ is autonomous. The function
$G$ shows that there is no dependence on
$t$ other than the dependence on
$y(t)$ and
$y'(t)$. This is an abstraction of a common way of defining autonomous: If we write
$y$ for
$y(t)$ and
$y'$ for
$y'(t)$ in the ODE, then the ODE is autonomous if the differential equation does not depend on
$t$; that is, in the example, there is no
$t$ in
$y' = 5+y$.
