Difference between $f(x(t))$ and $f(t,x)$

Question

Why is there a difference between the two differential equations:

$\overset{.}{x}(t)=f(x(t))$ and $\overset{.}{x}(t)=f(t,x)$ ?

Answer 1

When I see $x(t)$, I think of it as a path (or particle) moving depending on $t$ (time).

$f(x(t))$ is a function on the position of the particle only. Therefore, even if the particle returns to the same point more than once, the value of $f(x(t))$ is the same every time. An example of such a function is $f(x(t))=2x(t)$.

On the other hand, $f(t,x(t))$ means that $t$ is an input to the function as well. Therefore, if $x(t_1)=x(t_2)$, then $f(t_1,x(t_1))$ might not be the same as $f(t_2,x(t_2))$ because the function can depend on $t$. For example, you might have $f(t,x(t))=2tx(t)$.

Answer 2

So you have a parameter $t$, a function $x$ which depends on $t$. In your first equation you have a function $f$ which does something to $x$, which you can then express as a function of $t$ by substituting that in (although since you don't know $x(t)$, this is a moot point). In the second equation, you have a function of two variables, $x$ and $t$, and you can do anything you want to $x$ or $t$ (within reason).

This is quite important if you want to actually solve the equation. The first is only as hard to solve as $1/f(x)$ is to integrate. The second can be much harder to solve in general.

Answer 3

There is much difference. The first only depends on $x$ (which is parameterized by $t$), so $t$ does not appear explicitly in the equation. The second one can be time-dependent (if $t$ is interpreted as time).

In physics, the difference would be something like this:

The first equation describes a phenomenon, where the velocity of an object only depends on its position. The second equation can have arbitrary driving forces present, that change in time and space.

If there was a second derivative on the left, the difference would mean that in the first case, the energy is conserved and in the second it probably isn't (autonomous versus driven system).

Answer 4

As already pointed out in the comments, there is a difference between both expression. Here are two relevant links you can check out. But first let's state the following:

You can always transform a system of $\overset{.}{x}(t)=f(t,x(t))$ into a system of $\overset{.}{x}(t)=f(x(t))$ by going $1$ dimension higher, it is a quite basic procedure.

Your second one $\overset{.}{x}(t)=f(t,x(t))\equiv\overset{.}{x}(t)=f(t,x)$ describes a general ordinary differential equation of order $1$.

Your first one $\overset{.}{x}(t)=f(x(t))$ is a special case of the second one, a so called autonomous system or time-invariant system.