Generalize this notation for an ODE? I have

$$
\frac{dN(t)}{dt}=a(t)N(t) \tag 1
$$
where $N(0)=N_0 $ (constant).

Question 1:
From the information in $(1)$ I assume I have the functions $a, N:\mathbb R\rightarrow \mathbb R$? And $N_0\in \mathbb R$? Is that correct?
Suppose I now want to generalize the notation in $(1)$ by replacing the right hand side with a function $f$.
Question 2:
Does this mean I should write
$$
\frac{d N(t)}{dt}=f(t) \tag 2
$$
where $N, f:\mathbb R\rightarrow \mathbb R$ ?
Or maybe
$$
\frac{dN(t)}{dt}=f(a(t),N(t)) \tag 3
$$
where $f:\mathbb R \times \mathbb R \rightarrow \mathbb R$?
Or maybe
$$
\frac{dN(t)}{dt}=f(t,a(t),N(t)) \tag 4
$$
where $f:\mathbb R \times \mathbb R \times \mathbb R \rightarrow \mathbb R$ and $t \in \mathbb R$?
$\times$ is the Cartesian product.
Which one is correct? Thanks!
An explicit first order ODE has the following form (see wiki)
$$
\frac{dN(t)}{dt} = f(t, N(t)).
$$
The form (2) is too narrow, since does not allow dependence on $N(t)$ in the right hand side. Form (3) does not allow dependence on raw $t$, and the form (4) has extra $a(t)$ which may be easily built in the definition of $f(t, N)$ itself.
Few examples:

$\frac{dN(t)}{dt} = g(t)$ is a special case of $f(t, N) = g(t)$
$\frac{dN(t)}{dt} = a(t) N(t)$ is a special case of $f(t, N) = a(t) N$
$\frac{dN(t)}{dt} = \frac{a(t)}{N(t) + t}$ is a special case of $f(t, N) = \frac{a(t)}{N + t}$
An explicit first order ODE has the following form (see wiki)
$$
\frac{dN(t)}{dt} = f(t, N(t)).
$$
The form (2) is too narrow, since does not allow dependence on $N(t)$ in the right hand side. Form (3) does not allow dependence on raw $t$, and the form (4) has extra $a(t)$ which may be easily built in the definition of $f(t, N)$ itself.
Few examples:

$\frac{dN(t)}{dt} = g(t)$ is a special case of $f(t, N) = g(t)$
$\frac{dN(t)}{dt} = a(t) N(t)$ is a special case of $f(t, N) = a(t) N$
$\frac{dN(t)}{dt} = \frac{a(t)}{N(t) + t}$ is a special case of $f(t, N) = \frac{a(t)}{N + t}$