What is the type of the following coupled differential equation I want to learn more about systems described by differential equations of the form:
$$\begin{align}f'(x) & = c_0 + c_1 f(x) + c_2 f(x)^2 + \ldots \\ g'(x) &= d_0 \, f(x) + d_1 g(x) f(x) + d_2 f(x) g(x)^2  + \dots\end{align}$$
Both are nonlinear inhomogeneous differential equations, but the point is that the unknown function $g$ doesn't appear in the first equation.  So, it can be solved stepwise: integrate first equation, then substitute answer into second equation to get $g$.
I'd like to learn more about such systems, and how certain differential equations can be brought to this form, and a techincal name/property of this system would helpful.  I'm looking for a word that's something in between "coupled" and "uncoupled".
 A: Such a system is said to be in "triangular form", in analogy with the linear system $Ax=b$, where $A$ is a triangular matrix.  Any system of ordinary differential equations can be brought to this form using a change of variables, but it's not possible to find an explicit transformation in general.  (For PDEs, it's not even possible in principle.)  If $F$ and $G$ are sufficiently simple, you might get lucky.  I will illustrate the procedure below.
To bring a system
\begin{align}
f'(x)&=F(f(x),g(x)),\\
g'(x)&=G(f(x),g(x))
\end{align}
to this form, it suffices to find a change of variables $r=R(f,g),s=S(f,g)$ such that the equation
$$
r'(x)=A(r(x))
$$
is true for some $A$, since then the corresponding system for $r$ and $s$ will be triangular.   Substituting $r=R(f,g)$, we get:
$$
\frac{\partial R}{\partial f}f'(x)+\frac{\partial R}{\partial g}g'(x)=A(R(f,g)).
$$
The original differential equations imply
$$
\frac{\partial R}{\partial f}F(f,g)+\frac{\partial R}{\partial g}G(f,g)=A(R(f,g)).
$$
For a fixed $A$, this is a partial differential equation (PDE) for $R(f,g)$.  By the usual existence theorems, a local solution can be found in principle by the method of characteristics, but an explicit solution is, in general, not possible unless $F$ and $G$ are sufficiently simple.
For example, if $F(f,g)=pf+qg$ and $G(f,g)=uf+vg$ are linear functions, then choosing $R(f,g)=af+bg$ and $A(r)=\lambda r$ is equivalent to solving the matrix eigenvalue problem $$\begin{pmatrix}p&u\\q&v\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\lambda\begin{pmatrix}a\\b\end{pmatrix},$$
which just amounts to diagonalizing your linear system, or placing it in Jordan normal form.  
There is also a nice class of nonlinear examples that arises when demonstrating limit cycles in dynamical systems:
\begin{align}
F(f,g)=f\Phi(f^2+g^2)-g\Psi(f,g),\\
G(f,g)=g\Phi(f^2+g^2)+f\Psi(f,g),
\end{align}
where $\Phi$ and $\Psi$ are arbitrary.  Substituting $R(f,g)=f^2+g^2$ into the PDE gives:
$$
2(f^2+g^2)\Phi(f^2+g^2)=A(f^2+g^2),
$$
so $A(r)=2r\Phi(r)$.  We have thus shown that $r(x)=f(x)^2+g(x)^2$ solves the decoupled ODE:
$$
r'(x)=2r(x)\Phi(r(x)).
$$
Of course, $s(x):=\arctan(g(x)/f(x))$ solves something more complicated.
In this paper, which also discusses PDEs in triangular form, the phenomenon of "triangular form" is called "decoupling", but it's the same idea.   This paper shows how to solve some systems which are already in triangular form.
