Griffiths Intro to QM Section 9.1.2: What type of approximation is he using here and what is the justification for it? I really do not understand Griffiths logic in this section and was wondering if someone could help. This is basically a 1st order coupled system of  ordinary differential equations but I haven't seen an approximation like this before.
For brevity $f(t)=-\frac{i}{\hbar}H'_{ab}e^{-i\omega_0t}$ and $g(t)=-\frac{i}{\hbar}H'_{ba}e^{i\omega_0t}$ so the system of equations are (from Griffiths equation 9.13)
$$
\begin{align*}
\dot c_a &= f(t)\,c_b\\
\dot c_b &= g(t)\,c_a \hspace{10mm} [9.13]
\end{align*}
$$
Note both $c_a$ and $c_b$ are functions of time. Griffiths then goes on to express the derivative of a $n$th order approximation as proportional to the next lower order approximation of the other system variable. 
For example in equation 9.18 Griffiths states
$$
\frac{d c_a^{(2)}}{dt} = f(t) \, c_b^{(1)}  \hspace{10mm} [9.18]
$$
I just don't see how to justify this? (Note: Griffith says his superscript in parentheses indicates the order of the approximation.)
So from my reading of this $c_a^{(n)}$ and $c_b^{(n)}$ are just $n$th order expansions of $c_a$ and $c_b$. So in my mind I am thinking if we are approximating each out to $n$ terms them something like equation 9.18 above should instead be
$$
\frac{d c_a^{(2)}}{dt} = f(t) \, c_b^{(2)}  \hspace{10mm}
$$
In other words we are taking the derivative of $c_a$ and just approximating it out to say 2 terms, then shouldn't we use the same order approximation for $c_b$ in the system of equations [9.13]?
So why can Griffiths do this? Why can we just stick in the lower order approximation to solve for the next one up?
 A: This is a typical perturbative expansion, although presented in a more pedestrian manner. 
What is usually done for convenience of expansion, is to attach to $H'$ a (time-independent) coupling or scale constant, say $H' \rightarrow \lambda H'$, and to make explicit the assumption that solutions are sought as perturbative expansions in $\lambda$:
$$
c_a(t) = c_a^{(0)}(t) + \lambda c_a^{(1)}(t) + \lambda^2 c_a^{(2)}(t) + \dots\\
c_b(t) = c_b^{(0)}(t) + \lambda c_b^{(1)}(t) + \lambda^2 c_b^{(2)}(t) + \dots
$$
When substituted in your eqs.(9.13), these expansions generate a hierarchy of differential equations from the demand that the polynomial expansions hold for arbitrary $\lambda$. That is, first obtain
$$
\dot c_a^{(0)}(t) + \lambda \dot c_a^{(1)}(t) + \lambda^2 \dot c_a^{(2)}(t) + \dots = \lambda f(t)\left[ c_b^{(0)}(t) + \lambda c_b^{(1)}(t) + \lambda^2 c_b^{(2)}(t) + \dots \right]\\
\dot c_b^{(0)}(t) + \lambda \dot c_b^{(1)}(t) + \lambda^2 \dot c_b^{(2)}(t) + \dots = \lambda g(t)\left[ c_a^{(0)}(t) + \lambda c_a^{(1)}(t) + \lambda^2 c_a^{(2)}(t) + \dots \right]
$$
then identify the coefficients of successive powers of $\lambda$:
$\lambda^0$:
$$
\dot c_a^{(0)}(t) = 0\\
\dot c_b^{(0)}(t) = 0
$$
$\lambda$:
$$
\dot c_a^{(1)} = f(t) c_b^{(0)}\\
\dot c_b^{(1)} = g(t) c_a^{(0)}
$$
$\lambda^2$:
$$
\dot c_a^{(2)} = f(t) c_b^{(1)}\\
\dot c_b^{(2)} = g(t) c_a^{(1)}
$$
In general, for $k\ge 1$, 
$$
\dot c_a^{(k)} = f(t) c_b^{(k-1)}\\
\dot c_b^{(k)} = g(t) c_a^{(k-1)}
$$
The rest follows from successively solving lower order eqs. and substituting in the next order set.
