Solution
= Solution
Expand the secular <Poynting–Robertson drag> rates using $(1-e^2)^{-3/2}=1+\tfrac32e^2+O(e^4)$ and $(1-e^2)^{-1/2}=1+O(e^2)$. Through second order in <orbital eccentricity>,
$$
\boxed{\dot a_{\rm pr}=-\frac{2A}{a}(1+3e^2)+O(e^4),\qquad
\frac{d(e^2)}{dt}\bigg|_{\rm pr}=-\frac{5A}{a^2}e^2+O(e^4).}
$$
The second formula follows from $d(e^2)/dt=2e\dot e$. The cubic correction in $\dot e$ contributes only at fourth order in $d(e^2)/dt$.