For $t\gg\tau$, the free term has disappeared. Exactly at $A=g_1$, the resonant contribution tends to
$$
\boxed{z_1(t)=i\tau\nu_1e^{i(g_1t+\beta_1)}},
$$
with finite eccentricity $|\nu_1|\tau$ and a quarter-cycle phase shift. Without damping, exact resonance instead gives
$$
z_1(t)=e^{iAt}(C+i\nu_1t e^{i\beta_1}),
$$
whose <secular resonance> amplitude grows linearly in the ideal linear theory.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 3 / vi / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-316/3/vi/solution