Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 2 / 7B / Solution

Write $A=aI+B$, where $B=\left(\begin{array}{cc}1& -2\\ 1& -1\end{array}\right)$ has $B^2=-I$. One eigenpair isBecause $A$ is real, conjugating $Ae=\ell e$ gives $A\bar{e}=\bar{\ell }\bar{e}$. The eigenvector matrix $(e,\bar{e})$ has determinant $2i\ne 0$, so the eigenvectors form a basis. Reality of $z$ then forces conjugate coefficients, giving $z=\alpha e+\bar{\alpha }\bar{e}$.

Set $c=(1-i)/2$. The forcing has the same basis decomposition,Equating the coefficient of $e$ in the linear system of differential equations yieldsFor $a>0$, an integrating factor gives $\alpha =(c/a)e^{-it}+C{e}^{-(a+i)t}$. The zero initial vector means $\alpha (0)=0$, soTaking the real vector combination givesExact secular resonance occurs at $a=0$: the forcing frequency matches the undamped homogeneous mode. For positive $a$, damping prevents secular growth, and the late-time periodic amplitude is proportional to $1/a$. Thus there is a resonant small-damping limit, but no unbounded resonant response at fixed $a>0$.

Since $(1-e^{-at})/a\to t$ for each fixed $t$, the requested limit givesEquivalently $\alpha =ct{e}^{-it}$ at zero damping; its linear prefactor displays the resonance explicitly.