Solution. \(\Phi (t) = \exp \int _0^ta\), so \(\rho = \exp \int _0^Ta\) and \(\mu = \bar a = \frac 1T\int _0^Ta\), the mean of \(a\). Then \(p(t) = \exp \bigl (\int _0^ta - \bar at\bigr )\), which is \(T\)-periodic because \(\int _t^{t+T}a = T\bar a\). A non-zero periodic solution exists if and only if \(\rho = 1\), that is the mean of \(a\) is zero; stability depends only on the sign of the mean.

Solution. \(y_1 = \cos \omega t\), \(y_2 = \sin (\omega t)/\omega \), so \(\Delta = \cos 2\pi \omega + \cos 2\pi \omega = 2\cos 2\pi \omega \), and \(\rho = e^{\pm 2\pi i\omega }\). Since \(|\Delta | \leq 2\), the equation is always stable, with \(|\Delta | = 2\) exactly when \(2\omega \) is an integer, \(\delta = \omega ^2 = n^2/4\). There the multipliers are \(\pm 1\), double, and an arbitrarily small periodic perturbation \(\varepsilon \cos t\) can move them off the unit circle onto the real axis, which opens a tongue.

Solution. If \(e^{BT} = M\), then \(B' = B + \frac {2\pi i}{T}kI\) also satisfies \(e^{B'T} = Me^{2\pi ik} = M\), and its eigenvalues are shifted by \(2\pi ik/T\). The corresponding \(P'(t) = \Phi (t)e^{-B't} = P(t)e^{-2\pi ikt/T}\) is still \(T\)-periodic. So the decomposition \(\Phi = Pe^{Bt}\) trades a factor \(e^{2\pi ikt/T}\) between the periodic part and the exponential part. The real parts of the exponents, which govern stability, are unique.

Solution. By variation of parameters, \(x(T) = Mx_0 + c\) with \(c = \int _0^T\Phi (T)\Phi (s)^{-1}g(s)\,ds\). The solution is \(T\)-periodic if and only if \(x(T) = x_0\): by uniqueness it then repeats, since the equation is \(T\)-periodic. So we need \((I - M)x_0 = c\), which has the unique solution \(x_0 = (I - M)^{-1}c\) when \(1\) is not an eigenvalue of \(M\). If all multipliers lie inside the unit circle, this periodic solution attracts every other solution, so a stable periodically forced system settles into a periodic response.

Solution. The condition \(a^2\omega ^2 > 2g\ell \) gives \(\omega ^2 > 2(9.81)(0.2)/10^{-4} = 39\,240\), so \(\omega > 198\) rad/s, about \(31.5\) Hz. Then \(\varepsilon = a/\ell = 0.05\) and \(\delta = -g/(\ell \omega ^2) = -9.81/(0.2 \times 39\,240) = -0.00125 = -\varepsilon ^2/2\), on the boundary, as it should be. Any faster shaking moves \(\delta \) closer to zero, into the stable strip.

Solution. The trace is \(\tr A = -2 + \frac 32(\cos ^2t + \sin ^2t) = -\frac 12\). For the determinant, write \(c = \cos t\), \(s = \sin t\): \begin {align*} \det A &= \bigl (-1 + \tfrac 32c^2\bigr )\bigl (-1 + \tfrac 32s^2\bigr ) + \bigl (1 - \tfrac 32cs\bigr )\bigl (1 + \tfrac 32cs\bigr )\\ &= 1 - \tfrac 32 + \tfrac 94c^2s^2 + 1 - \tfrac 94c^2s^2 = \tfrac 12 . \end {align*}

Both are independent of \(t\), and the eigenvalues solve \(\lambda ^2 + \frac 12\lambda + \frac 12 = 0\), so \(\lambda = \frac 14(-1 \pm i\sqrt 7)\), with negative real parts. For the solution, the first component of \(Ax\) is \[ e^{t/2}\bigl [(-1 + \tfrac 32c^2)(-c) + (1 - \tfrac 32cs)s\bigr ] = e^{t/2}\bigl (c + s - \tfrac 32c\bigr ) = e^{t/2}\bigl (-\tfrac 12c + s\bigr ) , \] which is \(\frac {d}{dt}(-e^{t/2}\cos t)\); the second component is similar. The solution grows like \(e^{t/2}\) although every frozen-time eigenvalue has negative real part: stability of a time-varying system cannot be read from the eigenvalues of \(A(t)\).