Solution. \(A^2 = -I\), so \(A^{2k} = (-1)^kI\) and \(A^{2k+1} = (-1)^kA\). Splitting the series, \(e^{At} = \cos t\,I + \sin t\,A = \begin {bmatrix} \cos t & \sin t\\ -\sin t & \cos t\end {bmatrix}\), a rotation by angle \(-t\). The solutions move clockwise around circles centred at the origin with period \(2\pi \): a centre.

Solution. The system matrix \(\begin {bmatrix} 0 & 1\\ -k & -b\end {bmatrix}\) has \(\tau = -b\) and \(\Delta = k > 0\). If \(b = 0\) the origin is a centre (undamped oscillation). If \(0 < b^2 < 4k\) it is a stable spiral: the underdamped case, decaying oscillations. If \(b^2 = 4k\) it is a degenerate stable node, critical damping, the fastest return without oscillation. If \(b^2 > 4k\) it is a stable node: overdamped, a slow return without oscillation.

Solution. \(A^2 = B^2 = 0\), so \(e^A = I + A = \begin {bmatrix} 1 & 1\\ 0 & 1\end {bmatrix}\) and \(e^B = \begin {bmatrix} 1 & 0\\ 1 & 1\end {bmatrix}\), with \(e^Ae^B = \begin {bmatrix} 2 & 1\\ 1 & 1\end {bmatrix}\). Next \(A + B = \begin {bmatrix} 0 & 1\\ 1 & 0\end {bmatrix}\) has \((A + B)^2 = I\), so \(e^{A+B} = \cosh 1\,I + \sinh 1\,(A + B) = \begin {bmatrix} 1.543 & 1.175\\ 1.175 & 1.543\end {bmatrix}\). The two differ; indeed \(AB - BA = \begin {bmatrix} 1 & 0\\ 0 & -1\end {bmatrix} \neq 0\).

Solution. For \(A = 0\) every solution is constant, hence bounded. For the Jordan block, \(e^{At} = I + tA\), and the solution from \((0, 1)\) is \((t, 1)\), unbounded. This is Theorem 5.3.2(iv): the eigenvalue \(0\) has a Jordan block of size two.

Solution. \(\tau = 2\), \(\Delta = -3 + 8 = 5\), \(\tau ^2 - 4\Delta = -16\): an unstable spiral with eigenvalues \(1 \pm 2i\). An eigenvector for \(1 + 2i\) solves \((2 - 2i)v_1 - 2v_2 = 0\), so \(v = (1, 1 - i)\), with real part \(u = (1, 1)\) and imaginary part \(w = (0, -1)\). The real solutions are \(e^t(u\cos 2t - w\sin 2t) = e^t(\cos 2t, \cos 2t + \sin 2t)\) and \(e^t(u\sin 2t + w\cos 2t) = e^t(\sin 2t, \sin 2t - \cos 2t)\). At \(t = 0\) these are \((1, 1)\) and \((0, -1)\), so \(x(0) = (1, 0)\) is their sum, and \(x(t) = e^t(\cos 2t + \sin 2t,\ 2\sin 2t)\). Check: \(x'(0) = (1 + 2, 4) = (3, 4) = A(1, 0)\).

Solution. With \(\omega _0 = 2\), \[ y = \frac {1}{4 - 3.61}(\cos 1.9t - \cos 2t) = 2.564(\cos 1.9t - \cos 2t) = 5.128\sin (0.05t)\sin (1.95t) . \] The envelope \(5.128|\sin 0.05t|\) has period \(\pi /0.05 = 62.8\), and the maximum amplitude is \(5.13\). At \(\omega = 2\), \(y = \frac t4\sin 2t\), which grows without bound.