Solution. \(p(\lambda ) = \lambda ^4 - 2\lambda ^2 + 1 = (\lambda ^2 - 1)^2\), with double roots \(\pm 1\): \(y = (c_1 + c_2t)e^t + (c_3 + c_4t)e^{-t}\). For the second, \(\lambda = 2 \pm 3i\): \(y = e^{2t}(c_1\cos 3t + c_2\sin 3t)\).

Solution. \(y = t^r\) gives \(r(r - 1) - 3r + 4 = r^2 - 4r + 4 = (r - 2)^2\), a double root \(r = 2\). So \(y = c_1t^2 + c_2t^2\log t\).

Solution. For the first, \(\mu = 2i\) is a simple root of \(\lambda ^2 + 4\) and \(q = t\) has degree \(1\), so \(y_p = t\bigl ((A_1t + A_0)\cos 2t + (B_1t + B_0)\sin 2t\bigr )\). For the second, \(p(\lambda ) = \lambda (\lambda - 1)(\lambda + 1)\); the constant \(3\) corresponds to \(\mu = 0\), a simple root, giving \(At\), and \(e^t\) to \(\mu = 1\), a simple root, giving \(Bte^t\); so \(y_p = At + Bte^t\).

Solution. \((s^2 + 1)Y - s = e^{-\pi s}/s\), so \(Y = \frac {s}{s^2 + 1} + e^{-\pi s}\Bigl (\frac 1s - \frac {s}{s^2 + 1}\Bigr )\). Inverting with Theorem 6.4.2(ii), \(y = \cos t + H(t - \pi )\bigl (1 - \cos (t - \pi )\bigr ) = \cos t + H(t - \pi )(1 + \cos t)\). So \(y = \cos t\) for \(t < \pi \) and \(y = 1 + 2\cos t\) for \(t > \pi \): switching on a constant force at the moment of maximum displacement in the negative direction (\(y(\pi ) = -1\)) triples the amplitude of the oscillation about the new equilibrium \(1\).

Solution. By Theorem 6.3.1 with \(G(t) = \sin (\omega t)/\omega \), for \(t > \varepsilon \), \(y_\varepsilon (t) = \frac 1\varepsilon \int _0^\varepsilon \frac {\sin \omega (t - s)}{\omega }\,ds\), the average of \(G(t - s)\) over \(s \in [0, \varepsilon ]\). By continuity of \(G\) this tends to \(G(t)\). In transform terms, \(\frac {1 - e^{-\varepsilon s}}{\varepsilon s} \to 1 = \mathcal L[\delta ]\).

Solution. \(H(s) = 1/(s^2 + 2\zeta \omega _0s + \omega _0^2)\), with poles \(s = -\zeta \omega _0 \pm i\omega _d\) for \(\zeta < 1\). Completing the square, \(H = \frac {1}{\omega _d}\cdot \frac {\omega _d}{(s + \zeta \omega _0)^2 + \omega _d^2}\), which by Theorem 6.4.2(i) and (v) is the transform of \(e^{-\zeta \omega _0t}\sin (\omega _dt)/\omega _d\). For \(\zeta \geq 1\) the poles are real and negative, \(-\omega _0(\zeta \pm \sqrt {\zeta ^2 - 1})\): the response is non-oscillatory, a double pole at \(-\omega _0\) when \(\zeta = 1\).