6(b)

We take $y = \mbox{Im} z$ where $z'' + 2 z' + 2 z = 5 \exp(it)$. The trial solution $z = C \exp(it)$ results in $(-1 + 2 i + 2) C \exp(it) = 5 \exp(it)$, so $C = 5/(1+2i)$. We write this as $C = A \exp(-i\phi)$ where $A = \vert C\vert = 5/\sqrt{1^2 + 2^2} = \sqrt{5}$, and $\exp(i\phi) = A/C = (1+2i)/\sqrt{5}$. Thus $\cos \phi = 1/\sqrt{5}$ and $\sin \phi = 2/\sqrt{5}$, so $\phi = \arcsin(2/\sqrt{5}) \approx 1.107148718$. The steady-state solution is $y = \sqrt{5} \sin(t-\phi)$, with amplitude $\sqrt{5}$.

Alternatively: if the inhomogeneous term was $g(t) = 5 \cos t$, the steady-state solution would be $y_p(t) = A \cos(t-\phi)$. Our inhomogeneous term is $5 \sin t = 5 \cos(t - \pi/2) = 5 g(t-\pi/2)$, so our steady-state solution will be $y_p(t-\pi/2) = A \cos(t-\pi/2 - \phi) = A \sin(t-\phi)$. We have $\omega_0 = \sqrt{k/m} = \sqrt{2}$, $\Delta = \sqrt{m^2 (\omega_0^2 - \omega^2)^2 + \gamma^2 \omega^2} = \sqrt{5}$, $\sin \phi = \gamma \omega/\Delta = 2/\sqrt{5}$ and $\cos \phi = m(\omega_0^2 - \omega^2)/\Delta = 1/\sqrt{5}$, which leads to the same solution as above.