The final result is correct, but there are multiple mistakes in the derivation of the peak time. The derivation should be:
d d t y ( t ) = ζ ω n e − ζ ω n t ( cos ω d t + ζ 1 − ζ 2 sin ω d t ) − e − ζ ω n t ( − ω d sin ω d t + ζ ω d 1 − ζ 2 cos ω d t ) {\displaystyle {\frac {d}{dt}}y(t)=\zeta \omega _{n}e^{-\zeta \omega _{n}t}\left(\cos \omega _{d}t+{\frac {\zeta }{\sqrt {1-\zeta ^{2}}}}\sin \omega _{d}t\right)-e^{-\zeta \omega _{n}t}\left(-\omega _{d}\sin \omega _{d}t+{\frac {\zeta \omega _{d}}{\sqrt {1-\zeta ^{2}}}}\cos \omega _{d}t\right)}
d d t y ( t ) = ζ ω n e − ζ ω n t ( cos ω d t + ζ 1 − ζ 2 sin ω d t ) − e − ζ ω n t ( − ω d sin ω d t + ζ ω n cos ω d t ) {\displaystyle {\frac {d}{dt}}y(t)=\zeta \omega _{n}e^{-\zeta \omega _{n}t}\left(\cos \omega _{d}t+{\frac {\zeta }{\sqrt {1-\zeta ^{2}}}}\sin \omega _{d}t\right)-e^{-\zeta \omega _{n}t}\left(-\omega _{d}\sin \omega _{d}t+\zeta \omega _{n}\cos \omega _{d}t\right)}
d d t y ( t ) = ( ζ 2 ω n 1 − ζ 2 + ω d ) sin ω d t . {\displaystyle {\frac {d}{dt}}y(t)=\left({\frac {\zeta ^{2}\omega _{n}}{\sqrt {1-\zeta ^{2}}}}+\omega _{d}\right)\sin \omega _{d}t.}
