Explain how to solve the following IVP.
\[ 0 = 3 \, {y''} + 27 \, {y} + 27 \, \mathrm{u}\left(t - 1\right) \hspace{2em} y(0)= 0 , y'(0)= -6 \]
Hint: \( \frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s} \).
Answer:
\[ \mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s} \]
\[ \mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9} \]
\[ {y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right) \]