Explain how to solve the following IVP.
\[ 3 \, {y''} = -27 \, {y} - 81 \, \mathrm{u}\left(t - 2\right) \hspace{2em} y(0)= -2 , y'(0)= 0 \]
Hint: \( \frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s} \).
Answer:
\[ \mathcal{L}\{y\}= -\frac{2 \, s}{s^{2} + 9} - \frac{27 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 9\right)} s} \]
\[ \mathcal{L}\{y\}= \frac{3 \, s e^{\left(-2 \, s\right)}}{s^{2} + 9} - \frac{2 \, s}{s^{2} + 9} - \frac{3 \, e^{\left(-2 \, s\right)}}{s} \]
\[ {y} = 3 \, \cos\left(3 \, t - 6\right) \mathrm{u}\left(t - 2\right) - 2 \, \cos\left(3 \, t\right) - 3 \, \mathrm{u}\left(t - 2\right) \]