0003.tex

\begin{exerciseStatement}


 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: \( \frac{1}{s^{3} + 9 \, s} = -\frac{s}{9 \, {\left(s^{2} + 9\right)}} + \frac{1}{9 \, s} \).


\end{exerciseStatement}
    
\begin{exerciseAnswer} 
\[
          \mathcal{L}\{y\}= -\frac{6}{s^{2} + 9} - \frac{9 \, e^{\left(-s\right)}}{{\left(s^{2} + 9\right)} s} \]\[
          \mathcal{L}\{y\}= \frac{s e^{\left(-s\right)}}{s^{2} + 9} - \frac{e^{\left(-s\right)}}{s} - \frac{6}{s^{2} + 9} \]\[ {y} = \cos\left(3 \, t - 3\right) \mathrm{u}\left(t - 1\right) - 2 \, \sin\left(3 \, t\right) - \mathrm{u}\left(t - 1\right) \]
\end{exerciseAnswer}