0000.tex

\begin{exerciseStatement}


 Explain how to solve the following IVP. 

\[ 2 \, {y} = -2 \, {y''} - 2 \, \mathrm{u}\left(t - 3\right) \hspace{2em}
          y(0)= 0 ,
          y'(0)= -2 \]

Hint: \( \frac{1}{s^{3} + s} = -\frac{s}{s^{2} + 1} + \frac{1}{s} \).


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