0011.tex

\begin{exerciseStatement}


 Explain how to solve the following IVP. 

\[ -12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em}
          y(0)= 5 ,
          y'(0)= 0 \]

Hint: \( \frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s} \).


\end{exerciseStatement}
    
\begin{exerciseAnswer} 
\[
          \mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s} \]\[
          \mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s} \]\[ {y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right) \]
\end{exerciseAnswer}