0007.tex

\begin{exerciseStatement}


 Explain how to solve the following IVP. 

\[ 18 \, \delta\left(t - 3\right) = -6 \, {y'} - 24 \, {y} + 3 \, {y''} \hspace{2em}
          y(0)= 0 ,
          y'(0)= -6 \]

Hint: \( \frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}} \).


\end{exerciseStatement}
    
\begin{exerciseAnswer} 
\[
          \mathcal{L}\{y\}= \frac{6 \, e^{\left(-3 \, s\right)}}{s^{2} - 2 \, s - 8} - \frac{6}{s^{2} - 2 \, s - 8} \]\[
          \mathcal{L}\{y\}= -\frac{e^{\left(-3 \, s\right)}}{s + 2} + \frac{e^{\left(-3 \, s\right)}}{s - 4} + \frac{1}{s + 2} - \frac{1}{s - 4} \]\[ {y} = e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - e^{\left(-2 \, t + 6\right)} \mathrm{u}\left(t - 3\right) - e^{\left(4 \, t\right)} + e^{\left(-2 \, t\right)} \]
\end{exerciseAnswer}