0002.tex

\begin{exerciseStatement}


 Explain how to solve the following IVP. 

\[ 4 \, {y} + 2 \, {y''} + 6 \, {y'} - 6 \, \delta\left(t - 3\right) = 0 \hspace{2em}
          y(0)= 0 ,
          y'(0)= 3 \]

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


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