0005.tex

\begin{exerciseStatement}


 Explain how to solve the following IVP. 

\[ -3 \, {y''} = 18 \, {y} - 15 \, {y'} - 6 \, \delta\left(t - 3\right) \hspace{2em}
          y(0)= 0 ,
          y'(0)= 4 \]

Hint: \( \frac{1}{s^{2} - 5 \, s + 6} = -\frac{1}{s - 2} + \frac{1}{s - 3} \).


\end{exerciseStatement}
    
\begin{exerciseAnswer} 
\[
          \mathcal{L}\{y\}= \frac{2 \, e^{\left(-3 \, s\right)}}{s^{2} - 5 \, s + 6} + \frac{4}{s^{2} - 5 \, s + 6} \]\[
          \mathcal{L}\{y\}= -\frac{2 \, e^{\left(-3 \, s\right)}}{s - 2} + \frac{2 \, e^{\left(-3 \, s\right)}}{s - 3} - \frac{4}{s - 2} + \frac{4}{s - 3} \]\[ {y} = 2 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) - 2 \, e^{\left(2 \, t - 6\right)} \mathrm{u}\left(t - 3\right) + 4 \, e^{\left(3 \, t\right)} - 4 \, e^{\left(2 \, t\right)} \]
\end{exerciseAnswer}