Explain how to solve the following IVP.
\[ 24 \, \delta\left(t - 1\right) = -3 \, {y''} + 9 \, {y} + 6 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= -16 \]
Hint: \( \frac{1}{s^{2} - 2 \, s - 3} = -\frac{1}{4 \, {\left(s + 1\right)}} + \frac{1}{4 \, {\left(s - 3\right)}} \).
Answer:
\[ \mathcal{L}\{y\}= -\frac{8 \, e^{\left(-s\right)}}{s^{2} - 2 \, s - 3} - \frac{16}{s^{2} - 2 \, s - 3} \]
\[ \mathcal{L}\{y\}= \frac{2 \, e^{\left(-s\right)}}{s + 1} - \frac{2 \, e^{\left(-s\right)}}{s - 3} + \frac{4}{s + 1} - \frac{4}{s - 3} \]
\[ {y} = -2 \, e^{\left(3 \, t - 3\right)} \mathrm{u}\left(t - 1\right) + 2 \, e^{\left(-t + 1\right)} \mathrm{u}\left(t - 1\right) - 4 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-t\right)} \]