Explain how to solve the following IVP.
\[ 4 \, {y'} + 16 \, {y} = 2 \, {y''} - 12 \, \delta\left(t - 2\right) \hspace{2em} y(0)= 0 , y'(0)= 12 \]
Hint: \( \frac{1}{s^{2} - 2 \, s - 8} = -\frac{1}{6 \, {\left(s + 2\right)}} + \frac{1}{6 \, {\left(s - 4\right)}} \).
Answer:
\[ \mathcal{L}\{y\}= \frac{6 \, e^{\left(-2 \, s\right)}}{s^{2} - 2 \, s - 8} + \frac{12}{s^{2} - 2 \, s - 8} \]
\[ \mathcal{L}\{y\}= -\frac{e^{\left(-2 \, s\right)}}{s + 2} + \frac{e^{\left(-2 \, s\right)}}{s - 4} - \frac{2}{s + 2} + \frac{2}{s - 4} \]
\[ {y} = e^{\left(4 \, t - 8\right)} \mathrm{u}\left(t - 2\right) - e^{\left(-2 \, t + 4\right)} \mathrm{u}\left(t - 2\right) + 2 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-2 \, t\right)} \]