Explain how to solve the following IVP.
\[ -12 \, {y} - 3 \, {y''} - 12 \, \mathrm{u}\left(t - 2\right) = 0 \hspace{2em} y(0)= 5 , y'(0)= 0 \]
Hint: \( \frac{1}{s^{3} + 4 \, s} = -\frac{s}{4 \, {\left(s^{2} + 4\right)}} + \frac{1}{4 \, s} \).
Answer:
\[ \mathcal{L}\{y\}= \frac{5 \, s}{s^{2} + 4} - \frac{4 \, e^{\left(-2 \, s\right)}}{{\left(s^{2} + 4\right)} s} \]
\[ \mathcal{L}\{y\}= \frac{s e^{\left(-2 \, s\right)}}{s^{2} + 4} + \frac{5 \, s}{s^{2} + 4} - \frac{e^{\left(-2 \, s\right)}}{s} \]
\[ {y} = \cos\left(2 \, t - 4\right) \mathrm{u}\left(t - 2\right) + 5 \, \cos\left(2 \, t\right) - \mathrm{u}\left(t - 2\right) \]