Explain how to solve the following IVP.
\[ -8 \, \delta\left(t - 3\right) = -2 \, {y''} - 24 \, {y} + 14 \, {y'} \hspace{2em} y(0)= 0 , y'(0)= 1 \]
Hint: \( \frac{1}{s^{2} - 7 \, s + 12} = -\frac{1}{s - 3} + \frac{1}{s - 4} \).
Answer:
\[ \mathcal{L}\{y\}= \frac{4 \, e^{\left(-3 \, s\right)}}{s^{2} - 7 \, s + 12} + \frac{1}{s^{2} - 7 \, s + 12} \]
\[ \mathcal{L}\{y\}= -\frac{4 \, e^{\left(-3 \, s\right)}}{s - 3} + \frac{4 \, e^{\left(-3 \, s\right)}}{s - 4} - \frac{1}{s - 3} + \frac{1}{s - 4} \]
\[ {y} = 4 \, e^{\left(4 \, t - 12\right)} \mathrm{u}\left(t - 3\right) - 4 \, e^{\left(3 \, t - 9\right)} \mathrm{u}\left(t - 3\right) + e^{\left(4 \, t\right)} - e^{\left(3 \, t\right)} \]