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