Some Answers and One Question

[This post was missing one of the solutions, it just got updated 12:38 AM. I think I'll call it a night.]

The Laplace transform is a linear operator.

So we can apply the Laplace transform to each term separately.

The first term is the transform of $t^{3}$ but shifted by $c=-2$.

Thus we have $\frac{3!}{(s+2)^{4}}$.

The second term uses the convolution.

The transform of the integral is the product of the transforms.

So $\mathcal{L}\{e^{2t}\}\cdot\mathcal{L}\{\sin3t\}=\frac{1}{s-2}\cdot\frac{3}{s^{2}+9}$

The last term can be done as follows:

$\mathcal{L}\{t^{2}\delta(t-5)\}=\int_{0}^{\infty}t^{2}\delta(t-5)e^{-st}dt=5^{2}e^{-5s}=25e^{-5s}$

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We see a product so the inverse will be the convolution of the two
different pieces.

$\mathcal{L}^{-1}\left\{ \frac{s-3}{(s-3)^{2}+4}\right\} =e^{3t}\cos2t$

$\mathcal{L}^{-1}$$\left\{ \frac{e^{-2s}}{s^{3}}\right\} =\mathcal{L}^{-1}\left\{ e^{-2s}\frac{1}{s^{3}}\right\} $

This first equality makes it look more like so I can use number 13.

$=\mathcal{L}^{-1}\left\{ \frac{1}{2!}e^{-2s}\frac{2!}{s^{3}}\right\} =\frac{1}{2}\mathcal{L}^{-1}\left\{ e^{-2s}\frac{2!}{s^{3}}\right\} =\frac{1}{2}u_{2}(t)(t-2)^{2}$

Thus the inverse transform is

$\int_{0}^{t}e^{3(t-\tau)}\cos2(t-\tau)\cdot\frac{1}{2}u_{2}(\tau)(\tau-2)^{2}d\tau$

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We're looking at \#4 and \#13 on the table on page 317.

You have to get used to using combination of rules like this and learning
how to multiply and divide by the constant you need. In this case
we multiply and divide by $\Gamma(\frac{1}{2})$

Then we have $\mathcal{L}^{-1}(\frac{e^{-2s}}{s^{1/2}})=\frac{1}{\Gamma(\frac{1}{2})}\mathcal{L}^{-1}(e^{-2s}\frac{\Gamma(\frac{1}{2})}{s^{1/2}})=\frac{1}{\Gamma(\frac{1}{2})}u_{2}(t)(t-2)^{-1/2}$

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Compute $e^{At}$ where $A=\left(\begin{array}{cc}
-2 & 1\\
1 & -2\end{array}\right)$