Some Questions and Some Review Part 1

I issue the warning to take what I say with a grain of salt.

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Find the Laplace transform of

$f(t)=t^{3}e^{-2t}+\int_{0}^{t}e^{2s}\sin3(t-s)ds+t^{2}\delta(t-5)$

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Compute the inverse Laplace transform of

$\frac{s-3}{(s-3)^{2}+4}\cdot\frac{e^{-2s}}{s^{3}}$

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Compute the inverse Laplace transform of

$\frac{e^{-2s}}{s^{1/2}}$
[CORRECTION: It should be -2s instead of 2s]

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Practice using Laplace transforms to solve initial value problems:
pg 320 \#11-23, pg 343 \#1-12, etc.

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The key relationship regarding the fundamental matrix $\Phi$ with
an arbitrary fundamental matrix $\Psi$ is $\Phi(t)=\Psi(t)\Psi^{-1}(t_{0})$.

Thus if you need to compute $\Phi$, simply compute $\Psi(t_{0})$,
then compute it's inverse $\Psi^{-1}(t_{0})$, and then comptue the
product, $\Psi(t)\Psi^{-1}(t_{0})$.

For a two-by-two matrix, you can remember the formula for the inverse
of a matrix $\left(\begin{array}{cc}
a & b\\
c & d\end{array}\right)$ is $\frac{1}{ad-bc}\left(\begin{array}{cc}
d & -b\\
-c & a\end{array}\right)$.

In general you can consider the augmented matrix and row reduce. For
the two-by-two case we have:

$\left(\begin{array}{ccccc}
a & b & | & 1 & 0\\
c & d & | & 0 & 1\end{array}\right)$ which when we row reduce the left block to the identity we get the
inverse on the right block.

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I don't know if the problem will come up, but if asked, what is $\exp(At)$?

Then the answer will be the fundamental matrix $\Phi$ that satisfies
$\Phi^{\prime}=A\Phi$, $\Phi(0)=I$ (page 417).

Find this $\Phi$by first getting $\Psi$ from the solutions for $x^{\prime}=Ax$.
Then $\Phi(t)=\Psi(t)\Psi^{-1}(0)$

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Let's talk about two-by-two matrices $A$. If $A$ has a repeated
eigenvalue $\lambda$ with two eigenvectors implies $A$ is of the
form $\left(\begin{array}{cc}
\lambda & 0\\
0 & \lambda\end{array}\right)$. For a repeated eigenvalue $\lambda$ with only one eigenvector ,
then we have two solutions to the equation $x^{\prime}=Ax$, one of
the form $\xi e^{2t}$ and the second of the form $\xi te^{2t}+\eta e^{2t}$
where $(A-\lambda I)\xi=0$ and $(A-\lambda I)\eta=\xi$.

Note: $(A-\lambda I)\xi=0$ implies that $(A-\lambda I)\eta=\xi$
has a solution.

Sidenote: What I was thinking about in class was related, but not
useful for computation. $(A-\lambda I)^{2}\xi=(A-\lambda I)\eta=0$.

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I'm gonna be back in about four hours, so until then, hopefully you
can give yourself problems.