Relax and Other

Hey All,

So it was brought to my attention the page loads slowly. This was my mistake as I forgot to cut the posts short when they're viewed from the main page. So seven long posts are trying to load. I've added the HTML so that you have to click on the post if you want to see the full post. This will help with load times.

In any case, relax a little before the test and listen to some Demetri Martin (comedian):


And Flight of the Conchords:

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:

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

Page 409 Problem 1

Page 409 Problem 1

$\left(\begin{array}{cc}
3 & -2\\
4 & -1\end{array}\right)$ gives $\lambda^{2}-(2)\lambda+(-3-(-8))=\lambda^{2}-2\lambda+5$

Solving for the roots we have $\lambda=\frac{2\pm\sqrt{4-20}}{2}=1\pm2i$

Page 320 Problem 17

page 320 problem 17