# DIFFEQ - Discontinuous Forcing Functions (should be an easy question)

1. Oct 16, 2005

Ok, we just started this chapter, and I am slightly confused with one specific aspect of the info... I'll just go through an example, it's the best way to explain it IMHO.
I have to find the Laplace transform of the following function.
The table of transforms that I can use are (sorry about the formatting, I know they are not equal to each other):
$$u_c(t) = \frac{e^{-cs}}{s}$$
$$u_c(t)f(t-c) = e^{-cs}F(s)$$
$$t^n = \frac{n!}{s^{n+1}}$$

$$f(t)=$$ is defined as a system of equations (sorry I don't know the LaTeX formatting for it).
$$f(t)=0|t<1$$
$$f(t)=t^2-2t+2|t\geq1$$

So $$f(t)$$ can be rewritten as:
$$f(t) = u_1(t)(t^2-2t+2)$$

Ok, so now this is where I get confused. I have to do the Laplace transform of $$f(t) = u_1(t)(t^2-2t+2$$. But the only table value I have is:
$$u_c(t)f(t-c) = e^{-cs}F(s)$$

But, this doesn't actually match what I have. Since, f(t) is not of the form f(t-c). So if anyone could just explain this part better to me... that would be awesome. My thought process here is that I have to change f(t-c) to be f(t).
So:
$$(t-1)^2 = t^2-2t+1$$
$$(t-1)^2 +1 = f(t)$$
This would allow me to use the rule right?

So I would then have:

$$F(s) = e^{-cs}/s L((t-1)^2+1) = e^{-cs}/s [L(t^2)+L(-2t)+L(2)]$$

Is this idea even right? I guess I just don't understand what is really going on here.
On a second note what the hell is going on with the latex formatting? Is anyone else having troubles previewing their changes?

Last edited: Oct 16, 2005
2. Oct 16, 2005