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

by FrogPad
Tags: diffeq, discontinuous, forcing, functions
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 P: 835 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?
 P: 835 Nevermind... I figured it out. Thanks though :) If anyone wants to elaborate, be my guest.

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