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

by FrogPad
Tags: diffeq, discontinuous, forcing, functions
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FrogPad
#1
Oct16-05, 05:06 PM
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):
[tex] u_c(t) = \frac{e^{-cs}}{s} [/tex]
[tex] u_c(t)f(t-c) = e^{-cs}F(s) [/tex]
[tex] t^n = \frac{n!}{s^{n+1}} [/tex]

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

So [tex] f(t) [/tex] can be rewritten as:
[tex] f(t) = u_1(t)(t^2-2t+2) [/tex]

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

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:
[tex] (t-1)^2 = t^2-2t+1 [/tex]
[tex] (t-1)^2 +1 = f(t) [/tex]
This would allow me to use the rule right?

So I would then have:

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

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?
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FrogPad
#2
Oct16-05, 05:47 PM
P: 835
Nevermind... I figured it out. Thanks though :)
If anyone wants to elaborate, be my guest.


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