DIFFEQ  Discontinuous Forcing Functions (should be an easy question)by FrogPad Tags: diffeq, discontinuous, forcing, functions 

#1
Oct1605, 05:06 PM

P: 838

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(tc) = 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)=0t<1 [/tex] [tex] f(t)=t^22t+2t\geq1 [/tex] So [tex] f(t) [/tex] can be rewritten as: [tex] f(t) = u_1(t)(t^22t+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^22t+2 [/tex]. But the only table value I have is: [tex] u_c(t)f(tc) = e^{cs}F(s) [/tex] But, this doesn't actually match what I have. Since, f(t) is not of the form f(tc). 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(tc) to be f(t). So: [tex] (t1)^2 = t^22t+1 [/tex] [tex] (t1)^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((t1)^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? 



#2
Oct1605, 05:47 PM

P: 838

Nevermind... I figured it out. Thanks though :)
If anyone wants to elaborate, be my guest. 


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