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

In summary, the conversation covers the confusion of finding the Laplace transform of a function, specifically when the given function does not match the form required for the Laplace transform table. The individual works through an example and explores different ways to manipulate the function to fit the table values. They also mention difficulties with the LaTeX formatting.
  • #1
FrogPad
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0
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?
 
Last edited:
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  • #2
Nevermind

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

What is a discontinuous forcing function in differential equations?

A discontinuous forcing function is a function that is defined differently at different points on its domain. This means that there is a sudden change or jump in the function's value at a specific point.

Why are discontinuous forcing functions important in differential equations?

Discontinuous forcing functions can represent real-world phenomena that involve sudden changes or jumps, such as a switch turning on or off. They allow us to model these situations mathematically and solve for the behavior of the system.

How do we handle discontinuous forcing functions in differential equations?

We can handle discontinuous forcing functions by breaking them into multiple pieces and solving for each piece separately. We can also use techniques such as the Heaviside step function to represent the discontinuity in the function.

What are some examples of situations that can be modeled using discontinuous forcing functions?

Some examples include electrical circuits with switches, predator-prey relationships in ecology, and population growth with sudden changes in resources or environmental conditions.

What are the limitations of using discontinuous forcing functions in differential equations?

Discontinuous forcing functions can make the equations more complex and difficult to solve. They also may not accurately represent the real-world situation if the sudden changes occur too frequently or are not well-defined.

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