How Do You Apply Laplace Transform to Shifted Functions in Calculus?

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Homework Help Overview

The discussion revolves around applying the Laplace transform to shifted functions, specifically focusing on the transformation of H(t-1)t^2. Participants are exploring how to express t^2 in terms of t-1 and the implications of using the Laplace transform in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find the Laplace transformation of H(t-1)t^2 and expresses uncertainty about rewriting t^2 in terms of t-1. Some participants suggest using a Taylor series expansion or a specific formula related to the Laplace transform of shifted functions.

Discussion Status

Participants are actively discussing different methods to approach the problem, including a request for proof of the formula mentioned. There is a focus on understanding the underlying principles rather than reaching a definitive solution.

Contextual Notes

Some participants note that the formula for the Laplace transform of shifted functions is not found in their textbooks, indicating a potential gap in resources or references for the original poster.

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Homework Statement



Find Laplace transformation
H(t-1)t^2

Homework Equations



The Attempt at a Solution



[2(e^-s)f(t^2)]/s^3

I'm not sure how to get t^2 in terms of t-1

I know the answer is (2+2s+s^2)/s^3
 
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Just express ##t^2## as a finite Taylor series about ##t=1##. Alternatively you can use this formula$$
\mathcal L(u(t-a)f(t)) = e^{-as}\mathcal L(f(t+a)$$which would give, in your case$$
\mathcal L(u(t-1)t^2)=e^{-s}\mathcal L((t+1)^2)
= e^{-s}\mathcal L(t^2+2t+1)$$
 
can you show me the proof on that formula because it is no where in my book
 
jhendren said:
can you show me the proof on that formula because it is no where in my book

$$\mathcal L(u(t-a)f(t)=\int_a^\infty e^{-st}f(t)dt$$Let ##v = t-a##$$
=\int_0^\infty e^{-s(v+a)}f(v+a)dv =e^{-as}\int_0^\infty e^{-sv}f(v+a)dv
=e^{-as}\mathcal L(f(t+a))$$This formula is handy because, as in your problem, the function f(t) isn't usually given in terms of the argument of the step function.
 

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