Diff Eq problem, Laplace Transform

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Homework Help Overview

The discussion revolves around finding the Laplace Transform of a function involving a Heaviside step function and a polynomial term, specifically H(t-1)t^2. Participants are exploring the implications of the notation and the translation of functions in the context of Laplace Transforms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses confusion about how to incorporate the Heaviside function with the polynomial term. Some participants question the notation used and seek clarification on the original function. Others suggest that if the original function is e^t*t^2, the translation on the s-axis should be considered. There is also mention of using properties of the Heaviside function in relation to Laplace Transforms.

Discussion Status

The discussion is active, with participants providing insights into the use of the Heaviside function and its implications for the Laplace Transform. Some guidance has been offered regarding the translation of functions and the application of relevant properties, though multiple interpretations of the problem are being explored.

Contextual Notes

Participants are navigating the complexities of function translation in the context of Laplace Transforms, with references to specific properties and potential misunderstandings regarding notation. There is an acknowledgment of the Heaviside function's role, but clarity on the original function remains a point of discussion.

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Homework Statement


Find the Laplace Transform of the given function
H(t-1)t^2

I'm not sure how to add (t-1) to the t^2 term to solve the problem

Any help would be greatly appreciated.
 
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That notation makes no sense, are you trying to translate the function on the s-axis? If you are what is the original function.
 
If the original function is e^t*t^2 the translation = F(s-1) and since f(t)=t^2 its transform is 2/s^3, therefore you substitute s-1 in for s and get 2/(s-1)^3. Unless you are doing an inverse translation on the t-axis.
 
My book states that it is a H(t) is a heavyside function and I'm suppose to use the proposition:

L{H(t-c)f(t-c)}(s)=e^(-cs)F(s)
 
Oh a unit step therefore you must translate both H(t) and F(t). LH(t-1)=e^s and f(t)=t^2, L(t^2)=2/s^3, so the transform is 2e^s/s^3
 

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