Laplace Transform Homework: Solving f(t) Piecewise

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SUMMARY

The discussion centers on solving the Laplace Transform of a piecewise function defined as f(t) = 0 for 0 ≤ t < 1 and f(t) = t*exp(2t) for t ≥ 1. The initial attempt incorrectly applied the Laplace transform formula for f(t) = te^{2t}, yielding F(s) = 1/(s-2)^2, which is not applicable. The correct approach involves integrating the function from 1 to infinity, leading to F(s) = ∫_1^∞ te^{(2-s)t}dt, which accurately represents the Laplace transform of the given piecewise function.

PREREQUISITES
  • Understanding of piecewise functions
  • Familiarity with the Laplace Transform definition
  • Knowledge of integration techniques
  • Experience with exponential functions in calculus
NEXT STEPS
  • Study the properties of the Laplace Transform
  • Learn integration techniques for improper integrals
  • Explore the application of the Laplace Transform in solving differential equations
  • Review examples of piecewise functions and their transforms
USEFUL FOR

Students in engineering or mathematics, particularly those studying differential equations and transforms, will benefit from this discussion. It is also valuable for educators teaching Laplace Transforms and integration methods.

Philip Wong
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Homework Statement


f(t) is a piecewise function:
{0 0<= t< 1
{t*exp(2t) t = >1

Homework Equations


F(s)= L{f(t)}

The Attempt at a Solution



F(s)= L{t*exp(2t)}

for this problem I just took the Laplace Transformer directly from the table which is: n!/ (s-a)^(n+1)

and after plucking in the relevant number I got:
1/(s-2)^2

Firstly I want to ask, is this correct? And have I missed any steps when setting up this problem?

Secondly if this is correct and I don't want to use the table to get the formula directly, can someone please show me other methods to come to the same answer.

thanks,
Phil
 
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What you have is incorrect. That is the Laplace transform of f(t)= te^{2t} which is NOT the same as the function given here.

Do not know the definition of the Laplace transform? It is
F(f)= \int_0^\infty e^{-st}f(t)dt

Integrate that.
The Laplace transform of this function is
\int_1^\infty e^{-st}te^{2t}dt= \int_1^\infty te^{(2- s)t}dt
 

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