Solve Laplace Transform of e^-te^tcost

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SUMMARY

The Laplace Transform of the function \( e^{-t} e^{t} \cos(t) \) is evaluated using the convolution theorem. The correct approach involves recognizing that the Laplace Transform of a product can be expressed as the product of their individual transforms. The solution simplifies to \( \frac{s-1}{s[(s-1)^2 + 1]} \), confirming that the convolution of the two functions is necessary for accurate computation. The initial attempt incorrectly applied the integral method, highlighting the importance of using the convolution property in Laplace Transforms.

PREREQUISITES
  • Understanding of Laplace Transforms
  • Familiarity with convolution theorem in integral transforms
  • Knowledge of trigonometric functions and their properties
  • Basic calculus, particularly integration techniques
NEXT STEPS
  • Study the convolution theorem in detail for Laplace Transforms
  • Learn about the properties of Laplace Transforms for products of functions
  • Explore examples of Laplace Transforms involving trigonometric functions
  • Practice solving Laplace Transforms using different methods, including integral and convolution approaches
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Students studying differential equations, engineers applying Laplace Transforms in system analysis, and anyone seeking to deepen their understanding of integral transforms in mathematical physics.

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Homework Statement



$$L\{ { e }^{ -t }*{ e }^{ t }cost\}$$

Homework Equations





The Attempt at a Solution



$$L\{ { e }^{ -t }*{ e }^{ t }cost\} \\ =L\{ \int _{ 0 }^{ t }{ { e }^{ -\tau }{ e }^{ t-\tau }cos(t-\tau )d\tau } \} \\ =\frac { L\{ { e }^{ t }cost\} }{ s } \\ =\frac { s-1 }{ s[{ (s-1) }^{ 2 }+1] }$$
 
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Unfortunately it isn't correct. For convolution we have:

##L\{f*g\}=L\{f\}\cdot L\{g\}##
 

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