# Solve Laplace Transform of e^-te^tcost

• Turion
In summary, the Laplace Transform of e^-te^tcost is equal to 1/(s+1)^2 + 1/(s^2+2s+2). To solve it, you can use the definition of the Laplace Transform and integrate by parts. It can also be simplified using partial fraction decomposition. The inverse Laplace Transform of e^-te^tcost is t^2e^-tcos(t). In real-world applications, it is commonly used in engineering and physics to solve differential equations, analyze systems with time-varying inputs, and in signal processing and control systems.

## Homework Statement

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

## The Attempt at a Solution

$$L\{ { e }^{ -t }*{ e }^{ t }cost\} \\ =L\{ \int _{ 0 }^{ t }{ { e }^{ -\tau }{ e }^{ t-\tau }cos(t-\tau )d\tau } \} \\ =\frac { L\{ { e }^{ t }cost\} }{ s } \\ =\frac { s-1 }{ s[{ (s-1) }^{ 2 }+1] }$$

Unfortunately it isn't correct. For convolution we have:

##L\{f*g\}=L\{f\}\cdot L\{g\}##

## 1. What is the Laplace Transform of e^-te^tcost?

The Laplace Transform of e^-te^tcost is equal to 1/(s+1)^2 + 1/(s^2+2s+2).

## 2. How do you solve the Laplace Transform of e^-te^tcost?

To solve the Laplace Transform of e^-te^tcost, you can use the definition of the Laplace Transform and integrate by parts.

## 3. Can the Laplace Transform of e^-te^tcost be simplified?

Yes, the Laplace Transform of e^-te^tcost can be simplified using partial fraction decomposition.

## 4. What is the inverse Laplace Transform of e^-te^tcost?

The inverse Laplace Transform of e^-te^tcost is equal to t^2e^-tcos(t).

## 5. How is the Laplace Transform of e^-te^tcost used in real-world applications?

The Laplace Transform of e^-te^tcost is commonly used in engineering and physics to solve differential equations and analyze systems with time-varying inputs. It is also used in signal processing and control systems.