Q7.3 - 9E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]

${\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{tsin}\mathbf{2}\mathbf{t}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{4s+4}{{\left[{(s+1)}^2+4\right]}^2}$.

Definition of Laplace transform

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transform. 
• Let f(t) be supplied for t(0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by $\mathcal{L}\left\{f\left(t\right)\right\}$ or F(s).

Determine the Laplace transform for the given equation

Given that, $e^{-t}t\sin 2t$

Let $f\left(t\right)=e^{-t}\sin 2t$

Find the Laplace transform of $f\left(t\right)=e^{-t}\sin 2t$ using $\mathcal{L}\left\{e^{at}sinbt\right\}=\frac{b}{{(s-a)}^2+b^2}$ as:

$\begin{array}{rcl}F\left(s\right)& =& \mathcal{L}\left\{e^{-t}\sin 2t\right\}\\ & =& \frac{2}{{(s+1)}^2+4}\end{array}$

Find the Laplace transform of the given function $e^{-t}t\sin 2t$ using ${\left(\frac{f}{g}\right)}^{\text{'}}=\frac{f{g}^{\text{'}}-g{f}^{\text{'}}}{f^2}$ and $\mathcal{L}\left\{tf\right(t\left)\right\}=(-1{)}^n\frac{d^{\left(n\right)}}{dt}F\left(s\right)$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{e^{-t}t\sin 2t\right\}& =& \mathcal{L}\left\{tF\right(s\left)\right\}\\ & =& (-1{)}^1\frac{d}{dt}F\left(s\right)\\ & =& -\frac{d}{dt}\left[\frac{2}{{(s+1)}^2+4}\right]\\ & =& -\frac{\left({(s+1)}^2+4\right)·0-2·\left(2\right(s+1\left)\right)}{{\left[{(s+1)}^2+4\right]}^2}\end{array}$

Simplify the equation as:

$\mathcal{L}\left\{e^{-t}t\sin 2t\right\}=\frac{4s+4}{{\left[{(s+1)}^2+4\right]}^2}$

Therefore, the Laplace transform for the given equation is $\frac{4s+4}{{\left[{(s+1)}^2+4\right]}^2}$$\frac{4s+4}{{\left[{(s+1)}^2+4\right]}^2}$