Q7.3 - 10E - 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.]

${\mathbf{te}}^{\mathbf{2}\mathbf{t}}\mathbf{cos}\mathbf{5}\mathbf{t}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{{(s-2)}^2-25}{{\left[{(s-2)}^2+25\right]}^2}$.

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 $t{e}^{2t}\cos 5t$.

Let $f\left(t\right)=e^{2t}\cos 5t$.

Find the Laplace transform of $f\left(t\right)=e^{2t}\cos 5t$ using $\mathcal{L}\left\{e^{at}\cos bt\right\}=\frac{s-a}{{(s-a)}^2+b^2}$ as:

$\begin{array}{rcl}F\left(s\right)& =& \mathcal{L}\left\{e^{2t}\cos 5t\right\}\\ & =& \frac{s-2}{{(s-2)}^2+25}\end{array}$

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 $\mathcal{L}\left\{t^{n}f\left(t\right)\right\}=(-1{)}^n\frac{d^n}{d{s}^n}[\mathcal{L}\{f\left\}\right(s\left)\right]$  as follows:

$\begin{array}{rcl}\mathcal{L}\left\{t{e}^{2t}\cos 5t\right\}& =& \mathcal{L}\left\{tF\right(s\left)\right\}\\ & =& (-1)\frac{d}{ds}F\left(s\right)\\ & =& -\frac{d}{ds}\left[\frac{s-2}{{(s-2)}^2+25}\right]\\ & =& -\frac{\left({(s-2)}^2+25\right)\times 1-(s-2)\times \left(2\right(s-2\left)\right)}{{\left[{(s-2)}^2+25\right]}^2}\\ & & \end{array}$

Simplify the equation as:

$\begin{array}{rcl}\mathcal{L}\left\{t{e}^{2t}\cos 5t\right\}& =& -\frac{{(s-2)}^2+25-2{(s-2)}^2}{{\left[{(s-2)}^2+25\right]}^2}\\ & =& -\frac{25-{(s-2)}^2}{{\left[{(s-2)}^2+25\right]}^2}\\ & =& \frac{{(s-2)}^2-25}{{\left[{(s-2)}^2+25\right]}^2}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{{(s-2)}^2-25}{{\left[{(s-2)}^2+25\right]}^2}$.