Question

In Problems 3–10, determine the Laplace transform of the given function.

$\mathbf{7}{\mathit{e}}^{\mathbf{2}\mathbf{t}}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{t}\mathbf{-}\mathbf{2}{\mathit{e}}^{\mathbf{7}\mathbf{t}}\mathit{s}\mathit{i}\mathit{n}\mathbf{5}\mathit{t}$

Short answer

$L\{7e^{2t}cos3t-2e^{7t}sin5t\}\left(s\right)=\frac{7(s-2)}{{(s-2)}^2+9}-\frac{10}{{(s-7)}^2+25}$

Step-by-step solution

Step 1:Given Information

The given function is$7e^{2t}cos3t-2e^{7t}sin5t$

Determining the Laplace transform

Using the following Laplace transform definition, to find the Laplace of the given function:

$\begin{array}{l}L\left\{e^{at}sinbt\right\}\left(s\right)=\frac{b}{{(s-a)}^2+b^2}\\ L\left\{e^{at}cosbt\right\}\left(s\right)=\frac{s-a}{{(s-a)}^2+b^2}\end{array}$

Considering the Laplace transform's linearity, we get

$\begin{array}{c}L\{7e^{2t}cos3t-2e^{7t}sin5t\}\left(s\right)=7L\left\{e^{2t}cos3t\right\}\left(s\right)-2L\left\{e^{7t}sin5t\right\}\left(s\right)\\ =7\frac{s-2}{{(s-2)}^2+3^2}-2\frac{5}{{(s-7)}^2+5^2}\\ =\frac{7(s-2)}{{(s-2)}^2+9}-\frac{10}{{(s-7)}^2+25}\end{array}$

Determining the Result

Thus, the required Laplace transform is

$L\{7e^{2t}cos3t-2e^{7t}sin5t\}\left(s\right)=\frac{7(s-2)}{{(s-2)}^2+9}-\frac{10}{{(s-7)}^2+25}$