Suggested languages for you:

Americas

Europe

Q8RP

Expert-verified
Found in: Page 415

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# In Problems 3-10, determine the Laplace transform of the given function.${\mathbf{\left(}\mathbf{t}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbf{\left(}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{3}\mathbf{\right)}}^{{\mathbf{2}}}$

Therefore, the solution is$\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}$.

See the step by step solution

## Step 1: Given Information

The given value is ${\left(\text{t+3}\right)}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}$

## Step 2: Determining the Laplace transform

Using the following Laplace transform property, to find the Laplace of given integral:

$\text{L}\left\{{\text{e}}^{\text{at}}\right\}\text{=}\frac{\text{1}}{\text{s-a}}$

$\text{L}\left\{{\text{t}}^{\text{n}}\right\}\text{=}\frac{\text{n!}}{{\text{s}}^{\text{n+1}}}$

Apply the Laplace transform property, we get:

$\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=L}\left\{{\text{(t+3)}}^{\text{2}}\right\}\text{-L}\left\{{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\text{+6t+3}\right\}\text{-L}\left\{{\text{e}}^{\text{2t}}{\text{+6e}}^{\text{t}}\text{+3}\right\}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}+3L{1}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\text{-3L{1}}\\ \text{=L}\left\{{\text{t}}^{\text{2}}\right\}\text{+6L{t}-L}\left\{{\text{e}}^{\text{2t}}\right\}\text{+6L}\left\{{\text{e}}^{\text{t}}\right\}\end{array}$

Simplify further as follows

$\begin{array}{c}\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2!}}{{\text{s}}^{\text{3}}}\text{+6}\frac{\text{1!}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-6}\frac{\text{1}}{\text{s-1}}\\ \text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}\end{array}$

Therefore,$\text{L}\left\{{\text{(t+3)}}^{\text{2}}\text{-}{\left({\text{e}}^{\text{t}}\text{+3}\right)}^{\text{2}}\right\}\text{=}\frac{\text{2}}{{\text{s}}^{\text{3}}}\text{+}\frac{\text{6}}{{\text{s}}^{\text{2}}}\text{-}\frac{\text{1}}{\text{s-2}}\text{-}\frac{\text{6}}{\text{s-1}}$