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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.${{\mathbit{e}}}^{\mathbf{2}\mathbf{t}}{\mathbf{-}}{{\mathbit{t}}}^{{\mathbf{3}}}{\mathbf{+}}{{\mathbit{t}}}^{{\mathbf{2}}}{\mathbf{-}}{\mathbit{s}}{\mathbit{i}}{\mathbit{n}}{\mathbf{5}}{\mathbit{t}}$

$L\left\{{e}^{2t}-{t}^{3}+{t}^{2}-sin5t\right\}=\frac{1}{s-2}-\frac{6}{{s}^{4}}+\frac{2}{{s}^{3}}-\frac{5}{{s}^{2}+25}$

See the step by step solution

## Step 1:Given Information

The given function is .${e}^{2t}-{t}^{3}+{t}^{2}-sin5t$

## Step 2: Determining the Laplace transform:

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

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

Considering the Laplace transform's linearity, we get

$\begin{array}{c}L\left\{{e}^{2t}-{t}^{3}+{t}^{2}-sin5t\right\}\left(s\right)=L\left\{{e}^{2t}\right\}\left(s\right)-L\left\{{t}^{3}\right\}\left(s\right)+L\left\{{t}^{2}\right\}\left(s\right)-L\left\{sin5t\right\}\left(s\right)\\ =\frac{1}{s-2}-\frac{3!}{{s}^{4}}+\frac{2!}{{s}^{3}}-\frac{5}{{s}^{2}+{5}^{2}}\\ =\frac{1}{s-2}-\frac{0}{{s}^{4}}+\frac{z}{{s}^{3}}-\frac{5}{{s}^{2}+25}\end{array}$

## Step 3: Determining the Result

Thus, the required Laplace transform is .$L\left\{{e}^{2t}-{t}^{3}+{t}^{2}-sin5t\right\}=\frac{1}{s-2}-\frac{6}{{s}^{4}}+\frac{2}{{s}^{3}}-\frac{5}{{s}^{2}+25}$ ### Want to see more solutions like these? 