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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 1 and 2, use the definition of the Laplace transform to determine ${\mathbit{L}}\mathbf{\left\{}\mathbf{f}\mathbf{\right\}}$.${\mathbit{f}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}{\mathbf{=}}\mathbf{\left\{}\begin{array}{l}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{2}\\ \mathbf{6}\mathbf{-}\mathbf{t}\mathbf{,}\mathbf{2}\mathbf{<}\mathbf{t}\end{array}$

$L\left\{f\left(t\right)\right\}\left(s\right)=\frac{3}{s}+\frac{{e}^{-2s}}{s}-\frac{{e}^{-2s}}{{s}^{2}}$

See the step by step solution

## Step 1:Given Information

The given function is $f\left(t\right)=\left\{\begin{array}{l}3,0\le t\le 2\\ 6-t,2

## Step 2: Determining the L{f}

Using the Laplace transform definition, we get

$\begin{array}{c}L\left\{f\left(t\right)\right\}={\int }_{0}^{\infty }{e}^{-st}f\left(t\right)dt\\ ={\int }_{0}^{2}3{e}^{-st}dt+{\int }_{2}^{\infty }\left(6-t\right){e}^{-st}dt\\ =3{\left[-\frac{{e}^{-st}}{s}\right]}_{0}^{2}+\underset{N\to \infty }{\mathrm{lim}}{\int }_{2}^{N}\left(6-t\right){e}^{-st}dt\\ =\frac{3}{s}\left(1-{e}^{-2s}\right)+\underset{N\to \infty }{\mathrm{lim}}{\int }_{2}^{N}\left(6-t\right){e}^{-st}dt\end{array}$

Letrole="math" localid="1664044470656" $\left|\begin{array}{c}\begin{array}{ll}6-t=u& {e}^{-st}dt=dv\\ -dt=du& v=-\frac{{e}^{-st}}{s}\end{array}\\ \end{array}\right|$ in second integral, then we can write as:

$\begin{array}{c}L\left\{f\left(t\right)\right\}=\frac{3}{s}\left(1-{e}^{-2s}\right)+\underset{N\to \infty }{\mathrm{lim}}\left({\left[-\left(6-t\right)\frac{{e}^{-st}}{s}\right]}_{2}^{N}-{\int }_{2}^{N}\frac{{e}^{-st}}{s}dt\right)\\ =\frac{3}{s}\left(1-{e}^{-2s}\right)+\underset{N\to \infty }{\mathrm{lim}}\left(\frac{4{e}^{-2s}}{s}-\frac{\left(6-N\right){e}^{-sN}}{s}+{\left[\frac{{e}^{-st}}{{s}^{2}}\right]}_{2}^{N}\right)\\ =\frac{3}{s}\left(1-{e}^{-2s}\right)+\underset{N\to \infty }{\mathrm{lim}}\left(\frac{4{e}^{-2s}}{s}-\frac{\left(6-N\right){e}^{-sN}}{s}+\frac{{e}^{-sN}}{{s}^{2}}-\frac{{e}^{-2s}}{{s}^{2}}\right)\\ =\frac{3}{s}\left(1-{e}^{-2s}\right)+\underset{N\to \infty }{\mathrm{lim}}\frac{4{e}^{-2s}}{s}-\underset{N\to \infty }{\mathrm{lim}}\frac{\left(6-N\right){e}^{-sN}}{s}+\underset{N\to \infty }{\mathrm{lim}}\frac{{e}^{-sN}}{{s}^{2}}-\underset{N\to \infty }{\mathrm{lim}}\frac{4{e}^{-2s}}{{s}^{2}}\end{array}$

Simplify further as:

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

## Step 3: Determining the Result

Thus, the required Laplace transform is $L\left\{f\left(t\right)\right\}\left(s\right)=\frac{3}{s}+\frac{{e}^{-2s}}{s}-\frac{{e}^{-2s}}{{s}^{2}}$