Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:Given Information

The given function is $f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$

Step 2: Determining the L{f}

Using the Laplace transform definition, we get

$\begin{matrix} L {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t \\ = \int_{0}^{2} 3 e^{- s t} d t + \int_{2}^{\infty} (6 - t) e^{- s t} d t \\ = 3 {[- \frac{e^{- s t}}{s}]}_{0}^{2} + \lim_{N \to \infty} \int_{2}^{N} (6 - t) e^{- s t} d t \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} \int_{2}^{N} (6 - t) e^{- s t} d t \end{matrix}$

Letrole="math" localid="1664044470656" $|\begin{matrix} \begin{array}{ll} 6 - t = u & e^{- s t} d t = d v \\ - d t = d u & v = - \frac{e^{- s t}}{s} \end{array} \end{matrix}|$ in second integral, then we can write as:

$\begin{matrix} L {f (t)} = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} ({[- (6 - t) \frac{e^{- s t}}{s}]}_{2}^{N} - \int_{2}^{N} \frac{e^{- s t}}{s} d t) \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} (\frac{4 e^{- 2 s}}{s} - \frac{(6 - N) e^{- s N}}{s} + {[\frac{e^{- s t}}{s^{2}}]}_{2}^{N}) \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} (\frac{4 e^{- 2 s}}{s} - \frac{(6 - N) e^{- s N}}{s} + \frac{e^{- s N}}{s^{2}} - \frac{e^{- 2 s}}{s^{2}}) \\ = \frac{3}{s} (1 - e^{- 2 s}) + \lim_{N \to \infty} \frac{4 e^{- 2 s}}{s} - \lim_{N \to \infty} \frac{(6 - N) e^{- s N}}{s} + \lim_{N \to \infty} \frac{e^{- s N}}{s^{2}} - \lim_{N \to \infty} \frac{4 e^{- 2 s}}{s^{2}} \end{matrix}$

Simplify further as:

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

Step 3: Determining the Result

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$

Step by Step Solution

Step 1:Given Information

Step 2: Determining the L{f}

Step 3: Determining the Result

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems 1 and 2, use the definition of the Laplace transform to determine L{f}.f(t)={3,0≤t≤26-t,2<t

Step by Step Solution

Step 1:Given Information

Step 2: Determining the L{f}

Step 3: Determining the Result

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In Problems 1 and 2, use the definition of the Laplace transform to determine $L {f}$ .
$f (t) = {\begin{cases} 3, 0 \leq t \leq 2 \\ 6 - t, 2 < t \end{cases}$