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 3-10, determine the Laplace transform of the given function.
${(t + 3)}^{2} - {(e^{t} + 3)}^{2}$

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

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

Step 2: Determining the Laplace transform

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

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

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

Apply the Laplace transform property, we get:

$\begin{matrix} L {{(t+3)}^{2} - {(e^{t} +3)}^{2}} =L {{(t+3)}^{2}} -L {{(e^{t} +3)}^{2}} \\ =L {t^{2} +6t+3} -L {e^{2t} {+6e}^{t} +3} \\ =L {t^{2}} +6L{t}+3L{1}-L {e^{2t}} +6L {e^{t}} -3L{1} \\ =L {t^{2}} +6L{t}-L {e^{2t}} +6L {e^{t}} \end{matrix}$

Simplify further as follows

$\begin{matrix} L {{(t+3)}^{2} - {(e^{t} +3)}^{2}} = \frac{2!}{s^{3}} +6 \frac{1!}{s^{2}} - \frac{1}{s-2} -6 \frac{1}{s-1} \\ = \frac{2}{s^{3}} + \frac{6}{s^{2}} - \frac{1}{s-2} - \frac{6}{s-1} \end{matrix}$

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

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Given Information

Step 2: Determining the Laplace transform

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

In Problems 3-10, determine the Laplace transform of the given function.(t+3)2-(et+3)2

Step by Step Solution

Step 1: Given Information

Step 2: Determining the Laplace transform

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