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. $\frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)}$

Therefore, the solution is $L^{- 1} {\frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)}} (t) = 2 e^{t} + 2 e^{- 2 t} \cos 3 t + e^{- 2 t} \sin 3 t$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

The given value is $\frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)}$

Step 2: Use partial fractions

Make partial fraction of the given function, as:

$\begin{matrix} \frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)} = \frac{4 s^{2} + 13 s + 19}{(s - 1) ({(s + 2)}^{2} + 3^{2})} \\ = \frac{A}{s - 1} + \frac{B (s + 2) + 3 C}{{(s + 2)}^{2} + 3^{2}} \end{matrix}$

On comparing the numerator of above equation we get:

$4 s^{2} + 13 s + 19 = (A + B) s^{2} + (4 A + B + 3 C) s + 13 A - 2 B - 3 C$

On comparing the coefficients, we get the system

${\begin{cases} A + B = 4 \\ 4 A + B + 3 C = 13 \\ 13 A - 2 B - 3 C = 19 \end{cases}$

On simplifying further, we get

${\begin{cases} A = 4 - B \\ C = B - 1 \\ - 18 B = - 36 \end{cases}$

On solving the equations, we get

${\begin{cases} A = 2 \\ B = 2 \\ C = 1 \end{cases}$

Thus, the partial fractions are obtained as:

$\frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)} = \frac{2}{s - 1} + \frac{2 (s + 2) + 3}{{(s + 2)}^{2} + 3^{2}}$

Step 3: Take Inverse Laplace transform

Take inverse Laplace transform using $L^{- 1} {\frac{s - a}{{(s - a)}^{2} + b^{2}}} (t) = e^{a t} \cos b t$ and $L^{- 1} {\frac{b}{{(s - a)}^{2} + b^{2}}} (t) = e^{a t} \sin b t$ as:

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

Hence, the required inverse Laplace transform is

$L^{- 1} {\frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)}} (t) = 2 e^{t} + 2 e^{- 2 t} \cos 3 t + e^{- 2 t} \sin 3 t$

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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. $\frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)}$

Step by Step Solution

Step 1: Given Information

Step 2: Use partial fractions

Step 3: Take Inverse 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.4s2+13s+19(s-1)(s2+4s+13)

Step by Step Solution

Step 1: Given Information

Step 2: Use partial fractions

Step 3: Take Inverse Laplace transform

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In Problems 3-10, determine the Laplace transform of the given function. $\frac{4 s^{2} + 13 s + 19}{(s - 1) (s^{2} + 4 s + 13)}$