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

Use Heaviside's expansion formula derived in Problem 40 to determine the inverse Laplace transform of
$F (s) = \frac{3 s^{2} - 16 s + 5}{(s + 1) (s - 3) (s - 2)}$

$2 e^{- t} - 4 e^{3 t} + 5 e^{2 t}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Define Inverse Laplace transform

Given a function $F (s)$ , if there is a function $f (t)$ that is continuous on

$[0, \infty)$ and satisfies $L {f} = F$ ,then we say that $f (t)$ is the inverse Laplace transform of $F (s)$ and employ the notation $f = L^{- 1} {F}$

Non-repeated Linear Factors

If $Q (s)$ can be factored into a product of distinct linear factors,

$Q (s) = (s - r_{1}) (s - r_{2}) \dots (s - r_{n}),$

where the $r_{i}$ 's are all distinct real numbers, then the partial fraction expansion has the form

$\frac{P (s)}{Q (s)} = \frac{A_{1}}{s - r_{1}} + \frac{A_{2}}{s - r_{2}} + \dots + \frac{A_{n}}{s - r_{n}},$

where the $A_{i}$ 's are real numbers. There are various ways of determining the constants $A_{1}, \dots, A_{n}$ . In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If $s - r$ is a factor of $Q (s)$ and $(s - r)^{m}$ is the highest power of $s - r$ that divides $Q (s)$ , then the portion of the partial fraction expansion of $P (s) / Q (s)$ that corresponds to the term $(s - r)^{m}$ is

$\frac{A_{1}}{s - r} + \frac{A_{2}}{{(s - r)}^{2}} + \dots + \frac{A_{m}}{{(s - r)}^{m}},$

where the $A_{i}$ 's are real numbers.

Quadratic Factors

If $(s - α)^{2} + β^{2}$ is a quadratic factor of $Q (s)$ that cannot be reduced to linear factors with real coefficients and $m$ is the highest power of $(s - α)^{2} + β^{2}$ that divides $Q (s)$ , then the portion of the partial fraction expansion that corresponds to $(s - α)^{2} + β^{2}$ is

$\frac{C_{1} s + D_{1}}{{(s - α)}^{2} + β^{2}} + \frac{C_{2} s + D_{2}}{{[{(s - α)}^{2} + β^{2}]}^{2}} + \dots + \frac{C_{m} s + D_{m}}{{[{(s - α)}^{2} + β^{2}]}^{m}} .$

it is more convenient to express $C_{i} s + D_{i}$ in the form $A_{i} (s - α) + β B_{i}$ when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

$\frac{A_{1} (s - α) + β B_{1}}{{(s - α)}^{2} + β^{2}} + \frac{A_{2} (s - α) + β B_{2}}{{[{(s - α)}^{2} + β^{2}]}^{2}} + \dots + \frac{A_{m} (s - α) + β B_{m}}{{[{(s - α)}^{2} + β^{2}]}^{m}} .$

Step 2: Find the factor of the denominator

There we have

$P (s) = 3 s^{2} - 16 s + 5 Q (s) = (s + 1) (s - 3) (s - 2)$

$Q^{'} (s) = (s - 3) (s - 2) + (s + 1) (s - 2) + (s + 1) (s - 3)$

Now we have

$Q (- 1) = 12 P (- 1) = 24 Q^{'} (3) = 4 P (3) = - 16 Q^{'} (2) = - 3 P (2) = - 15$

Using the equation from Problem 40 we get

$L^{- 1} \{\frac{3 s^{2} - 16 s + 5}{(s + 1) (s - 3) (s - 2)}\} (t) = \frac{P (- 1)}{Q^{'} (- 1)} e^{- t} + \frac{P (3)}{Q^{'} (3)} e^{3 t} + \frac{P (2)}{Q^{'} (2)} e^{2 t} = \frac{24}{12} e^{- t} + \frac{- 16}{4} e^{3 t} + \frac{- 15}{- 3} e^{2 t} = 2 e^{- t} - 4 e^{3 t} + 5 e^{2 t}$

Thus, $L^{- 1} \{\frac{3 s^{2} - 16 s + 5}{(s + 1) (s - 3) (s - 2)}\} (t) = 2 e^{- t} - 4 e^{3 t} + 5 e^{2 t}$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Use Heaviside's expansion formula derived in Problem 40 to determine the inverse Laplace transform of
$F (s) = \frac{3 s^{2} - 16 s + 5}{(s + 1) (s - 3) (s - 2)}$

Step by Step Solution

Step 1: Define Inverse Laplace transform

Step 2: Find the factor of the denominator

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Use Heaviside's expansion formula derived in Problem 40 to determine the inverse Laplace transform ofF(s)=3s2-16s+5(s+1)(s-3)(s-2)

Step by Step Solution

Step 1: Define Inverse Laplace transform

Step 2: Find the factor of the denominator

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Use Heaviside's expansion formula derived in Problem 40 to determine the inverse Laplace transform of
$F (s) = \frac{3 s^{2} - 16 s + 5}{(s + 1) (s - 3) (s - 2)}$