28E In Problems 21-30, determine L-1... [FREE SOLUTION]

91影视

Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 7: 28E (page 350)

In Problems 21-30, determine $L^{- 1} \{F\}$
$s^{2} F (s) + s F (s) - 6 F (s) = \frac{s^{2} + 4}{s^{2} + s}$

Short Answer

Expert verified

$L^{- 1} \{\frac{s^{2} + 4}{s (s + 1) (s^{2} + s - 6)}\} = - \frac{2}{3} + \frac{5}{6} e^{- t} - \frac{13}{30} e^{- 3 t} + \frac{4}{15} e^{2 t}$

Step by step solution

Find the factor of the denominator

From the given equation we get

$F (s) = \frac{s^{2} + 4}{s (s + 1) (s^{2} + s - 6)} = \frac{s^{2} + 4}{s (s + 1) (s + 3) (s - 2)}$

Using partial fractions, we get

$\frac{s^{2} + 4}{s (s + 1) (s + 3) (s - 2)} = \frac{A}{s} + \frac{B}{s + 1} + \frac{C}{s + 3} + \frac{D}{s - 2}$

from where it follows

role="math" localid="1664441613707" $s^{2} + 4 = A (s + 1) (s + 3) (s - 2) + B s (s + 3) (s - 2) + C s (s + 1) (s - 2) + D s (s + 1) (s + 3)$

Using $s = 0, - 1, - 3, 2,$ respectively we get

$s = 0 : 4 = - 6 A 鈬� A = - \frac{2}{3} s = - 1 : 5 = 6 B 鈬� B = \frac{5}{6} s = 2 : 8 = 30 D 鈬� D = \frac{4}{15} T h e r e f o r e, \frac{s^{2} + 4}{s (s + 1) (s^{2} + s - 6)} = - \frac{2}{3 s} + \frac{5}{6 (s + 1)} - \frac{13}{30 (s + 3)} + \frac{4}{15 (s - 2)}$

Find the inverse

Find the inverse Laplace transform:

$L^{- 1} \{\frac{s^{2} + 4}{s (s + 1) (s^{2} + s - 6)}\} = L^{- 1} \{- \frac{2}{3 s}\} + L^{- 1} \{\frac{5}{6 (s + 1)}\} + L^{- 1} \{\frac{13}{30 (s + 3)}\} + L^{- 1} \{\frac{4}{15 (s - 2)}\} = - \frac{2}{3} L^{- 1} \{\frac{1}{s}\} + \frac{5}{6} L^{- 1} \{\frac{1}{s + 1}\} - \frac{13}{30} L^{- 1} \{\frac{1}{s + 3}\} + \frac{4}{15} L^{- 1} \{\frac{1}{s - 2}\} = - \frac{2}{3} + \frac{5}{6} e^{- t} - \frac{13}{30} e^{- 3 t} + \frac{4}{15} e^{2 t} T h e r e f o r e L^{- 1} \{\frac{s^{2} + 4}{s (s + 1) (s^{2} + s - 6)}\} = - \frac{2}{3} + \frac{5}{6} e^{- t} - \frac{13}{30} e^{- 3 t} + \frac{4}{15} e^{2 t}$

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91影视

In Problems 21-30, determine $L^{- 1} \{F\}$
$s^{2} F (s) + s F (s) - 6 F (s) = \frac{s^{2} + 4}{s^{2} + s}$

Short Answer

Step by step solution

Find the factor of the denominator

Find the inverse

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91影视

In Problems 21-30, determine L-1Fs2Fs+sFs-6Fs=s2+4s2+s

Short Answer

Step by step solution

Find the factor of the denominator

Find the inverse

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Mechanics Maths

Logic and Functions

Calculus

Geometry

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 21-30, determine $L^{- 1} \{F\}$
$s^{2} F (s) + s F (s) - 6 F (s) = \frac{s^{2} + 4}{s^{2} + s}$