Chapter 7: 33E (page 350)

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as
$L^{- 1} \{\frac{d^{n} F}{d s^{n}}\} (t) = {(- t)}^{n} f (t),$
where, $f = L^{- 1} \{F\}$ .Use this equation in Problems 33-36 to compute $L^{- 1} \{F\} .$ $F (s) = l n (\frac{s + 2}{s - 5})$

Short Answer

Expert verified

$L^{- 1} (F) = \frac{1}{t} (e^{5 t} - e^{2 t})$

Step by step solution

Simplify the function and find derivative

Using the property of function $\ln t$ we get:

$F (s) = \ln (\frac{s + 2}{s - 5}) F (s) = \ln (s + 2) - \ln (s - 5)$

Find the derivative of F with respect to s:

$\frac{d F}{d s} = \frac{d}{d s} (\ln (s + 2)) - \frac{d}{d s} (\ln (s - 5)) \frac{d F}{d s} = \frac{1}{s + 2} - \frac{1}{s - 5}$

Find the Laplace inverse

From the given condition, we have $L^{- 1} (F) = \frac{1}{{(- t)}^{n}} L^{- 1} \{\frac{d^{n} F}{d s^{n}}\}$ ,apply this to find the Laplace inverse as:

$L^{- 1} (F) = \frac{1}{{(- t)}^{1}} L^{- 1} \{\frac{d F}{d s}\} L^{- 1} (F) = \frac{1}{(- t)} L^{- 1} (\frac{1}{s + 2} - \frac{1}{s - 5}) = \frac{1}{t} (- L^{- 1} (\frac{1}{s + 2}) + L^{- 1} (\frac{1}{s - 5})) = \frac{1}{t} (e^{5 t} - e^{2 t}) T h e r e f o r e, L^{- 1} (F) = \frac{1}{t} (e^{5 t} - e^{2 t})$

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Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as
$L^{- 1} \{\frac{d^{n} F}{d s^{n}}\} (t) = {(- t)}^{n} f (t),$
where, $f = L^{- 1} \{F\}$ .Use this equation in Problems 33-36 to compute $L^{- 1} \{F\} .$ $F (s) = l n (\frac{s + 2}{s - 5})$

Short Answer

Step by step solution

Simplify the function and find derivative

Find the Laplace inverse

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Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform asL-1dnFdsnt=-tnft,where,f=L-1F.Use this equation in Problems 33-36 to computeL-1F.Fs=lns+2s-5

Short Answer

Step by step solution

Simplify the function and find derivative

Find the Laplace inverse

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Decision Maths

Calculus

Applied Mathematics

Discrete Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as
$L^{- 1} \{\frac{d^{n} F}{d s^{n}}\} (t) = {(- t)}^{n} f (t),$
where, $f = L^{- 1} \{F\}$ .Use this equation in Problems 33-36 to compute $L^{- 1} \{F\} .$ $F (s) = l n (\frac{s + 2}{s - 5})$